## Algebra Solutions: Hungerford III.1

3) A ring R such that $a^2 = a \forall a \in R$ is called a Boolean ring. Prove that every Boolean ring R is commutative and $a+a = 0 \forall a\in R$.
Proof-
Let $a,b \in$. Since R has an underlying additive group structure we know that $a\ +b \in R \Rightarrow (a+b)^2 = a + b$. But also, $(a+b)(a+b) = a^2 + ab + ba + b^2 = a + ab + ba + b$. Thus we have $a + b = a + b + ab + ba \Rightarrow ab + ba = 0 \Rightarrow ab = -ba$. Then we have$a-b = (a-b)^2 \Rightarrow a-b = a - ab - ba + b = a - ab + ab + b = a + b \Rightarrow b = -b$ Thus $ab = -ba = ba$ so that R is commutative. Lastly, $a + a = (a+a)^2 = a^2 + a^2 + a^2 + a^2 = a + a + a + a \Rightarrow a + a = 0$.

5) If A is the abelian group $\mathbb{Z} \bigoplus \mathbb{Z}$, then End A is a noncommutative ring.
Proof-
Recall that End A is defined to be the set of all group homomorphisms from A to A with addition being pointwise and multiplication being composition. Let $f\,g \in End A$ and $a,b\in A$. Then we have $(f+g)(a+b) = f(a+b) + g(a+b) = f(a) + f(b) + g(a) + g(b) = f(a) + g(a) + f(b) + g(b) = (f+g)(a) + (f+g)(b)$ which shows that addition is well defined. Composition is similarly well defined. Addition is clearly associatve and commutative, with additive identity the 0 map and additive inverse of f being $-f$. Composition is also clearly associative and distributive over addition so that End A is a ring. Now we show that End A is noncommutative. Consider the function $f:A \rightarrow A$ defined by $f(a,b) = (b,a)$ and the function $g:A\rightarrow A$ defined by $g(a,b) = (2a,b).$ It is easily checked that they are both homomorphisms. We have that $fg(a,b) = f(2a,b) = (b,2a)$ while $gf(a,b) = g(b,a) = (2b,a)$ so that End A is not commutative.

6) A finite ring with more than one element and no zero divisors is a division ring.
Proof –
Let $a\in R$ be nonzero. Since we have a finite ring, there exists natural numbers $i>j$ such that $a^i = a^j$. First we show that there is a multiplicative identity element. Since we are in a ring we have additive inverses so we can say $a^i + (-a^j) = a^j + (-a^j) = 0$. Then we can rewrite $a^i - a^j = a\cdot a^{j-1}\cdot a^{i-j} - a\cdot a^{j-1} = a^{j-1} ( a\cdot a^{i-j} - a) = 0.$ Lastly, because we have no zero divisors one of the two factors must be 0, but a was assumed to be nonzero so we have $a\cdot a^{i-j} - a = 0 \Rightarrow a\cdot a^{i-j} = a$ which gives us the right identity. We can similarly get the left identity so that R has a multiplicative identity, call it 1. Now we must show that is a a unit, i.e. that it has an inverse. Again we can write $a^i = a^j \Rightarrow a^ja^{i-j} - a^j = 0 \Rightarrow a^j( a^{i-j} -1) = 0 \Rightarrow a^{i-j} = 1$. We can write this as $a \cdot a^{i-j-1} = 1$ so that the right inverse of a is $a^{i-j-1}$. Similarly it is the left inverse of a so that we have a is invertible and thus R is a division ring.

7) Let R be a ring with more than one element such that for each nonzero $a\in R$ there is a unique $b\in R$ such that $aba = a$. Prove:
a)R has no zero divisors
b) $bab= b$
c) R has an identity
d) R is a division ring
Proof-
a) Assume that $a\in R$ is nonzero and that $c\in R$ and assume $ac=0$. Then we can say $aca = 0$. Let $b\in R$ such that $aba = a$. Then we have $aca + aba = 0 + a = a \Rightarrow a(c+b)a = a$. However we know b is the unique element with this property so that $c + b = b \Rightarrow c=0$. Thus R has no zero divisors.
b)$aba = a \Rightarrow abab = ab \Rightarrow abab - ab = 0$. This gives $0 = a(bab-b) \Rightarrow bab - b = 0 \Rightarrow bab = b$ since a was nonzero.
c) $aba = a \Rightarrow abac = ac \Rightarrow bac = c$. This gives the element ba as the left identity. Then $bab = b \Rightarrow cbab = cb \Rightarrow cba = c$. This gives that the element ba is the right identity, and thus is the identity, call it 1.
d) $ba = 1$ gives that b is a left inverse of a and $aba = 1 \cdot a \Rightarrow ab = 1$ gives that b is also a right inverse. Thus a is a unit.

11) Let R be a commutative ring with identity of prime characteristic p. If $a,b \in R$ then $(a\pm b)^{p^n} = a^{p^n} \pm b^{p^n}$ for all integers $n\geq 0$.
Proof-
Since R is commutative with identity we have that $(a+b)^{p^n} = \sum_{k=0}^{p^n} \dbinom{p^n}{k} a^{p^n-k}b^k$. Then since $p | \dbinom{p^n}{k}$ for all $a\leq k \leq p^n-1$ and R has characteristic p, we know that $\dbinom{p^n}{k}a^{p^n-k}b^k = 0 \forall 1 \leq k \leq p^n-1$. Therefore $(a+b)^{p^n} = \dbinom{p^n}{0}a^{p^n-0}b^0 + \dbinom{p^n}{p^n}a^{p^n-p^n}b^{p^n}$. Thus $(a+b)^{p^n} = a^{p^n} + b^{p^n}$. Also $(a-b)^{p^n} = a^{p^n} + (-b)^{p^n}$. When p = 2 we have that $(-b)^{p^n} = b^{p^n}$, otherwise $(-b)^{p^n} =-b^{p^n}$. This completes the proof.

12) An element of a ring is nilpotent if $a^n=0$ for some n. Prove that in a commutative ring a+b is nilpotent if a and b are.
Proof-
Let a and b be two nilpotent elements such that $a^r = b^s = 0$ for some r,s. Take $n = r + s$ and consider $(a+b)^n= \sum_{k=0}^n \dbinom{n^k}{k} a^{n-k}b^k$. We have two cases. If $k \geq s$ then $b^k = 0$. If $k\leq s$ then $s-k \geq 0 \Rightarrow n - k = r + s-k \geq r \Rightarrow a^{n-k} = 0$. Thus $(a+b)^n = 0$ so that a + b is nilpotent.

13) In a ring R, the following are equivalent.
(a) R has no nonzero nilpotent elements
(b) If $a\in R$ and $a^2 = 0$ then $a = 0$.
Proof-
$( \Rightarrow )$ Assume that R has no nonzero nilpotent elements. If $a\in R$ and $a^2 = 0$ then a is nilpotent, but by assumption there are no nonzero nilpotent elements so that a must be 0.
$(\Leftarrow )$ Assume that if $a\in R$ and $a^2=0$ then $a= 0$. Want to show that there are no nonzero nilpotent elements. We do this by showing that $a^n = 0 \Rightarrow a=0 \forall n \in \mathbb{N}$ using induction. If n is one then it is trivially true. Assume that $a^k = 0 \Rightarrow a = 0$, for some $k \geq 1$. Then if $a^{k+1} = 0 \Rightarrow a^{k+1}a^{k-1} = 0 a^{k-1} \Rightarrow a^{2k} = 0\Rightarrow (a^k)^2 = 0 \Rightarrow a^k = 0 \Rightarrow a = 0$ by assumption and by inductive hypothesis. Thus any element which is nilpotent is zero, so the claim is true.

14) Let R be a commutative ring with identity and prime characteristic p. The map $R \rightarrow R$ given by $r \mapsto r^p$ is a homomorphism of rings called the Frobenius homomorphism.
Proof-
Call the map f. Let $a,b \in R$. Then to check that f is a ring homomorphism we note that $f(a+b) = (a+b)^p = a^p + b^p = f(a) + f(b)$ by exercise 11. Also, $f(ab) = (ab)^p = a^pb^p = f(a)f(b)$.

15)(a) Give an example of a nonzero homomorphism $f: R \rightarrow S$ of rings with identity such that $f(1_R) \neq 1_S$.
(b) If $f:R \rightarrow S$ is an epimorphism then $f(1_r) = 1_S$.
(c) If $f:R \rightarrow S$ is a homomorphism of rings with identity and u is a unit in R such that $f(u)$ is a unit in S, then $f(1_R) = 1_S$ and $f(u^{-1}) = f(u)^{-1}$. Give an example to show that $f(u)$ need not be a unit in S even though u is a unit in R.
Proof-
a) Consider $f: \mathbb{Z}_3 \rightarrow \mathbb{Z}_6$ by $f(k) = 4k$. Then for $a,b\in \mathbb{Z}$ we have $f(a+b) = 4(a+b) = 4a +4b = f(a) + f(b)$ and $f(a)f(b) = 4a4b = 16ab = 4ab \pmod{6} = f(ab)$ but note that $f(1) = 4 \neq 1$ in $\mathbb{Z}_6$.

b) Let $s\in S$. Since f is surjective we have that $s = f(r)$ for some $r\in R$. Then $f(1_R) s = f(1_r) f (r) = f(r) = s$ and similarly $sf(1_R) = s$. Since the identity element is unique and $f(1_R)$ is acting as the identity we must have that $f(1_R) = 1_S$.

c) First note that $f(u) = f(1_R u ) = f(1_R) f(u)$. Then since $f(u)$ is given to be a unit we can multiply both sides by the inverse so that $1_s = f(1_R)$. Then using this fact we have $1_S = f(1_R) = f(uu^{-1}) = f(u) f(u^{-1})$ and similarly $1_S = f(u^{-1}) f(u)$. Then since inverses are unique and $f(u^{-1})$ is acting as the inverse of $f(u)$ we must have that $f(u^{-1}) = f(u)^{-1}$. Lastly, use the same map as in part a) for the counterexample and note that 2 is a unit $\mathbb{Z}_3$, but $f(2)=2$ is not a unite in $\mathbb{Z}_6$.

16) Let $f: R \rightarrow S$ be a homomorphism of rings such that $f(r) \neq 0$ for some nonzero $r\in R$. If R has identity and S has no zero divisors, then S is a ring with identity $f(1_R)$.
Proof-
Let $r\in R$ be nonzero such that $f(r) \neq 0$. Then we have $0 = f(r)s-f(r)s = f(r1_R)s - f(r)s = f(r)f(1_R)s-f(r)s = f(r)(f(1_R)s-s)$. Then since S has no zero divisors and it was assumed that $f(r) \neq 0$ we must have that $f(1_R)s -s = 0 \Rightarrow f(1_R)s = s$. Similarly we can get that $s f(1_R) = s$ which means that $f(1_R)$ must be the multiplicative identity element in S, $1_S$.

18) Let $\mathbb{Q}$be the field of rational numbers and R any ring. If $f,g : \mathbb{Q} \rightarrow R$ are ring homomorphisms such that $f|_{\mathbb{Z}} = g|_{\mathbb{Z}}$, then f = g.
Proof –
Let x be an arbitrary rational number so that we can write $x = \frac{m}{n}$ where m and n are integers and n is nonzero. Then note that $g(1) = f(1) = f(\frac{1}{n} n) = f(\frac{1}{n} ) f( n) = f(\frac{1}{n} ) g( n)$. Then also $g(\frac{1}{n} ) = g(1 \frac{1}{n} ) = g(1) g( \frac{1}{n} ) =f(\frac{1}{n} ) g( n) g( \frac{1}{n} ) = f(\frac{1}{n} ).$ Using this finally we have that $f(x) = f(\frac{m}{n}) = mf(\frac{1}{n}) = mg(\frac{1}{n}) = g(\frac{m}{n}) = g(x)$. Thus f and g agree on the rationals.