Modern Algebra – Math 542 – Spring 2007 – R. PollackSolutions for HW #5

1. Which of the following are examples of ring homomorphisms? Explain!

(a) φ : R → R defined by φ(x) = 2x.

This is not a ring homomorphism. For example,

φ(1) · φ(1) = 2 · 2 = 4,

whileφ(1 · 1) = φ(1) = 2.

(b) φ : Z5 → Z5 defined by φ(x) = x5.

This is a ring homomorphism. Indeed x5 = x (mod 5) by Fermat’slittle theorem so this map is simply the identity map.

(c) φ : Q⊕Q → Q defined by φ((a, b)) = a.

This is a ring homomorphism. The proof is a direct computation.

(d) φ : C(R, R) → R defined by φ(f(x)) = f(1).

This is a ring homomorphism. The proof is a direct computation.

(e) φ : M2(Z) → Z defined by φ

((a bc d

))= a.

This is not a ring homomorphism. For example,

φ

((1 11 1

)·(

1 11 1

))= φ

((2 22 2

))= 2,

while

φ

((1 11 1

))· φ

((1 11 1

))= 1 · 1 = 1

(f) φ : C → M2(R) defined by φ(a + bi) =(

a b−b a

).

This is a ring homomorphism. The proof is a direct computation.

2. Let φ : R → S be a ring homomorphism. Prove that φ(0) = 0.

On the one handφ(0 + 0) = φ(0)

as 0 is the additive identity. On the other hand,

φ(0 + 0) = φ(0) + φ(0)

since φ is a homomorphism. Thus,

φ(0) = φ(0) + φ(0).

Adding −φ(0) to both sides then gives

φ(0) = 0.

3. Let φ : R → S be a ring homomorphism. Prove that φ(−x) = −φ(x).

We haveφ(0) = φ(x +−x) = φ(x) + φ(−x)

since φ is a homomorphism. By the previous question, φ(0) = 0. Thus,

φ(x) + φ(−x) = 0.

Adding −φ(x) to both sides then gives

φ(−x) = −φ(x).

4. Let φ : R → S be a ring homomorphism. Recall that

φ(R) = {s ∈ S : s = φ(r) for some r ∈ R}.

Prove that φ(R) is a subring of S.

Using the subring test, let s1, s2 ∈ φ(R). Then there exists r1, r2 ∈ R suchthat φ(r1) = s1 and φ(r2) = s2. We have

s1 − s2 = φ(r1)− φ(r2) = φ(r1 − r2)

which by definition is in φ(R). The check for multiplication follows thealong the same line.

5. Let φ : R → S be a ring homomorphism. If A ⊆ R is an ideal, is it truethat φ(A) is an ideal of S? Prove this or give a counter-example.

This is false. Consider R = Z and S = C. We can then take φ to be theinclusion of Z into C. If we take A = Z, then φ(A) is again Z viewed asa subset of C. Note that Z is not an ideal of C as 1 ∈ Z and i ∈ C, but1 · i = i /∈ Z.

6. Let φ : R → S be a ring homomorphism. Let B ⊆ S be an ideal. Wedefine

φ−1(B) = {r ∈ R : φ(r) ∈ B}.

Is it true that φ−1(B) is an ideal of R? Prove this or give a counter-example.

This is true. To prove it recall the following defintion.

x ∈ φ−1(B) if and only if φ(x) ∈ B.

Let x, y ∈ φ−1(B). We must show that the x−y ∈ φ−1(B) or equivalentlythat φ(x− y) ∈ B. Since x, y ∈ φ−1(B), we have φ(x) ∈ B and φ(y) ∈ B.Thus

φ(x− y) = φ(x)− φ(y) ∈ B

as B is an ideal. Therefore x− y ∈ φ−1(B).

Now let x ∈ φ−1(B) and r ∈ R. We must check that rx ∈ φ−1(B) orequivalently that φ(rx) ∈ B. Since x ∈ φ−1(B), we have φ(x) ∈ B. Thus

φ(rx) = φ(r)φ(x) ∈ B

as B is an ideal and φ(r) ∈ S. Therefore, rx ∈ φ−1(B).

Chapter 15:

21. Determine all ring homomorphisms from Z to Z.

Let φ : Z → Z be a ring homomorphism. Note that for n ∈ Z,

φ(n) = nφ(1).

Thus φ is completely determined by its value on 1.

Since 1 is an idempotent in Z (i.e. 12 = 1), we know by question #24 that φ(1)is again idempotent.

Let’s determine all of the idempotents of Z. To this end, take x ∈ Z such thatx2 = x. Thus

x2 − x = x(x− 1) = 0.

Since Z is an integral domain, we deduce that either x = 0 or x = 1. Thus thecomplete list of idempotents of Z are 0 and 1.

Thus φ(1) being idempotent implies that either φ(1) = 0 or φ(1) = 1. In thefirst case, φ(n) = 0 for all n and in the second case φ(n) = n for all n. Thus,the only ring homomorphisms from Z to Z are the zero map and the identitymap.

22. Suppose φ is a ring homomorphism from Z ⊕ Z to Z ⊕ Z. What are thepossibilities for φ((1, 0))?

Note that (1, 0)2 = (12, 0) = (1, 0) and thus (1, 0) is idempotent. By question#24, we then have that φ((1, 0)) is idempotent.

So let’s determine all idempotents of Z ⊕ Z. So take (x, y) ∈ Z ⊕ Z such that(x, y)2 = (x, y). Then x2 = x and y2 = y and by the solution to #21, we havethat x = 0 or 1 and y = 0 or 1.

Thus the only possibilities for φ((1, 0)) are (0, 0),(1, 0),(0, 1) and (1, 1).

23. Determine all ring homomorphisms from Z⊕ Z to Z⊕ Z.

A ring homomorphism from Z⊕Z to Z⊕Z is determined by its values on (1, 0)and (0, 1). Indeed,

(a, b) = a(1, 0) + b(0, 1)

and thusφ((a, b)) = aφ((1, 0)) + bφ((0, 1)).

By question #22 the only possible values for φ((1, 0)) are (0, 0),(1, 0),(0, 1) and(1, 1). Similarly, the only possible values for φ((0, 1)) are these same 4 values.Thus, in total there are at most 16 possible ring homomorphisms from Z⊕Z toZ⊕ Z.

However, not all of these 16 maps are ring homomorphisms. Let’s make a list:

1. φ((1, 0)) = (1, 0), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (a + b, 0).

2. φ((1, 0)) = (0, 1), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (b, a).

3. φ((1, 0)) = (1, 1), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (a + b, a).

4. φ((1, 0)) = (0, 0), φ((0, 1)) = (1, 0). Thus φ((a, b)) = (b, 0).

5. φ((1, 0)) = (1, 0), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (a, b).

6. φ((1, 0)) = (0, 1), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (0, a + b).

7. φ((1, 0)) = (1, 1), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (a, a + b).

8. φ((1, 0)) = (0, 0), φ((0, 1)) = (0, 1). Thus φ((a, b)) = (0, b).

9. φ((1, 0)) = (1, 0), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (a + b, b).

10. φ((1, 0)) = (0, 1), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (b, a + b).

11. φ((1, 0)) = (1, 1), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (a + b, a + b).

12. φ((1, 0)) = (0, 0), φ((0, 1)) = (1, 1). Thus φ((a, b)) = (b, b).

13. φ((1, 0)) = (1, 0), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (a, 0).

14. φ((1, 0)) = (0, 1), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (0, a).

15. φ((1, 0)) = (1, 1), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (a, a).

16. φ((1, 0)) = (0, 0), φ((0, 1)) = (0, 0). Thus φ((a, b)) = (0, 0).

Note that if φ is going to be a ring homomorphism, it must send (1, 1) to anidempotent (as (1, 1) is clearly idempotent). Looking at the above 16 maps, wehave that maps 1,3,6,7,9,10 and 11 don’t have this property and thus can’t bering homomorphisms. (For instance, the first map sends (1, 1) to (2, 0) whichisn’t an idempotent.)

This leaves us with the 9 maps:

φ((a, b)) = (a, b), (b, a), (b, 0), (0, b), (b, b), (a, 0), (0, a), (a, a) or (0, 0)

which one can verify are in fact all ring homomorphisms.

24. Prove that a ring homomorphism carries an idempotent to an idempotent.

Let a ∈ R be an idempotent. This means that a2 = a. We need to show thatφ(a) is an idempotent. We have

φ(a)2 = φ(a2) = φ(a)

and thus φ(a) is idempotent.