Introduction to Ring Theory. Math 228

Unless otherwise stated, homework problems are taken from Hungerford, Abstract Algebra,Second Edition.

Homework 6 - due March 9

3.3.9: If f : Z → Z is an isomorphism, prove that f is the identity map. [Hint: Whatare f(1), f(1 + 1) . . .?]

Solution: We know that f(1) = 1 and f(0) = 0, since f is an isomorphism. Alsof(1 + 1) = f(1) + f(1) = 1 + 1 = 2. We claim that f(n) = n for n > 0. We provethis by induction. For n = 0 it is true. Assume it is true for n = k, then f(k) = k.Now f(k + 1) = f(k) + f(1) = k + 1, therefore it is true for n = k + 1.

For negative numbers −n (with n > 0), we have that f(−n) = −f(n) = −n,therefore f must be the identity.

3.3.10 c,d,e: Which of the following functions are isomorphisms? Prove your statements.

(c) g : Q→ Q, defined by g(x) = 1x2+1

.

(d) h : R→M2(R), defined by h(a) =

(−a 0a 0

).

(e) f : Z12 → Z4, defined by f([x]12) = [x]4, where [u]n denoted the class of the integeru in Zn.

Solution:

(c) It is not an isomorphism, since g(1) = 12

and isomorphisms among rings withunity send 1 to 1.

(d) It is not an isomorphism, since it is not surjective, for example,

(1 00 1

)is not

in the image of h.

(e) It is not an isomorphism, since it is not injective. f([1]12) = [1]4 = f([5]12).

3.3.19: Let Z∗ denote the ring of integers with the ⊕ and � operations defined in3.1.18 (homework 5). Prove that Z is isomorphic to Z∗. [Hint: Consider the functionf : Z→ Z∗ defined by f(x) = 1− x.]

Solution: First notice that f is injective: f(x) = f(y) implies 1− x = 1− y, whichimplies x = y.

f is surjective: If y ∈ Z∗, then y = 1− (1− y) = f(1− y).

f is a homomorphism: f(x) ⊕ f(y) = f(x) + f(y) − 1 = (1 − x) + (1 − y) − 1 =1− x− y = 1− (x + y) = f(x + y).

f(x)� f(y) = f(x) + f(y)− f(x)f(y) = 1− x + 1− y− (1− x)(1− y) = 2− x− y−1 + x + y − xy = 1− xy = f(xy).

3.3.32 a: If f : R → S is an isomorphism of rings, which of the following properties arepreserved by this isomorphism? Justify your answers.

(a) a ∈ R is a zero divisor.

Solution:

(a) If a is a zero divisor, then a 6= 0R and there is a b ∈ R 6= 0R such that ab = 0R

or ba = 0R. But then f(a)f(b) = 0S or f(b)f(a) = 0S. Moreover, f(a) 6= 0S andf(b) 6= 0S because of injectivity. Then f(a) is a zero divisor.

3.3.33 f: Show that the first ring is not isomorphic to the second.

(f) Z4 × Z4 and Z16.

Solution: (f) Both rings have unity: (1, 1) and 1 respectively. If there were anisomorphism f , we would have that f((1, 1)) = 1. Then f((2, 2)) = f((1, 1) +(1, 1)) = f((1, 1)) + f((1, 1)) = 1 + 1 = 2. Then f((4, 4)) = f((2, 2) + (2, 2)) =f((2, 2)) + f((2, 2)) = 2 + 2 = 4. But (4, 4) = (0, 0) in Z4×Z4, and 4 6= 0 in Z16. Sowe get a contradiction.

4.1.3 a: List all polynomials of degree 3 in Z2[x].

Solution: Any polynomial of degree 3 in Z2[x] has the form f(x) = a0+a1x+a2x2+

a3x3, where a3 = 1 in Z2 and a0, a1, a2 may be 0 or 1 in Z2. Therefore, there are 8

possibilities:

x3, x3 + x2, x3 + x, x3 + 1, x3 + x2 + x, x3 + x2 + 1, x3 + x + 1, x3 + x2 + x + 1

4.1.10: If f(x), g(x) ∈ R[x] and f(x) + g(x) 6= 0R, show that

deg(f(x) + g(x)) ≤ max{deg(f(x)), deg(g(x))}

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Solution: Write f(x) = a0 +a1x+ . . .+anxn with ai ∈ R for 0 ≤ i ≤ n and an 6= 0R,

and g(x) = b0 + b1x + . . . + bmxm with bi ∈ R for 0 ≤ i ≤ m and bm 6= 0R. Thendeg(f(x)) = n and deg(g(x)) = m. From the definition of addition,

f(x) + g(x) = (a0 + b0) + (a1 + b1)x + (a2 + b2)x2 + . . . + (aM + bM)xM

where M = max{n, m} and we take ak = 0R for k > n and bk = 0R for k > m. Thendeg(f(x) + g(x)) ≤M = M = max{deg(f(x)), deg(g(x))}.Notice that it can only be smaller than the maximum if there is some cancelation.This can only happen if n = m and an = −bm.

4.1.18: Let D : R[x]→ R[x] be the derivative map defined by

D(a0 + a1x + a2x2 + . . . + anx

n) = a1 + 2a2x + 3a3x2 + . . . + nanx

n−1.

Is D a homomorphism of rings? An isomorphism?

Solution: If D is a homomorphism, it has to satisfy D(fg) = D(f)D(g). Takef(x) = x and g(x) = 1R. Then D(fg) = D(x) = 1R. But D(f)D(g) = D(x)D(1R) =1R0R = 0R. Then D is not a homomorphism, and it can not be an isomorphism either.

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