Spencer Leonardis4-7-2015

Discussion 01AMath 111B

2. Let R be a ring. Prove that if a,b ∈ R, then (a+b)2 = a2 +ab+ba+b2.Proof. We know that in a ring that a(b+ c)= ab+ac and (b+ c)a = ba+ ca. Applying the secondequality (right distribution law), we obtain (a+b)2 = (a+b)(a+b)= a(a+b)+b(a+b). We then leftdistribute a and b to obtain a(a+b)+b(a+b)= a2 +ab+ba+b2, which is the desired result. 2

4. If every x ∈ R satisfies x = x2, prove that R must be commutative.Proof. We know that for a,b ∈ R one has (a+b)2 = a2 +ab+ba+b2. Suppose that every a ∈ R satisfiesa = a2. Then a2 +ab+ba+b2 = a+ab+ba+b = a+b+ab+ba and since a+b ∈ R, we have(a+b)2 = a+b = a+b+ab+ba. Subtracting (a+b) from both sides of the preceding equality yields0= ab+ba ⇐⇒ (?) −ab = ba. Since a+a ∈ R, it follows that

a+a = (a+a)2 = a(a+a)+a(a+a)= a2 +a2 +a2 +a2 = a+a+a+a.

Thus a+a = a+a+a+a. Subtracting both sides of this equation by a+a, we see that0= a+a ⇐⇒−a = a. Thus (?) becomes −ab = ab = ba. 2

6. If D is an integral domain with Char(D)<∞, then Char(D) is prime.Proof. Suppose to the contrary that the Char(D)= x = ab for some 1< a,b < x. If we define a ringhomomorphism ψ :Z→ D, k 7→ k1D , then ψ(a) and ψ(b) are both nonzero andψ(a)ψ(b)= a1D b1D = ab1D =ψ(ab)=ψ(x)= x1D = 0 so that ψ(a) and ψ(b) are both zero divisors. Thiscontradicts our hypothesis that D is an integral domain. 2

8. If D is an integral domain and if na = 0 for some 0 6= a ∈ D and some integer n 6= 0, prove that D isof finite characteristic.Proof. We first prove that n(bc)= (nb)c holds for every b, c ∈ D and n ∈N. If n = 0 then clearly theequality holds. Observe that ((n+1)b)c = (nb+b)c = (nb)c+bc = n(bc)+bc = b(nc+ c)= b(c(n+1)). Weknow that na = 0, so (na)b = 0. Since a 6= 0, it must be the case that nb = 0, proving that D has finitecharacteristic. 2

10. Show that the commutative ring D is an integral domain if and only if for a,b, c ∈ D with a 6= 0 therelation ab = ac implies that b = c.Proof. Suppose D has the left cancellation law ab = ac ⇒ b = c. If ab = 0 for a 6= 0 then ab = 0= a0which implies that b = 0. It follows that D has no zero-divisors meaning that D is integral domain.Conversely, if D is an integral domain and ab = ac for some 0 6= a ∈ D and b, c ∈ D, then0= ab−bc = a(b− c)⇒ b− c = 0⇒ b = c, demonstrating that D cancels on the left. 2

11. Prove that Lemma 3.2.2 is false if we drop the assumption that the integral domain is finite.Proof. The ring Z is an example of an infinite integral domain that is not a field. The multiplication ofZ does not form a group since almost all the elements of Z do not have integer multiplicative inverses(ie 3−1 6∈Z since there is no a′ in Z such that 3a′ = a′3= 1 ∈Z).

1