Brian Bi
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Return to table of contents for Brian's unofficial solutions to Artin's Algebra

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 \(R/J\).