**1) The set of all nilpotent elements in a commutative ring forms an ideal.
Proof-** Let R be a commutative ring and let . Since 0 is in N we know that N is nonempty. From exercise 3.1.12 we know that the sum of two nilpotent elements is nilpotent. Let , then say that . Therefore is nilpotent. Thus for we have that being a sum of nilpotent elements. Lastly, let . Then we must show that . Note that . Thus so that N is an ideal by the ideal test theorem.

**2) Let I be an ideal in a commutative ring R and let Rad(I) . Show that Rad(I) is an ideal.
Proof-** 0 is in Rad(I) trivially so that Rad(I) is nonempty. Let and with . We want to show that . Consider then:

. If . On the other hand if then . Thus we have that . Thus . Lastly since I is an ideal. Thus . So once again by the ideal test, Rad(I) is an ideal.

**3) If R is a ring and , then is a left ideal and is a right ideal in R.
Proof-** 0 is obviously in J so J is nonempty. Let and . Then . Then so that . Also, . Thus J is a left ideal. The proof for K is exactly the same but on the right side.

**4) If I is a left ideal of R, then is an ideal in R.
Proof-** Since we have that so that A(I) is nonempty. Now let and . Then Thus . Therefore . Lastly, and since because and I is an ideal. Thus so that A(I) is an ideal in R.

**5) If I is an ideal in a ring R, let . Prove that is an ideal of R which contains I.
Proof-** First note that so that . Thus is nonempty. Now let and . Then . Then . Therefore . Also because and . Thus . Similarly so that is an ideal. Lastly, let . Then because I is an ideal. Thus so that .

**10) (a) Show that is a principle ideal ring.
(b) Every homomorphic image of a principle ideal ring is also a principle ideal ring.
(c) is a principle ideal ring for every .
Proof- a)** Let I be an ideal in . Then being a subring of the integers it must have a lowest positive integer, call it a. Claim that . Obviously so we must show that . Let . Then we can write by the division algorithm, , where . Then . We have that , and since I is an ideal and . This gives that . But a was the least positive integer in I, so that r must be 0. Thus . Thus and since I was arbitrary, any ideal of must be principle.

**b)** Let R be a principle ideal ring and be a homomorphism. Want to show that is a principle ideal ring, i.e. that every ideal in is principle. Let J be an ideal in . Then consider . Since R is principle, and the inverse image of an ideal is an ideal, we have that I is a principle ideal, so we can write, . Since f is onto the image we know that f((a)) is an ideal in S. We have that so that:

since f is surjective onto the image.

Want to show that . Since we have that . Now let . Then since f is surjective onto the image, there exist such that . In fact, . Thus we can write . Then . Therefore so that Im(f) is a principle ideal ring.

**c)** Consider the canonical homomorphism by . Since this is a surjective homomorphism and since the integers are a principle ideal ring, from part b we have that is a principle ideal ring.

**13) Let be a homomorphism of rings, I an ideal in R, and J an ideal in S.
(a) is an ideal in R that contains Kerf.
(b) If f is an epimorphism then f(I) is an ideal in S. If f is not surjective then f(I) need not be an ideal in S.
Proof- a**) Let and . Then we have that . Also, because J is an ideal. Therefore and so is an ideal. Now let . This gives us that . So that the Kernel is contained in .

**b) **Now we assume that f is onto. Let , and . Since f is onto there exists, such that . Since I is an ideal, and then . Also, since f is onto, there exists an such that . Since I is an ideal, . Then . Similarly, , so that is an ideal in S.

**17) Let be an epimorphism of rings with kernel K.
(a) If P is a prime ideal in R that contains K, then is a prime ideal in S.
(b) If Q is a prime ideal in S, then is a prime ideal in R that contains K.
(c) There is a 1-1 correspondence between the set of all prime ideals in R that contain K and the set of all prime ideals in S, given by .
(d) If I is an ideal in a ring R, then every prime ideal in is of the form where P is a prime ideal in R that contains I.
Proof- a)** Let P be a prime ideal in R that contains K. Then since f is an epimorphism we know that is an ideal in S. Let and assume that . Since f is onto there exist such that . Then . Since is an ideal this also tells us that so that there exists such that . Then , by assumption. Therefore there is some such that . Since P is a prime ideal this tells us that or . Therefore or so that f(P) is prime.

**b)** Let Q be a prime ideal in S. Then we have that is an ideal in R that contains K. Let and assume that . Then or since Q is a prime ideal. Therefore or which tells us that is prime.

**c)** Let X be the set of all prime ideals in R that contain K and Y be the set of all prime ideals in S. Defined by . Claim that is bijective. Let . and assume that . Then, . Let . Then gives us that there exists a such that . Then . Which means that . So then , and we can similarly show that so that . Therefore is one to one. Next, let . From part b we know that is a prime ideal so it is in the set X. Then so that is surjective. Thus we have a one to one correspondence.

**d)** Let be a prime ideal in . Consider given by . Since is an epimorphism with Kernel I we have that is a prime ideal in R which contains I by part (b). Then . Then call so that , where P has the wanted properties.