Brian Bi
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## Section 11.6. Product Rings

Exercise 11.6.1 It is clear that $$\varphi(f) = (f(1), f(i))$$. So $$\ker f$$ is the set of real polynomials such that $$f(1) = f(i) = 0$$, which is the ideal $$(x - 1) \cap (x^2 + 1)$$. Since these two polynomials have no common factors, this ideal is equivalently generated by the single element $$(x - 1)(x^2 + 1) = x^3 - x^2 + x - 1$$. The image of $$\varphi$$ is obviously contained in $$\R \times \C$$, since $$f(1) \in \R$$ for all $$f \in \R[x]$$. Suppose $$a \in \R, b \in \C$$. Let $$f(x) = (a/2) (x^2 + 1) + (x - 1) (\Re(z) + x \Im(z))$$ where $$z = b/(i - 1)$$. Then $$f(1) = a$$ and $$f(i) = b$$, and $$f$$ has real coefficients, so indeed $$\im \varphi = \R \times \C$$.

Exercise 11.6.2 The ring $$\Z/(6)$$ is isomorphic to $$\Z/(3) \times \Z/(2)$$ and we can prove it using Proposition 11.6.2. Let $$e = 3$$; then $$e$$ is idempotent in $$\Z/(6)$$. Let $$e' = 1 - e = -2$$; we then have $$\Z/(6) \cong (\Z/(6))/(3) \times (\Z/(6))/(-2) \cong \Z/(3) \times \Z/(2)$$.

The ring $$\Z/(8)$$ is not isomorphic to $$\Z/(2) \times \Z/(4)$$. One way to see this is to note that 0 and 1 are the only idempotent elements of $$\Z/(8)$$. Another way is to observe that in $$Z/(8)$$ there are elements satisfying $$x^2 \ne 0, x^3 = 0$$, namely 2 and 6, but there are no such elements in $$\Z/(2) \times \Z/(4)$$.

Exercise 11.6.3 Let $$R$$ be a ring of order 10. The group $$R^+$$ is abelian and has order 10, so it must be isomorphic to $$C_{10}$$. Thus, $$R$$ has characteristic 10, and each element of $$R$$ is of the form $$1 + \ldots + 1$$. Thus, $$R \cong \Z/10\Z$$.

Exercise 11.6.4 By Proposition 11.5.5(a), all of these rings consist of the four elements $$0, 1, \alpha, \alpha + 1$$.

1. Let $$R$$ denote the specified ring. Here $$\alpha(\alpha + 1) = \alpha^2 + \alpha = -1 = 1$$, so the elements $$\alpha$$ and $$\alpha + 1$$ are inverses, so all nonzero elements of $$R$$ have an inverse; thus, $$R$$ is a finite field of order 4. (We could also see that this is the case using the fact that $$\alpha^2 + \alpha + 1$$ is irreducible over $$\Z/2\Z$$.)
2. Let $$R$$ denote the specified ring. Unlike in case (a), the element $$\alpha + 1$$ has no inverse, so $$R$$ isn't a field; furthermore, 0 and 1 are the only idempotent elements of $$R$$, so $$R$$ isn't a nontrivial internal direct product either. The best we can do is observe that $$R$$ is isomorphic to a ring with a slightly simpler definition: $$R = (\Z/2\Z)[x]/ (x^2 + 1) \cong (\Z/2\Z)[y]/(y^2)$$ where $$y = x + 1$$.
3. Let $$R$$ denote the specified ring. The element $$\alpha$$ is idempotent in $$R$$; by Proposition 11.6.2, $$R \cong \alpha R \times (1 - \alpha)R \cong \Z/2\Z \times \Z/2\Z$$.

Exercise 11.6.5 Let $$R$$ denote the specified ring, $$R = \R[x]/(x^2 - 1)$$, with $$\alpha$$ being the residue of $$x$$. Let $$e = \frac{1}{2}(1 + \alpha)$$ and $$e' = \frac{1}{2}(1 - \alpha)$$. By Proposition 11.5.5(a), the elements $$1, \alpha$$ form a basis of $$R$$. The element $$r_0 + r_1 \alpha$$ can be written in the form $$ce + c'e'$$, with $$c, c' \in \R$$ in exactly one way, namely $$r_0 + r_1 \alpha = (r_0 + r_1)e + (r_0 - r_1)e'$$. Finally, it is easy to verify that $$e^2 = e, e'^2 = e', ee' = 0$$, so $$R \cong \R \times \R$$ when we identify $$ce + c'e'$$ with $$(c, c')$$.

Exercise 11.6.6 The ring we want to describe is given by the quotient $$\R^2[x]/I$$ where $$I$$ is the principal ideal generated by $$(2, 0)x - (1, 1)$$. Multiplying by $$(0, 1)$$, we see that this ideal contains the element $$(0, 1)$$. Let us consider the map $$\pi : \R^2[x] \to \R^2[x]/((0, 1))$$. By the Correspondence Theorem, we have $$\R^2[x]/I \cong \R^2[x]/((0, 1))/\pi(I)$$. But quotienting out $$(0, 1)$$ from $$\R^2[x]$$ gives a ring isomorphic to $$\R[x]$$, by taking only the first element from each coefficient, in which the residue of $$(2, 0)x - (1, 1)$$ is then $$2x - 1$$. So $$\R^2[x]/I \cong \R[x]/(2x - 1) = \R[1/2]$$.

Exercise 11.6.7 In $$\Z[x]$$, a polynomial is divisible by $$2x$$ if and only if it is divisible by both $$2$$ and $$x$$, so it is clear that $$(2x) = (2) \cap (x)$$.

Consider the natural projection maps $$\pi_2 : \Z[x] \to \Z[x]/(2)$$ and $$\pi_x : \Z[x] \to \Z[x]/(x)$$. They can be equivalently viewed as maps from $$\Z[x]$$ to $$\mathbb{F}_2[x]$$ and $$\Z$$, where $$\pi_x$$ acts by the substitution $$x \leftarrow 0$$. The product map $$\varphi : \Z[x] \to \mathbb{F}_2[x] \times \Z$$ given by $$\varphi(f) = (\pi_2(f), \pi_x(f))$$ is also a homomorphism. Evidently $$\ker \varphi = (2) \cap (x) = (2x)$$. Note that $$\pi_2(f)(0) = \pi_2(f(0))$$ for each $$f$$, so whenever $$(\overline{f}, n) \in \im \varphi$$, we have that $$\overline{f}(0) \equiv n \pmod 2$$. Furthermore, each $$(\overline{f}, n)$$ of this form is the image of some $$f$$, constructed by changing the constant term of $$\overline{f}$$ to $$n$$. So $$\im \varphi$$ consists of exactly those pairs $$(\overline{f}, n)$$ with $$f(0) \equiv n \pmod 2$$. By the first isomorphism theorem, this subring of $$\mathbb{F}_2[x] \times \Z$$ is isomorphic to $$\Z[x]/(2x)$$.

Exercise 11.6.8

1. From Exercise 11.3.13 we know that $$IJ \subseteq I \cap J$$, so we only need to prove that $$I \cap J \subseteq IJ$$. Since $$I + J = R$$, write $$1 = i + j$$ where $$i \in I, j \in J$$. Let $$x \in I \cap J$$. Write $$x = 1x = (i + j)x = ix + xj$$. Since $$x \in J$$ and $$x \in I$$, it follows that $$ix + xj \in IJ$$, so $$x \in IJ$$.
2. Write $$a - b = j - i$$ where $$i \in I, j \in J$$ (we can always do this simply by taking $$a - b = i' + j'$$ and letting $$i = -i', j = j'$$). Obtain the desired solution by setting $$x = a + i = b + j$$.
3. Let $$\pi_I : R \to R/I$$ and $$\pi_J : R \to R/J$$ be the canonical projection maps. The product map $$\varphi : R \to (R/I) \times (R/J)$$ is then clearly a homomorphism. We must show that $$\varphi$$ is surjective and injective.

Let $$y \in R/I$$ and $$z \in R/J$$ be given; they are respectively additive cosets $$a + I$$ and $$b + J$$ for some $$a, b \in R$$. By part (b), there is some $$x \in R$$ such that $$x \in a + I$$ and $$x \in b + J$$. Therefore $$\varphi(x) = (y, z)$$. This establishes that $$\varphi$$ is surjective.

Now suppose $$\varphi(x) = (0, 0)$$. Then $$x \in I$$ and $$x \in J$$, that is, $$x \in I \cap J = IJ$$. So $$x = 0$$. This establishes that $$\varphi$$ is injective. So $$\varphi$$ is an isomorphism.

4. By part (b), there is an element $$e \in R$$ such that $$e \equiv 1 \pmod I$$ and $$e \equiv 0 \pmod J$$, and as we argued in part (c), this element is unique when $$IJ = 0$$. Since $$e^2 \equiv 1 \pmod I$$ and $$e^2 \equiv 0 \pmod J$$ as well, it follows that $$e = e^2$$, so $$e$$ is one of the idempotent elements we are looking for. The other is $$e' = 1 - e$$, which satisfies $$e' \equiv 0 \pmod I$$ and $$e' \equiv 1 \pmod J$$ and is similarly seen to be idempotent.

Let $$x \in R$$ and write $$x = i + j = i' + j'$$ with $$i, i' \in I, j, j' \in J$$. Then $$i - i' = j' - j$$, and since $$I \cap J = IJ = 0$$, this implies $$i - i' = j' - j = 0$$. So each $$x \in R$$ has a unique decomposition.

If $$x \in I$$, so that $$x = i + 0$$, then $$x \equiv 0 \pmod I$$ and $$x \equiv i \pmod J$$, so that $$ex \equiv 0 \pmod I$$ and $$ex \equiv 0 \pmod J$$, implying $$ex = 0$$. Conversely, if $$x = i + j$$ and $$ex = 0$$, then we have $$x \equiv j \pmod I$$, so that $$ex \equiv j \pmod I$$, so $$j \in I$$, so $$j = 0$$, so $$x \in I$$. This establishes that the kernel of the map $$x \mapsto ex$$ is $$I$$, so one of the factor rings is isomorphic to $$R/I$$. Analogously we can show that the kernel of the map $$x \mapsto e'x$$ is $$J$$, so the other factor ring is isomorphic to $$R/J$$.