## My Experience Teaching Differential Equations

This summer I taught lower division differential equations. The purpose of this post is to share my experiences and recommendations for teaching mathematics with differential equations as an example.

In summary I emulated the things that I’ve seen John Baez do in classes I’ve taken from him. I consider the most important lesson I learned to be the way in which he does homework. Briefly, homework should be like a story that leads the reader into learning something new. I’ve attached my homework assignments at the bottom of the page for anybody who wants to see how it works without having to read my entire post where I’ll go into more detail.

I had been a teaching assistant for this class so I knew a little bit about what troubled students and about what I should cover. Additionally I had some discussions with another graduate student, Andrew Walker, who taught this course the previous summer. From our conversations and my personal experience I believe the following things to be useful notes.

1) Instructors artificially increase the difficulty of this class by making problems that require difficult integration techniques.

In my opinion the point of a basic differential equations class is to learn how to solve some differential equations, not how to solve difficult integrals. Difficulty can be increased in other more productive ways that deal with concepts strictly having to do with differential equations. Students get so bogged down that they leave the class without ever grasping what they were supposed to. It seems like many students finish without understanding what it means to solve a differential equation, what a solution is, or even what a differential equation is.

The solution to this issue is simple- when integrals are necessary for solving a technique make sure problems being done have easy integrals. It’s a bit painstaking to create problems which satisfy this condition, but I believe the payoff is absolutely worth it. (Note: Rarely tough integrals are unavoidable!).

2) Students only write down what you write down.
Even the best students can’t write as fast as someone can talk. Math instructors are too often guilty of haphazardly writing down concepts while verbally explaining important points or sloppily writing down a solution without a single word on the board. The verbal explanation can be enlightening and perfect, but if the students can’t copy it down then they probably won’t remember it!

While teaching differential equations I wrote down every single concept that I wanted students to understand in complete sentences and in complete detail. I wrote down every single example I did as complete thoughts and sentences which brings me to the next point.

3) Writing complete sentences to explain math is a must.
It’s best to illustrate this with an example. I may eventually upload my notes, but for now the following will suffice. Suppose we had a question that said, “Solve the differential equation $\frac{dy}{dx} = \frac{x}{y}.$

Many professors will teach this in the following way. They will say (but not write) that this differential equation is separable and explain things (verbally) about separable equations while simply writing on the board-

$dy/dx = x/y$
$ydy = xdx$
$\int ydy = \int xdx$
$\frac{y^2}{2} = \frac{x^2}{2} + C.$

The student will then write the same 4 lines in their notes. Instead the instructor should write the following on the board-

Since,

$\frac{dy}{dx} = \frac{x}{y}$

is a separable equation, we can “multiply” to get

$ydy=xdx$

then we can integrate both sides to get

$\int ydy = \int xdx$

so that

$\frac{y^2}{2} = \frac{x^2}{2} + C.$

This problem is simple so it’s possible a student can have a perfect understanding of what happened without the words, but difficulty increases it becomes more necessary to write complete thoughts and sentences so you may as well make it a habit during the easier topics.

Writing out problems like this serves a few purposes-

-By doing this enough times they come to understand mathematics is not just lists of equations. It’s about presenting something true in a very logical way that flows from a clear beginning to a clear end. They work the same way sentences do. Sure I could read a sentence with a bunch of things missing and get the gist of it, but whoever wrote it doesn’t really understand how sentences are used to express their ideas. Getting the final answer correct is much less important than having an understanding of the concepts!

-When problems are more complex students have notes that make enough sense to be useful. Words as simple as “since, then, thus, if” etc. increase understanding by a huge amount because the notes become readable. When looking back at notes they can easily follow along from start to finish instead of just seeing a jumbled mess of equations. The reality is that most will never open a textbook so this is their textbook.

-When they do problems on homework, quizzes, and tests, their solutions look much better because they start to solve problems in the same logical way. Their thoughts are clearly expressed. The lack of complete thoughts being taught to them from elementary school onward is evident in the work they turn in. It looks exactly like what they are told to do. That is, write down messes of equations that make no sense on their own and then a box around some final answer.

Besides writing nice details one should also reiterate additional details (like context) verbally while writing down the nice solutions, but whatever the additional details are should have been written down somewhere earlier when the concepts were being taught. The verbal repetition is just there to nail the ideas (which had already been written down!) into their head.

4) Assign written homework which acts like a story.

Traditionally, instructors will teach a section and assign a huge amount of problems which require the student to do the same thing over and over that they learned in the section. The idea is to get some process stuck in their heads.

The problem here is that different students need different amounts of problems until things click. Thus assigning a certain fixed amount is usually very annoying for students. The ones who already get it will be annoyed that they have to keep doing something that know how to do (over and over and over). Often they’ll not really know how to do it and just keep doing problems incorrectly which creates a bad habit! The ones who don’t get it will be frustrated that they can’t even start the assignment and still have so many more problems left to do.

It’s pointless because once you can do a few problems in a section that usually means you can do all of them.

I’ll try to explain my solution to this issue, but it’s better to see an example in the links at the bottom.

Assignments should remind students about what they’ve been doing in class with a brief explanation. Questions should lead students through the step by step inner workings of solving a problem without telling them too much. After they’ve gone through questions the assignment should explain to them what they just did and how it relates to other things they know.

This type of homework is more engaging and interesting without being tedious and repetitive. Unfortunately though repetition is also necessary at some point.  For this reason I also added some practice problems which are not required to be turned. My hope is without the burden of a due date the problems will seem easier. I also asked the teaching assistant to go over these during the discussion section in class to give them more help with the boring repetition.

I believe using the discussion section like this was very helpful, but I realize not all people have this at their schools. Speaking of discussion sections, I had them do 1 question quizzes at the end of discussion sections in order to force attendance. They were supposed to be similar to practice problems that the teaching assistant went over, but also way easier. The students knew it was coming so hopefully would pay attention to the practice problems and then be able to do a simple problem.

Finally, here are all of the assignments that I wrote. They work better with my specific notes because they tie into things that I did in class, but my notes probably won’t ever be typed up.

It’s possible that there are typos which went unfixed and I’m sure things could be explained better by someone more experienced. Additionally I would have liked to cover more topics, but it was my first time running a class so I didn’t have a good sense of how much time I had to work with.

If you are interested I can send you the latex files.

homework-1

homework-2

homework-3

homework-4

homework-5

homework-6

homework-7

homework-8

homework-9

homework-10

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## Category Theory for the Working non-Mathematician 2

Last time I talked about the very basics of Category Theory, objects and morphisms and what exactly a Category consisted of. If nothing else sticks just remember that categories consist things (objects) and ways to go between things (morphisms) where you always have the ability to go from something to itself by doing nothing and you can go from thing to thing to thing via arrows composed together having a common middle object. The next step is to figure out how to go between two different categories. At the moment our view of the world is like a human stuck on the earth. There are places on earth and there are ways that we can go from place to place (routes) and we can compose our traveling methods to get from one place to another and then to another. So a category is much like a planet. However humans are curious. We look up at the stars and we want to venture to other planets. We are also very fragile creatures and we can’t just go to to any planet all willy nilly, we have to go there in a very specific way so that we don’t die. Not only that but the planet we go to has to have similar features to earth, so that again we don’t die. I know how earth looks and so the best way to survive is to correspond this new planet to the one I already know. This very restrictive way of going from one planet to another is similar to the nature of going from one category to another.

Let’s think about what it would mean to go from a category C to a category D. We start out in C, and there is some set of objects. Well D also has objects so the most logical thing to do is to have something in D correspond to something in C. The mathy way to write this is that for any object X in C, there is an object F(X) in D. Simple, elegant, no restrictions. Now, there are also morphisms (ways to go between objects) living in the category C and D. Any morphism in C goes between two objects in C and I’ve already said that the objects in C MUST be associated to some objects in D. The most logical thing to do once again would be to associate a given morphism from X to Y in C to a morphism between F(X) and F(Y) over in the category D. This part is a bit trickier because I want the morphism in C to be associated to the morphism in D in a way that makes sense. If I have a way to go from California to New York and then I go to Earth 2.0 discover California 2.0 and New York 2.0 then I’m going to want the way to go between the originals to correspond to the way to go between the new ones rather than some completely different route.

How is this going to work? First of all we need better notation. We are going to start calling our space travel from Earth to Earth 2.0, a functor. A functor $F:C\rightarrow D$ is a map from the category C to the category D that does the following. For any object X in C, there is an object F(X) in D. For any morphism between objects X,Y in C, $f:X\rightarrow Y$ there is a corresponding morphism $F(f):F(X)\rightarrow F(Y)$ which obey some rules to be explained later. For now just see how this functor is taking objects to CORRESPONDING objects in D, and paths between objects to paths between CORRESPONDING objects in D. What other rules do I need? Well if you will recall from the previous article, a category is not just objects and morphisms. It had some special rules about these things. Remember that for any object X in a category C there has to be at least this one special morphism; the identity morphism $id_X : X\rightarrow X$ which was like taking the path from X to X by not moving. In the current analogy, I can go from California to California by doing nothing at all. Since C and D are both categories, they both have these special “do nothing” paths and so logically it would be best if we took the one from C and sent it to the one in D. So our first condition for the functor F is that $F(id_X) = id_{F(X)}$. This is merely saying that F takes in the identity morphism from X to X, and sends it to the identity morphism from F(X) to F(X), which is the object in D that corresponded to X in C via the functor.

The last condition we need involves the other property that was forced upon a category in the previous post. In any category we had to be able to put two morphisms together, i.e. compose two morphisms $f:X\rightarrow Y$ and $g:Y\rightarrow Z$ to make one longer morphism $g\circ f : X\rightarrow Z$. So let’s see how this works with our functor going from one category to another. First, this requires 3 objects in C. Call them X,Y, Z. If we go from X to Y via the morphism f we can use the functor to make a corresponding path from F(X) to F(Y) via F(f) and if we go from Y to Z via g we can make a corresponding path from F(Y) to F(Z) via F(g).

Over in the land of C we were able to put f and g together to make $g\circ f:X\rightarrow Z$ so what we really want is to be able to put the corresponding morphisms F(f) and F(g) together in a way that makes sense. Since $g\circ f:X\rightarrow Z$ is itself just a morphism in C we know that $F:C\rightarrow D$ can transfer it over to a corresponding morphism in D. The functor F takes a morphism between two objects to a morphism between the two corresponding objects. So F is going to take $g\circ f$ to a functor between F(X) and F(Z). We can then write that $F(g\circ f ): F(X) \rightarrow F(Y)$. But now think about what F does to f and g separately. We have $F(f):F(X)\rightarrow F(Y)$ and $F(g):F(Y)\rightarrow F(Z)$. These are two morphisms with a common middle object, F(Y), in the category D, so of course we are allowed to stick them together to get $F(g) \circ F(f) : F(X) \rightarrow F(Z)$. We now have two morphisms living in D that go from F(X) to F(Z). Would it make sense that they are secretly the same? Yes my dear Watson it would.

Consider the states (objects) California, Texas, and New York on Earth with a route to go from California to Texas then to New York. We can use our spaceship (functor) to go over to Earth 2.0 that has corresponding states and routes. Specifically we make corresponding states, California 2.0, Texas 2.0, and New York 2.0 with corresponding routes. You and I are very different people though. You prefer a bunch of information at once and hate wasting precious time so you write down the entire damn route in a little notebook. From California to Texas then to New York. You fly over to Earth 2.0 and then you draw the corresponding entire route on Earth 2.0. However I am a simple minded peasant and need bits and pieces at a time. So first I draw the route from California to Texas and fly (using my spaceship called functor) on over to Earth 2.0, then draw it there between the California 2.0 and Texas 2.0. Then I take another trip and functor over the route between Texas and New York. The first trip ended at Texas 2.0 and the second one began at Texas 2.0, so I decide I can just glue these two routes together in my notebook. Well what difference does it make if we do it my way or your way? If there is to be any sense in the universe these two routes will be the same.

That is my long winded explanation that a functor needs to “preserve” composition. We had composition in the category C and then functored it over to D in the only logical way to do it. First composing two morphisms, then sending the resulting morphism over through the functor is exactly the same as first sending each morphism through the functor and then composing them afterwards. The next step is to go very meta which I will probably not do for another month or so. This may seem very meta already, but it isn’t.

First we talked about objects and ways to go between objects. Then we talked about a way to go between two categories, which consist of objects and ways to go between objects. When it comes down to it though, categories are themselves still sort of object like (thus far) and so going from one to another isn’t too much trouble after a lot of practice. After we achieve space travel going from one planet to another won’t be that complicated. The point is that in real life it’s typically easy to go from point A to point B. But what if I have two ways to go from A to B and because going from A to B is so easy, I want to know what it means to go between the two paths? Going from a path to a path turns out to be a lot more difficult than going from a thing to a thing.

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## Category Theory for the Working non-Mathematician 1

After another long break from writing about math in order to write about other math, I am back to write about math. Eventually I should talk about my own research so in order to get to that point I am going to attempt to explain category theory to non-mathies.

This should be easier than explaining it to math people because math people already have a stereotype about the subject. It’s way too hard! That’s what I used to think too, but like anything else all it takes is an amazing speaker to spark interest and keep attention. Unfortunately for you the reader, I am not that person, so you’ll have to deal with my mediocrity.

Math people are good at studying dots or points, but suddenly you throw in a bunch of arrows and it is too much to handle. Dots, and arrows between dots are the easiest way to start talking about category theory. Category theory is the study of categories and a category is nothing more than a bunch of dots and arrows between them that satisfy certain rules. Math needs rules, just like people. Otherwise the civilization of dots and arrows would break down into chaos!

Somehow a category is also an abstraction of all other types of math. Everything is just a bunch of dots and arrows! So lets stop procrastinating and define a category. A category consists of a bunch of dots, which we call objects and a bunch of arrows that go from an object to another object, which we fancily call morphisms . In order to actually be a category we need that for any given dot we have an arrow from that dot to itself.

What if I have an arrow going from dot A to dot B and then an arrow from dot B to dot C? Well I have to be able to put them together to have one big arrow from dot A to dot C. This is called composition of arrows. There is one more rule that will be easier to see after we have an example. So here is one in the form of a picture.

The objects consist of A, B and C. The arrows are $f$ and $g$, where $f$ goes from A to C and $g$ goes from C to B. Notice that each object has an arrow from it to itself. This is a necessary for a reason not shown in this example. It boils down to needing an arrow that does nothing. In multiplication we have the number 1, which does nothing. With addition we have 0, which does nothing.

Lastly, I can go from A to B by means of going through $f$ and then $g$. Confusingly, mathematicians write the composition of $f$ followed by $g$ as $gf$.

The final rule that I left out comes into play when we have more than two arrows in a row. It’s called associativity and may seem like a really stupid rule because most things obey it so you never actually think about it. For numbers and addition it just means something like this: $(4+3) + 5 = 4 + (3+5)$. The idea is simple, we can add the first two numbers together, then add that result with the third or we can add the second and third together and put that with the first. It also works with multiplication: $(a \times b ) \times c = a \times (b\times c)$.

Not everything is associative though! For example if you need to bake a cake. Mixing wet ingredients with dry ingredients followed by baking will give you a much tastier result than baking dry ingredients and then mixing that result with wet ingredients. $( \text{Wet} + \text{Dry} ) + \text{Baking } \neq \text{ Wet} + (\text{Dry} + \text{Baking} )$

For a category we need to do the same idea with our arrows. That is, if we have an arrow $f$ from A to B, $g$ from B to C, $h$ from C to D, it shouldn’t matter how we make up the big arrow from A to D. Putting $f$ and $g$ together, then taking that to go from A to C followed by going through $h$ to get to D is the same as putting $g$ and $h$ together first and using $f$ to get to B then taking that grouping from B to D. Here is a pretty picture to clear that long sentence up.

The dotted lines represent the composition of two arrows. Remember we can turn any two arrows that are next to each other into one long arrow (dotted arrow) by means of composition.

So lets turn the route from A to C into one arrow, the dotted arrow $gf$. Again, mathematicians annoyingly write $gf$ as meaning to first go through the arrow $f$ and then the arrow $g$. Then we can use that long arrow (dotted arrow) and the leftover arrow, $h$, to get from A to D.

However we could have went to D from A through a different route. Since the arrows $g$ and $h$ are next to each other we could have also put them together first. This is the dotted line, $hg$. Then we can go from A to D by first going through $f$ and then the dotted arrow $hg$.

Associativity is simply a complicated way of saying both of these routes are the secretly the same. (Same being a very weird idea which is left for another time.)

To recap, a category needs to have some objects, some arrows between those objects, an arrow from any object to itself, composition of arrows, and associativity of composition. I never mentioned how this abstracts all of math. The way that happens is just by thinking of objects which are more complicated than dots, and arrows with cool properties instead of just a simple arrow.

For example we could talk about sets and functions between sets as the category called Set. There are mathematical objects called groups, so we could think about the dots as groups and the arrows as the special kinds of maps that happen to go from group to group called group homomorphisms.

Some more examples are even cooler. You can think about propositions and proofs which go from proposition to proposition as a category. In computer science you have data types and programs that go from one to another. Theories and experiments can be thought of a category in the same way that propositions and proofs are. Basically anything that involves a state and a way to go from state to state is a category. Then later you see what kinds of properties these categories have.

One can even think of a category as an object and then think about what kind of arrows can go between categories. That’s for a much later time. The point is that all the cool things that math people think of can be made into a category. So instead of studying some property of a specific kind of that object, we can study what all of them have in common.

Posted in Category Theory | 2 Comments

## Is Math the same on other planets?

I went to an amazing wedding this pi day for my cousin Jonathan and his now wife Kd where the following question was posed on the dance floor by Kd’s friend Jade, “Would math be the same on Jupiter?” I don’t know if it was the occasion, the alcohol, or the dancing, but I was inspired to finally write again after getting some moderate research done over the quarter.

The real gist of the question is whether or not math is something subjective which would be completely different on some alien world. Her answer to the original question was a big “no,” that is, that math would be different on Jupiter and hence is subjective like all other things.

I agreed with Jade that everything is subjective, but said math is the only true exception! This is what makes math the coolest. I’ll try to explain why math is the same everywhere as best as I can.

I first need to mention science because a lot of people mistakenly believe that math and science are heavily intertwined. In the past this was true because scientists often came up with math accidentally while trying to explain things, but in modern times math is so specialized that only mathematicians come up with the abstract nonsense. Sometimes scientists will use modern math but typically old math is sufficient so they rarely come up with new ideas. Their relationship is better described as parasitic now. Science the parasite uses math everywhere, but math can exist on it’s own without science. Scientists observe things, make guesses, and try to show that their guesses are correct in the best way possible. The guesses and theories they make can be heavily influenced by location (but some physical truths are universal).

For example, in X amount of years which I can’t be bothered to Google, there will be no stars in our sky because they will all be too far away to observe. A scientist of that time would have no ability to observe celestial bodies and might never figure out what we’ve figured out because of that. Yes, the truth is the same either way, but a scientist cannot figure out that truth without something to study.

Math is a different beast. Math is not done through experiments first of all, we have no fancy labs to work in, we just sit there and think for a while. Then we jot something down, cross it out because it was the dumbest thing a human has ever written down, and keep thinking. Math is all about thinking and proving your thoughts with logical rules. Proofs are just collections of statements which start at some assumed statement and end at a conclusion. In between those two things you are only allowed to say objectively true facts to get to the end of the proof. You cannot use things that you believe are true, only things that are known to be true.

For example I proved in another post that there were an infinite number of prime numbers. See here for details. To prove the statement you have to assume several things, like that we have a number system, and that there is a definition of a prime number along with some more subtle assumptions that go way too deep to mention. I use these facts to eventually conclude that the statement was true.

This leads me to the next big point, assumptions.

It was long ago realized in math that you have to start somewhere. The late 1800’s saw a surge in Foundational Mathematics. People wanted to redo all of math starting from the bare minimum. The problem is that you cannot define everything because that would require an infinite amount of words. For example, if you go to a dictionary and look up a word, then you look up the words in that definition, then keep doing the same with each of those words and so on, eventually you are going to end up going in circles. Not every word can be defined by other words. Math is similar.

If I tell you that a “Set” in math is defined to be a collection of elements, you might ask, well what does a collection mean? What does an element mean? And I’ll tell you that a collection is just a grouping and an element is a thing. Then you ask what is a grouping or what is a thing? And it never ends so you simply accept the intuitive idea of what it is supposed to mean! How definitions interact with each other is what really matters! In this example, the fact that a Set consists of Elements is what matters. What we do in math is start from some beginning definitions like this, assumptions, or axioms and see what pops out when natural questions are raised.

Here is a link to David Hilbert’s axioms of Geometry. Axiom is just another word for foundational assumption. It is the most basic thing you can possibly assume because it doesn’t follow from a different assumption. All the things about geometric shapes that you know and love can be proven from the starting points in the above link and if I take these assumptions to Jupiter, all my proofs will still be true. Universal truths simply pop out of assumptions and our job is to find them. If the universe ends, all the facts about shapes will still be true starting from these axioms regardless of whether or not someone is alive to say so. The shapes don’t even have to exist for the proofs to be true. There is no such thing as infinity in our universe but we still have tons of theorems about infinity that are true regardless.

Now the cool part is that you and I can start from different axioms and get different but also true statements. With the above axioms and a lot of free time you can prove the Pythagorean Theorem as it is commonly known ($a^2+b^2 = c^2$) for the sides of a right triangle. If you change one axiom, the Parallel axiom, to say that parallel lines are allowed to cross (like on a sphere),

you get a completely different version of geometry with a different Pythagorean Theorem and they don’t contradict each other because they started from different assumptions.

This kind of stuff also allows new math to be created from minimal things. We don’t need a universe or observations to make math. We just assume some arbitrary stuff and see what happens. The assumptions don’t have to go along with physical facts, they can be whatever so long as they don’t contradict each other and are foundational. Obviously assuming some stuff and not others makes for more interesting questions and answers.

You might exclaim here that I was incorrect! An alien could come up with different theorems because they used some other axioms, I even admitted it! Yes that is absolutely true, however it doesn’t make their math different from our math. Our math is still true on their planet and theirs on ours. If an alien starts with the same assumptions, they cannot contradict what we’ve figured out. They will get the exact same results. Getting different results from different axioms is totally fine and doesn’t make our maths any different in the sense that people usually think about. One plus one is not suddenly going to equal three anywhere in the universe.

“It does not matter if we call the things chairs, tables and beer mugs or points, lines and planes.” – David Hilbert when referring to Geometry.

## Calculus: The Derivative

Since I am finally TAing for a calculus class it seemed like a good time to write about this topic. Unfortunately I still can’t make nice pictures and a good visualization is extremely useful in intuitively understanding calculus.

Before taking calculus students are taught the equations which represent lines on a graph, i.e. $y = mx+b$. The letter “m” represents the important information, the slope of the line. The slope is just how quickly the line rises compared to how far across it goes. A line with slope 2, would go up 2 units for every 1 unit it moved to the right. A line with slope $\frac{5}{13}$ would move up 5 units for every 13 units it moves to the right. Hence the familiar idea of “rise over run.”

Higher slopes indicate steeper lines. This is because the line is rising much faster than it is moving to the right. Negative slope just means going to the down and to the right. A line with slope -5 will move down 5 units and to the right one unit.

After learning how to graph, students will typically learn a bunch of other seemingly random stuff, but slope is actually the most important idea. One of the random things is the mysterious function called the “difference quotient.” The difference quotient involves a starting function, f(x), and doing what seems like an arbitrary mash up of things.

Given a function f(x), the difference quotient of f(x) is the function, $\frac{f(x+h)-f(x)}{h}$. It looks like gibberish to the untrained eye but it is secretly the slope, and here’s why.

First, $f(x+h)$ means to replace x with “x+h” in the function f(x). For example, if the function was $f(x) = \sin(x)$ then $f(x+h) = sin(x+h)$. Replacing the input with “x+h” is supposed to represent plugging in slightly more than x into the function. More precisely, plugging in “h more than x”.

For example if $f(x) = x^2 + 5$ then $f(1) = 6$ and $f(1+h) = (1+h)^2 + 5 = 1 + 2h + h^2 + 5 = 6 + 2h + h^2$. The idea here is that adding just a little more to 1, namely h, affects the function much differently than simply adding h to the end result, 6. In fact we can see here that by adding h into the input of the function, we get in general $f(x+h) = x^2 + 2xh + h^2 + 5$. So again, adding h into the input, can have a drastic affect on the output.

Now that we understand $f(x+h)$ we can look closely at $f(x+h) - f(x)$. Again, starting with an example makes it easier to see.
Let $f(x) = x^2 + 5$ and lets use an actual x and an actual h. How about, $x = 5$ and $h = 1$. Then we get:
$f(x+h) -f(x)= f(5 + 1) -f(5)$ and stop there. We are just subtracting two outputs, aka two y-values. This should remind you of calculating the slope between two points, $\frac{f(x_1)-f(x_2)}{x_1-x_2}$. Here the first point is given, and the second point comes from adding a little bit more to the first point, (h more).

Right now we have $f(x+h)-f(x)$ and to get a slope equation we need to divide by the two inputs subtracted, just like with the slope between two points we subtract the two x-values in the denominator. Well what are the two inputs here? “x+h” and “x”. Therefore, the slope between the two points $(x,f(x))$ and $(x+h,f(x+h))$ is given by
$\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}$ This is the difference quotient! So this mysterious equation was secretly the slope equation written in a special way all along. It will become clear later why we write it in this way, but for now just know that it is the slope between the points, $(x,f(x))$ and $(x+h,f(x+h))$.

Students don’t typically dissect this formula in a good way so they just forget about it. Then much later they learn about the idea of limits, which really should be shown right after slope. Limits can take a lot of rigorous explaining which I won’t go through. Instead I will talk about them in a very informal way.

Think of a function, represented by a graph, as “moving” as you plug in values. As you plug in $x= 1,2,3,4...$ what is happening to the graph? For something simple like a line, $f(x) = 2x+1$ you can easily visualize as the input x moves left and right, that the line follows what it should. The idea of seeing what happens to the function as the input x moves around is what limits are all about.

We represent this with the following notation, $\lim_{x \rightarrow c} f(x)$. This means, “the limit of f(x) as x approaches c”. We are looking at what the function does, as the inputs approach some value c. Note that we don’t care what actually happens when plugging c into f(x), but rather we care about what the function is doing as the input gets close to the value c. The typical example is a piece-wise function, i.e. a function that consists of putting other functions together. We can have functions approach some limit, L, as the inputs approach something, c, but the actual value of the function at c, f(c), is not equal to L. This can be hard to grasp and leads to the idea of continuity, but that is a topic for another time. So instead of worrying about that, just think of limits of functions as what happens to outputs as the input goes toward a certain number.

So, they teach you this thing called the difference quotient, and then they later teach you this idea of limits. The logical step is to put the two topics together in a smart way.

What happens to $\frac{f(x+h)-f(x)}{h}$ when we move h around? Consider x as just some number, and we are not moving it around. It is immovable. Think of x as a fixed point and we add the value h to it. Well, we already know that the difference quotient will give us the slope between the two points $(x,f(x))$ and $(x+h,f(x+h))$. What happens when we make h get smaller and smaller? This means we are adding less and less to the fixed point x. This corresponds to the slope of two points, where the second point is getting closer to the stationary one. Now imagine we keep making h smaller and smaller, in fact we want to make it as small as possible. We can’t make it be exactly zero, or else we would be dividing by zero. However, we can take the limit of the function as h goes toward zero. This would be written as:
$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
When h goes to zero, we can think of the second point, $(x+h,f(x+h))$ as actually right on top of the point $(x,f(x))$. So if we could actually take this limit to when h becomes zero, we would get the slope of two points which are actually the same point. Thus we magically have the slope of a line going through a single point instead of two points! Note I said “if we could actually take this limit.” This is because the limit does not have to exist in the first place, but when it does we have this neat idea of a slope of a line going through a single point.

This slope is tied to the original function we started with. So it’s not exactly just a slope of a line going through some random point, but instead is the slope of a line related to f(x). We have a special way of saying what this thing is. It is the slope of the line tangent to the graph, at the point x. A tangent line is a line that barely touches the graph once (a topic for another day).

In practice what happens is that you are given some function. Then you get the difference quotient and if it is a “good” function you will be able to manipulate it in a way that gets rid of the h in the denominator. Assuming you do mathematically correct manipulations (multiply by 1 or add zero) the end result will have the same limit that the non manipulated equation had. Then because it no longer has the risk of zero being the denominator, you will be able to plug in $h = 0$ directly to find the exact limit as h approaches zero.

After doing this entire process we finally have what is known as the derivative of a function!

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## Algebra Qual Study Problems: 7

Qual 2011 #1 (Hungerford 4.5.9a): Let I be a two-sided ideal in a ring R and let IM be the abelian subgroup of an R-module M generated by all elements of the form $xm, x\in I, m \in M$.
(i) Show that IM is a left submodule of M
(ii) Describe the natural left R-module structure on $R/I \bigotimes_R M$
(iii) Show that $R/I \bigotimes_R M \cong M/IM$ as left R-modules.
Proof- i) We are given that $IM \leq M$ by definition. Let $r\in R, \sum_{i=1}^n x_{i}m_{i}\in IM$. Then $r(\sum_{i=1}^n x_{i}m_{i} ) = \sum_{i=1}^n r (x_{i}m_{i} ) = \sum_{i=1}^n (r x_{i})m_{i}$. Since I is an ideal, $rx_{i} \in I$ so that $\sum_{i=1}^n (r x_{i})m_{i} \in IM$. Thus IM is a left submodule of M.

ii) The natural left R-module structure on $R/I \bigotimes_R M$ is given by $r(s + I\bigotimes M) = r(s+I) \bigotimes M = rs+I \bigotimes M$.

iii) Consider the map $\phi M \rightarrow R/I \bigotimes M$ by $\phi (m) = (1+I) \otimes m$ Then we can check that $\phi$ is an R-module homomorphism with $IM \subset Ker(\phi )$. Thus by the first isomorphism theorem, there exists a unique $\bar{\phi }: M/IM \rightarrow R/I \bigotimes M$. Then consider the map $\psi R/I \times M \rightarrow M/IM$ by $(r+I,m) \mapsto rm + IM$. Then it can be checked that $\psi$ is a middle linear map and well defined. Therefore it induces the unique map $\bar{\psi } : R/I \bigotimes M \rightarrow M/ IM$. Lastly it can be easily checked that $\bar{\phi } \bar{\psi } = id_{R/I \bigotimes M}$ and that $\bar{\psi } \bar{\phi } = id_{M/IM}$

Facts and counter examples to know:
1) Free module implies projective module.
2) Projective does not imply free. $\mathbb{Z}_2$ is a projective $\mathbb{Z}_6$ module but not a free $\mathbb{Z}_6$ module.
3) Torsion-free does not imply projective. $\mathbb{Q}$ is torsion free as a $\mathbb{Z}$ module but not projective as a $\mathbb{Z}$ module.
4) Projective does not imply torsion free. $\mathbb{Z}_4$ is free over itself but not torsion free since $2 \cdot 2 = 0$ but $2 \neq 0$.
5) Torsion free does not imply free. $\mathbb{Q}$ is torsion free as a $\mathbb{Z}$ module but is not free.
6)Free does not imply torsion free. See number 4.

$\mathbb{Q}$ is not a projective $\mathbb{Z}$ module
Proof- Since $\mathbb{Q}$ is a module there exists a free $\mathbb{Z}$ module A and an epimorphism $g: A \rightarrow \mathbb{Q}$ Say that A has basis X. Consider also the identity map $1_{\mathbb{Q} } \mathbb{Q} \rightarrow \mathbb{Q}$. Suppose that the rationals are projective, then by definition there exists a map $h:\mathbb{Q} \rightarrow A$ sucht that $gh = 1_{\mathbb{Q} }$. Then $h (1) = \sum_{j=1}^n m_ix_i$ with $x_i\in X$ and $m_i\in \mathbb{Z}\backslash \{0\}$. Since this is a finite sum, there exists some nonzero integer N that does not divide some $m_i$, say $m_k$. But we have that $1 = N\frac{1}{N}$ which gives that $Nh(\frac{1}{N} ) = h(1) = \sum_{j=1}^n m_ix_i$. But since X is linearly independent we have that N must divide $m_i \forall i$, which is a contradiction. Thus $\mathbb{Q}$ is not projective.

Final Exam #9: Let K be a field. We say that a K[x]-Module M is nilpotent if $x^kM=0$ for some $k>0$. Describe isomorphism classes of all nilpotent K[x]-modules M such that $dim_KM = 5$.
Proof- Recall that if F is a field then F[x] is a PID (in fact Euclidean). Since M is nilpotent, M is a torsion module. If $dim_KM=5$ then M is a finitely generated torsion module. So we may apply the structure theorem of modules over a PID to classify all such modules. M must be the direct sum of a finite number of cyclic torsion modules. A cyclic torsion module $A = Ra$ is isomorphic to $R/(r)$ where $(r) = \theta_a = \{ r\in R | ra =0\}$. So a cyclic torsion module $K[x]a$ is of the form $K[x]/(f)$ with $(f) = \theta_a = \{ f\in K[x] | fa =0\}$. The element a of M then has order f. Note the order of any element $a\in M$ is of the form $x^n$ with $n\leq 5$ because M is nilpotent with $dim_KM=5$. Therefore f must be of this form. So we can say that any cyclic torsion module $K[x]a \cong K[x]/(x^n)$ with $n\leq 5$. Then by the structure theorem M is a direct sum of a finite number of these with the property that orders of the summands, $r_1,r_2,...r_t$ satisfy $r_1 | r_2 | ... |r_t$. Since the highest order is $x^5$ we can have the following for M:
$K[x]/(x^5)$
$K[x]/(x) \bigoplus K[x]/(x^4)$
$K[x]/(x) \bigoplus K[x]/(x)\bigoplus K[x]/(x^3)$
$K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x^2)$
$K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x) \bigoplus K[x]/(x)$
$K[x]/(x^2) \bigoplus K[x]/(x^3)$
$K[x]/(x) \bigoplus K[x]/(x^2)\bigoplus K[x]/(x^2)$
Note the powers are all partitions of 5.
This solution needs a ton of explanation.

1993 Qual #4: A submodule of a finitely generated module which is not finitely generated.
Proof- Let $R = \mathbb{Z}[x_1,x_2,\ldots]$ be a polynomial ring in infinite variables. Then R is a finitely generated module over itself because $R = \langle 1 \rangle$. Recall that $\langle X\rangle$ is the set of all linear combinations of $r_ia_i$ where $r_i\in R$ and $a_i \in X$. Any polynomial in R, can be written as itself with 1 multiplied in front of each term and thus satisfies the condition to be finitely generated by $\{1\}$. Now the ideal $I = \langle x_1,\ldots ,x_n,\ldots \rangle$ is not equal to R and is not finitely generated. Suppose that the set I was finitely generated by some set A. Then A is finite so every element of A must be a finite sum of terms in finitely many factors. Then there exists an index k such that $x_k$ does not appear as a factor of any terms in A. Also, A does not contain any elements with nonzero constant term. Therefore any element of $\langle A \rangle$ which has a term with $x_k$ as a factor necessarily has a term with degree at least 2. Thus $x_k\notin \langle A\rangle$ which is a contradiction since $x_k \in I \forall k$. Thus I is not finitely generated.

2009 Qual #4: Let $R = \mathbb{R}[x]$ and $A = Rz_1 \bigoplus Rz_2 \bigoplus Rz_3$ where
$ann(z_1) = ((x+1)^2(x^2+1))$
$ann(z_2) = ((x^2+1)^2)$
$ann(z_3) = ((x^4-1)) = ((x-1)(x+1)(x^2+1))$
Find the elementary divisors and invariant factors of A.
Proof- Since the real numbers are a field, R is a pid. Thus A is a finitely generated module over a PID. So by the structure theorem we know that each summand is cyclic and that $Rz_i \cong R/ ann(z_i)$. Thus $A \cong R/((x+1)^2(x^2+1)) \bigoplus R/((x^2+1)^2)) \bigoplus ((x-1)(x+1)(x^2+1))$. The elementary divisors are the prime power factors of the generators of $ann(z_i)$.
$(x+1)^2,(x^2+1), (x^2)^2, (x-1), (x+1), (x^2+1)$
To find the invariant factors we draw the power grid as follows. There are 3 distinct factors so there are 3 columns. The factor $x^2+1$ appears the most, 3 times, so there are 3 rows. Then we arrange the factors in the following way and multiply across to get the polynomials $f_i$.
$f_1 = (x^2+1)$
$f_2 = (x^2+1) (x+1)$
$f_3 = (x^2+1)^2 (x+1)^2 (x-1)$

2003 Qual #6: Let K be a field and let M, N be finitely generated modules over the polynomial ring $K[x]$. Suppose M has invariant factors: $(x-1), (x-1)(x-2)^2, (x-1)(x-2)^2(x-3)$ and that N has invariant factors: $(x-2)(x-3)^2, (x-2)^2(x-3)^2(x-5)$.
a) Give the elementary divisors of $M \bigoplus N$
List all the prime powers of M and N:
$(x-1), (x-1), (x-2)^2, (x-1),(x-2)^2, (x-3), (x-2),(x-3)^2, (x-2)^2, (x-3)^2, (x-5)$
b) Give all the invariant factors of $M\bigoplus N$
$f_1 = (x-2)$
$f_2 = (x-2)^2(x-1)(x-3)$
$f_2 = (x-2)^2(x-1)(x-3)^2$
$f_2 = (x-2)^2(x-1)(x-3)^2(x-5)$

2011 Qual #4 (Greeinstein): Let K be a field. We say that a $K[x]$ module M is nilpotent if for every non-unit $p\in K[x]$ we have $p^nM=0$ for $n>>0$. Prove that a finitely generated nilpotent indecomposable K[x] module is isomorphic to $K[x]/(x^k)$ for some $k>0$.
Proof- Since K is a field we have that $K[x]$ is a PID. So Then M is a finitely generated nilpotent (so torsion) module over a PID. Then if $x\in K[x]$ is irreducible we have that x is prime. In particular every element of M has order $x^k$ for some $k>0$ So by the structure theorem we have $M\cong K[x]/(x^{k_1}) \bigoplus \cdots \bigoplus K[x]/(x^{k_r})$ But M is indecomposible so $r = 1$.

2010 Qual #4: Let R be the polynomial ring $\mathbb{C} [t]$. Let I be the ideal generated by $t^2$. Let $M = R/I$. Prove that M has a proper nonzero submodule and that M cannot be written as a direct sum of proper nonzero submodules. Suppose now thtat we take J to tbe the ideal generated by $t(t-1)$. Prove that $N = R/J$ is isomorphic to the sum of two proper nonzero submodules.
Proof- First consider $R/(t^2)$. A nonzero submodule of M is $A = \{\bar{at} | a \in \mathbb{C}\}$. Since $t^2$ is the only elementary divisor it can’t factor any further. Also, $R/(t(t-1)) \cong R/(t) \bigoplus R/(t-1)$. Needs a ton of explanation. A submodule of M should be of the form R/(something), how is A even a submodule?

2013 Qual #6: Let K be a field. Describe all, up to an isomorphism, nilpotent $K[x]$ module structures on $K^5$ (that is, all elements of $xK[x]$ act nilpotently) and $End_{K[x]} (K[x]/(x^5))$.
Proof- K is a field so that K[x] is a PID. So by the classification theorem, any nilpotent K[x] module which is finitely generated, M, has the structure $M \cong \bigoplus_{i=1}^n K[x]/(x^{k_i})$ for $k_i > 0$. Therefore $K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x) \bigoplus K^5 \cong K[x]/(x)$ as K[x] modules and this isomorphism is the only such nilpotent K[x]-module structure of K^5\$. (WHY?)
Next we consider $End_{K[x]} (K[x]/(x^5))$. First note that this is a K[x] module, $K[x]/(x^5)$ has basis $\{i\}$ (WHY?). So any endomoprhism is determined by where $\{i\}$ is mapped. In paricular $\langle id \rangle = End_{K[x]} (K[x]/(x^5))$ as K[x] modules and $order(id) = x^5$. Thus $End_{K[x]} (K[x]/(x^5)) \cong K[x]/(x^5)$ by the classification theorem.
Needs a lot of explanation also.

## Algebra Qual Study Problems: 6

Theorem: The following are equivalent on an R-module P-
(i) $0 \rightarrow A \rightarrow B \rightarrow P\rightarrow 0$ is split exact.
(ii) P is a direct summand of a free module
(iii) P is projective.
Proof-
$( 3\Rightarrow 1)$ Assume that P is projective and let $0\rightarrow A \xrightarrow{f} B \xrightarrow{g} P\rightarrow 0$ be a short exact sequence. By the exactness we have that the map g is surjective, and for P we have the identity map from P to P. Since P is projective there exists a map $h:P\rightarrow B$ such that $gh=1_P$. By one of the equivalent properties we know that this gives that the sequence splits.
$(1\Rightarrow 2)$ Assume that every short exact sequence $0 \rightarrow A \rightarrow B \rightarrow P\rightarrow 0$ splits. Every module is the homomorphic image of a free module, so there exists some free module F and some surjection $f: F\rightarrow P$. Then we can construct a short exact sequence, $0 \rightarrow Ker(f) \xrightarrow{i} F \xrightarrow{f} P\rightarrow 0$ where the map $i:Ker(f) \rightarrow F$ is the inclusion map. Then by assumption the sequence splits so that $F = Ker(f) \bigoplus P$.
$(2\Rightarrow 3)$ Suppose that P is a direct summand of a free module F so that $F \cong K \bigoplus P$ for some R module K. Then we have the canonical projection and inclusion maps, $\pi:F \rightarrow P$ and $i: P\rightarrow F$. Let $g:B\rightarrow C \rightarrow 0$ and $\phi :P\rightarrow C$. Since F is free and any free module is projective, there exists a map $\psi :F \rightarrow B$ such that $g\psi = \phi \pi$. Then consider $\psi i :P \rightarrow B$. We have that $g ( \psi i ) = (g \psi ) i = (\phi \pi ) i = \phi (\pi i ) = \phi 1_P = \phi$. Therefore the diagram commutes so that P is projective.

2009 Qual #3: Let R be a ring. Let M be a finitely generated R-module and let N be a projective R -module. Prove that if $f:M\rightarrow N$ is a surjective homomorphism, then the $Ker(f)$ is finitely generated.
Proof-
First consider the sequence $0\rightarrow Ker(f) \xrightarrow{i} M \xrightarrow{f} N \rightarrow 0$. This is an exact sequence because f is surjective and $i :Ker(f) \rightarrow M$ is the canonical inclusion map which is injective. Since N is projective we know that the sequence splits so that $M \cong Ker(f) \bigoplus N$. Since M is finitely generated it must follow then that $Ker(f)$ is also finitely generated.

2010 Qual #7: Suppose that $P_1,P_2$ are R-modules. Prove that $P_1 \bigoplus P_2$ is projective iff $P_1,P_2$ are projective.
Proof-
$(\Rightarrow )$ Suppose that $P_1 \bigoplus P_2$ is projective. Then there eixsts a free module F and an R module K such that $F = K\bigoplus (P_1 \bigoplus P_2) = (K \bigoplus P_1 )\bigoplus P_2 = (K \bigoplus P_2 )\bigoplus P_1$. Then by the same reason we have that $P_1, P_2$ are projective.
$(\Leftarrow )$ Suppose that $P_1,P_2$ are projective. Then there exists free modules $F_1,F_2$ and R-modules $K_1,K_2$ such that $F_1 = P_1 \bigoplus K_1$ and $F_2 = P_2\bigoplus K_2$. Then we have that $F_1\bigoplus F_2 = (P_1 \bigoplus K_1)\bigoplus (P_2\bigoplus K_2) = (K_1 \bigoplus K_2)\bigoplus (P_1\bigoplus P_2)$. Since the direct sum of two free modules is still free this gives that $P_1 \bigoplus P_2$ is projective.

Lemma 4.3.9(general case): A R-module D is a divisible R module iff D is an injective R-module.
Proof-
$(\Leftarrow )$. Let D be an injective R-module. Take $y\in D$ and $r\in R\backslash \{0_R\}$. We want to find some $x\in D$such that $y = rx$. Consider the principle ideal $(r) = rR$. Then we have $0\rightarrow (r) \xrightarrow{i}R$ is an exact sequence where $i:(r) \rightarrow R$ is the inclusion map. Define $f:(r) \rightarrow D$ by $f(r) = y$. Then since D is injective there exists a map $g:R\rightarrow D$ such that $gi=f$. So we have that $g(i(r)) = f(r) \Rightarrow g(r) = rg(1) = f(r) = y$. Therefore we choose $x= g(1)$ and we are done.

2005 Qual #4: Decide whether they are injective $\mathbb{Z}$-modules.
a) $\mathbb{Q}$
Yes because it is a divisible $\mathbb{Z}-module$
b) $\mathbb{Z}_{12}$
No, it is not divisible because $2x\neq 1 \forall x\in \mathbb{Z} \backslash \{0\}$, since 2 is not a unit. Thus it is also not injective.

Final #5b (Hungerford 4.4.3): Let $\pi : \mathbb{Z} \rightarrow \mathbb{Z}_2$ be the canonical epimorphism. The induced map $\bar{\pi} : Hom(\mathbb{Z}_2 ,\mathbb{Z} ) \rightarrow Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 )$ is the zero map. Since $Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ) \neq 0$ we have that $\bar{\pi}$ is not an epimorphism.
(Give an example of a commutative ring R, R-modules M, M’, N and $f\in Hom_R(M,M')$ such that f is surjective, but $\bar{f} : Hom_R(N,M) \rightarrow Hom_R(N,M')$ is not surjective.)
Proof-
Let $f \in Hom(\mathbb{Z}_2 ,\mathbb{Z} )$. Suppose that $f(1) = k \neq 0$. Then $0 = f(0) = f(1+1) = f(1) + f(1) = 2k \Rightarrow k =0$. This is a contradiction, so that f must be the zero map and so $Hom(\mathbb{Z}_2 ,\mathbb{Z}) = 0$. Thus $\bar{\pi} : Hom(\mathbb{Z}_2 ,\mathbb{Z} ) = 0 \rightarrow Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 )$ must be the zero map. However we have that $Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 ) \neq 0$ because the $1_{\mathbb{Z}_2} \in Hom(\mathbb{Z}_2 ,\mathbb{Z}_2 )$. Thus $\bar{\pi}$ cannot be an epimorphism.

Final #6 (Hungerford 4.4.11(a)): Let P be a finitely generated projective left module. Prove that P* is a projective right R module.
Proof-
Let P be a finitely generated projective left R module. Then there exists some free module left module F and an epimorphism $f: F \rightarrow P$ since every module is the homomorphic image of a free R module. Since P is finitely generated, F can be chosen to be finitely generated. Also, there is some left R module K such that $F = P \bigoplus K$, since P is projective. Then we get that $F* = (P\bigoplus K )* \cong P* \bigoplus K*$. Since F is a free left module which is finitely generated we get that $F*$ is a free right R module. Thus $P*$ is a direct summand of a free module and so is projective.

Final #5a (Hungerford 4.5.7): The usual injection map $f:\mathbb{Z}_2 \rightarrow \mathbb{Z}_4$ is a monomorphism of abelian groups. Show that $1\otimes f : \mathbb{Z}_2 \bigotimes \mathbb{Z}_2 \rightarrow \mathbb{Z}_2 \bigotimes \mathbb{Z}_4$ is the zero map.
Proof-
Consider what the map does to each generator: $0\otimes 0\mapsto 0$ and $1\otimes 1 \mapsto 1\otimes 2 =1 \cdot 2 \otimes 1 = 2 \otimes 1 = 0 \otimes 1 = 0$. Thus it is indeed the zero map. Note that $\mathbb{Z}_2 \bigotimes \mathbb{Z}_2 \cong \mathbb{Z}_2$ and $\mathbb{Z}_2 \bigotimes \mathbb{Z}_4 \cong \mathbb{Z}_2$ by exercise 2b. Thus we have the zero map between two nonzero spaces so that it is not a monomorphism. This also tells us that tensoring two injective maps need not result in an injective map.

Qual 2013 #3a, Final #4: a) Give an example of a divisible $\mathbb{Z}$-module. Prove that if N is a torsion right R module, M is a divisible left R module, then $N \bigotimes_R M = 0$.
Proof-
Consider the element $n\otimes m \in N\bigotimes M$. Since N is torsion there is some $r\in R$ such that $n \cdot r = 0$. M is divisible so there exists some $m'\in M$ such that $m = r m'$. So then $n\otimes m = n \otimes rm' = nr \otimes m' = 0 \otimes m' = 0$. Thus an arbitrary generator is zero so that $N\bigotimes_R = 0$.

b) Let R be a commutative ring and let S be a multiplicative set. Prove that if the natural homomorphism of R-modules, $f: M\rightarrow S^{-1} M$ by $m\mapsto \frac{sm}{s}$ is an isomorphism, then M is divisible with respect to S. Conversely if M is divisible with respect to S then the map f is surjective.
Proof-
Suppose that f is an isomorphism. Let $m\in M$ and $s\in S$. Since f is an isomorphism we have that $f^{-1}$ is an isomorphism. $m = f^{-1}(\frac{sm}{s} )= sf^{-1} ( \frac{m}{s})$ and $f^{-1}(\frac{m}{s} ) \in M$ so that M is divisible with respect to S.
Now suppose that M is divisible with respect to S. Let $\frac{m}{s} \in S^{-1}M,$ then there exists some $m'\in M$ such that $m = sm'$. So then $\frac{m}{s} = \frac{sm'}{s} = f(m')$. Thus f is surjective.

Hungerford 4.5.3b: $\mathbb{Q} \bigotimes \mathbb{Q} \cong \mathbb{Q}$.
Proof-
Consider the short exact sequence $0\rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q} / \mathbb{Z} \rightarrow 0$. Since $\mathbb{Q}$ is flat we have that $0 \rightarrow \mathbb{Q} \bigotimes \mathbb{Z} \rightarrow \mathbb{Q} \bigotimes \mathbb{Q} \rightarrow \mathbb{Q} \bigotimes \mathbb{Q} / \mathbb{Z} \rightarrow 0$ is also exact. Then since $\mathbb{Q} / \mathbb{Z}$ is torsion and $\mathbb{Q}$ is a divisible $\mathbb{Z}-$ module we have that $\mathbb{Q} \bigotimes \mathbb{Q} / \mathbb{Z} = 0$. Thus $\mathbb{Q} \bigotimes \mathbb{Z} \cong \mathbb{Q} \bigotimes \mathbb{Q}$. And since $\mathbb{Q} \cong \mathbb{Q} \bigotimes \mathbb{Z}$ we are done.

2002 Qual #4(ii): Prove or disprove, $A \bigotimes B \cong A \bigotimes C \Rightarrow B\cong C$.
Proof-
False $\mathbb{Q} \cong \mathbb{Q} \bigotimes \mathbb{Z} \cong \mathbb{Q} \bigotimes \mathbb{Q}$, but obviously the rationals are not isomorphic to the integers.

2013 Qual #5: R a ring, $f\in Hom_R(M,M')$ and $g\in Hom_R(N,N')$. Is it always true that $Ker(f\otimes g) = M \bigotimes Ker(g) + Ker(f) \bigotimes N$.
Proof-
False, consider the maps $1:\mathbb{Z}_2 \rightarrow \mathbb{Z}_2$ and $f:\mathbb{Z}_2 \rightarrow \mathbb{Z}_4$. They are both injective however $1\otimes f$ is not injective so that $Ker(1\otimes f) \neq 0$ but $M \bigotimes Ker(g) + Ker(f) \bigotimes N = M \bigotimes 0 + 0 \bigotimes N = 0$.

Qual 2011 #3: Let R be a ring and let M, N be left R-modules. Construct a natural homomorphism of abelian groups $M* \bigotimes_R N \rightarrow Hom_R(M,N)$. If R = K is a field and M, N are finite dimensional K vector spaces prove that the natural homomorphism is in fact an isomorphism of K vector spaces.
Proof-
Define the map $\phi: M* \bigotimes_R N \rightarrow Hom_R(M,N)$ by sending the element $f\otimes n \mapsto \psi$ where $\psi (m) = f(m) n$. Check that this is a homomorphism of abelian groups.
Then, assuming that R = K is a field we show that $\phi$ is an isomorphism. First,
$dim(M* \bigotimes N ) = dim(M*) dim N$ by Corollary 4.5.12
$= dim(M) dim(N)$ by Thm 4.4.11(ii)
$= dim(Hom_R(M,N))$.
Thus it is sufficient to show that $\phi$ is surjective by previous exercise. To show that $\phi$ is surjective, let $\{m_i\}_1^s$ be a basis of M and $\{n_j\}_1^t$ be a basis of N. By theorem 4.4.11, for each $m_i \in \{m_i\}_1^s$ we have that $f_{m_i}: M \rightarrow R$ given by $f_{m_i}(y) = \delta_{m_iy}$. Then these maps form a basis for $M*$ since M is finitely generated. Now, $Hom_R(M,N)$ has a basis $\{g_{ij}\}_{i = 1, j = 1}^{s,t}$ where $g_{ij}(m_k) =\delta_{ik} n_j$ with $g_{ij}$ sending $m_i$ to $n_j$ and all other $m_k$ to zero. Then by definition of $\phi$ we have $\phi (f_{m_i} \otimes n_j) (m_k )= f_{m_i}(m_k) n_j = \delta_{m_i,m_k} n_j = \delta_{i,k} n_j$. This tells us that $\phi (f_{m_i} \otimes n_j) = g_{ij}$. So every basis element $g_{ij} \in Hom_R(M,N)$ gets hit by $\phi (f_{m_i} \otimes n_j) \in M*\bigotimes N$ through $\phi$. Therefore $\phi$ is onto so we are done.