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 havea-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.

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