Solutions of Assignment 8 Basic Algebra I

November 11, 2004

Solution of the problem 1. Recall that a field F has only two ideals: {0}, F. Also recall that the kernel of any ring homomorphism is an ideal. Now back to the problem, in order to show that f is injective, it is enough to show that ker(f) = {0}. If not, then ker(f) = F. So 1 ∈ker(f), i.e., f(1) = 0, which is a contradiction. Thus f is injective. The first isomorphism theorem now implies the other part of the problem:

F ≅ F/{0} ≅ F/ker(f) ≅ f(F).

Solution of the problem 2. Let f : ℤ ⟶ R be a ring homomorphism from ℤ to an arbitrary ring R. If ker(f) = {0}, then the argument given in the previous problem shows that ℤ is isomorphic to its image under f. And if ker(f) ≠ {0}, then it is of the form dℤ, for some d > 0. Once again, the first isomorphism theorem implies that the image of ℤ under f is isomorphic to ℤ/dℤ ≅ ℤ_{d}, which is a finite ring with d elements.

Solution of the problem 3. This is false. For example, ℤ is an integral domain, however, its quotient by the ideal 6ℤ, namely ℤ_{6}, is not an integral domain.

Solution of the problem 4. This is true. Let J be an ideal of R/I. Recall that the natural homomorphism π : R ⟶ R/I, π(a) = a + I, is a surjective ring homomorphism. We now claim that the inverse image π^{−1}(J) := {a ∈ R : π(a) ∈ J} is an ideal of R:

If a, b ∈ π^{−1}(J), then π(a + b) = π(a) + π(b) ∈ J, so a + b ∈ π^{−1}(J).

If a ∈ π^{−1}(J), r ∈ R, then π(ra) = π(r)π(a) ∈ J, so ra ∈ π^{−1}(J). Every ideal of R is assumed to be principal, so π^{−1}(J) = (a_{0}) = a_{0}R, for some a_{0} ∈ R. Now since π is onto, we conclude that

J = π(π^{−1}(J)) = π((a_{0}R)) = {π(a_{0}r) : r ∈ R} = {a_{0}r + I : r ∈ R}
= {(a_{0} + I)(r + I) : r ∈ R} = (a_{0} + I).

This means that J is generated by the element a_{0} + I. Done.

Solution of the problem 5. False. Let R = ℤ[x] and let I = (x), the ideal generated by x. We first claim that R/I ≅ ℤ. To see this, define

ϕ : R ⟶ ℤ, ϕ(f(x)) = f(0).

It is apparent that ϕ is a ring homomorphism. ϕ is also surjective (every integer can be regarded as a polynomial). Also note that

ker(ϕ) = {f(x) : ϕ(f(x)) = 0} = {f(x) : f(0) = 0} = {f(x) : x | f(x)} = I.

So, R/I ≅ ℤ, and the claim is proved.

Since every ideal of ℤ is principal, this in fact shows that every ideal of R/I is so. We now show that the same is not true for R by showing that the ideal J = {f(x) : 2 | f(0)} is not principal (it is left to you to check that J is in fact an ideal). On the contrary, suppose that J is generated by some polynomial g(x). Since 2, x ∈ J, we would have g(x) | 2, g(x) | x. So, g(x) = ±1, which is a contradiction (why?).

Solution of the problem 6. Our first claim is that for any prime p,

\frac{ℤ[x]}{(p, x^{2} + 1)} ≅ \frac{ℤ_{p}[x]}{(x^{2} + 1)}.

To see this, define ϕ : ℤ[x] ⟶ \frac{ℤ_{p}[x]}{(x^{2} + 1)} by the rule

ϕ(a_{0} + a_{1}x + ⋯ + a_{n}x^{n}) = ā_{0} + ā_{1}x + ⋯ + ā_{n}x^{n} + (x^{2} + 1),

where ā denotes the congruence class of a mod p. It is readily seen that ϕ is a surjective ring homomorphism (check this!). To find the kernel, notice that since any f(x) can be written as f(x) = a + bx + g(x)(x^{2} + 1) for some g(x) (division algorithm), so f(x) is in the kernel ⟺ ā + b̄x = 0 ⟺ p | a, p | b ⟺ f(x) ∈ (p, x^{2} + 1). The first isomorphism theorem now concludes the proof of our first claim.

Now we specialize to the case where p = 5 or p = 7.

(I) For p = 5, we have the factorization x^{2} + 1 = (x − 3)(x − 2). Let us now define

ψ : ℤ_{5}[x] ⟶ ℤ_{5} × ℤ_{5}, ψ(f(x)) = (f(3), f(2)).

ψ is clearly a ring homomorphism with the kernel

ker(ψ) = {f(x) : f(3) = f(2) = 0}
= {f(x) : x − 3 | f(x), x − 2 | f(x)}
= {f(x) : x^{2} + 1 | f(x)}
= (x^{2} + 1).

It remains to show that ψ is surjective. Given any (α, β) ∈ ℤ_{5} × ℤ_{5}, take f(x) = (3β − 2α) + (α − β)x. We then have

ψ(f(x)) = (f(3), f(2))
= (3β − 2α + 3α − 3β, 3β − 2α + 2α − 2β)
= (α, β).

Hence, by the first isomorphism theorem, we deduce that

\frac{ℤ[x]}{(5, x^{2} + 1)} ≅ \frac{ℤ_{5}[x]}{(x^{2} + 1)} ≅ ℤ_{5} × ℤ_{5}.

(II) Now suppose that p = 7. In contrast to 5, x^{2} + 1 does not factor in

ℤ_{7}[x], i.e., it is irreducible. Now we claim that \frac{ℤ_{7}[x]}{(x^{2} + 1)} is a field. To prove

this, we have to show that every nonzero class has an inverse. So, suppose that f(x) ∉ (x^{2} + 1). Thus gcd(f(x), x^{2} + 1) = 1, and since ℤ_{7} is a field, we can find g(x), h(x) ∈ ℤ_{7}[x] so that f(x)g(x) + h(x)(x^{2} + 1) = 1. Therefore (f(x) + (x^{2} + 1))(g(x) + (x^{2} + 1)) = 1 + (x^{2} + 1). In other words, the class g(x) + (x^{2} + 1) is the inverse of f(x) + (x^{2} + 1). And finally we count the

number of classes in \frac{ℤ_{7}[x]}{(x^{2} + 1)}. Since every class has a unique representative

of the form a + bx + (x^{2} + 1) with 0 ≤ a, b ≤ 6 (could you explain why?), we conclude that the total number of classes is 7 × 7 = 49. Done!

Extra Credit

Solution of the problem 7. As usual, we define the right map and will exploit it to conclude the desired result. So, consider the

ϕ : F [[x]] ⟶ F, ϕ(∑_{n=0}^{∞} a_{n}x^{n}) = a_{0}.

Now we check in details that ϕ is a surjective ring homomorphism.

(i) ϕ respects addition:

ϕ (∑_{n=0}^{∞} a_{n}x^{n} + ∑_{n=0}^{∞} b_{n}x^{n}) = ϕ (∑_{n=0}^{∞}(a_{n} + b_{n})x^{n})
= a_{0} + b_{0}
= ϕ (∑_{n=0}^{∞} a_{n}x^{n}) + ϕ (∑_{n=0}^{∞} b_{n}x^{n}) .

(ii) ϕ respects multiplication:

ϕ (∑_{n=0}^{∞} a_{n}x^{n} · ∑_{n=0}^{∞} b_{n}x^{n}) = ϕ (∑_{n=0}^{∞} (a_{0}b_{n} + a_{1}b_{n−1} + ⋯ + a_{n}b_{0}) x^{n})

= a_{0} · b_{0}
= ϕ (∑_{n=0}^{∞} a_{n}x^{n}) · ϕ (∑_{n=0}^{∞} b_{n}x^{n}) .

(iii) The identity element of the ring F [[x]] is the formal power series

1 = 1 + 0x + 0x^{2} + 0x^{3} + ⋯,

and we have ϕ(1) = 1.

(iv) ϕ is surjective: for any a ∈ F, we have

ϕ(a + 0x + 0x^{2} + 0x^{3} + ⋯) = a.

(v) The kernel of ϕ is the ideal generated by x:

f(x) = ∑_{n=0}^{∞} a_{n}x^{n} ∈ ker(ϕ) ⟺ a_{0} = 0 ⟺ f(x) = xg(x) ⟺ f(x) ∈ (x).

Therefore, the first isomorphism theorem implies that

R = \frac{F [[x]]}{(x)} ≅ F.

To prove the second part, just note that by the isomorphism established above, every element of R and off the ideal (x) corresponds to a nonzero element of a field, hence it is invertible. And now the last part is immediate: if an ideal of R is not contained in I = (x), it has to have an invertible element, hence it is the whole ring F [[x]]. Done!