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

Section 11.4. Quotient Rings

Exercise 11.4.1 The given homomorphism (let us denote it by \(\varphi\)) is surjective. The kernel \(K\) of \(\varphi\) is the ideal generated by \(x - 1\). The Correspondence Theorem asserts that ideals of \(\mathbb{Z}[x]\) that contain \(x - 1\) are in bijection with their images in \(\mathbb{Z}\), which are ideals of \(\mathbb{Z}\). By enumerating all ideals of \(\mathbb{Z}\), we can determine all ideals of \(\mathbb{Z}[x]\) that contain \(x - 1\). By Proposition 11.3.21, the ideals of \(\mathbb{Z}\) are principal ideals generated by some \(n \in \mathbb{Z}\). The inverse image \(\varphi^{-1}(n\mathbb{Z})\) consists of the polynomials \(f \in \mathbb{Z}[x]\) satisfying \(f(1) \in n\mathbb{Z}\). An equivalent condition is that polynomial division of \(f\) by \(x - 1\) yields \((x - 1)g + r\) where \(r\) is a constant divisible by \(n\). Therefore the ideal \(\varphi^{-1}(n\mathbb{Z})\) is \((x - 1, n)\), that is, all ideals of \(\mathbb{Z}[x]\) that contain \(x - 1\) are generated by \(x - 1\) together with a constant (possibly 0 in the case of \(K\) itself).

Exercise 11.4.2 Let \(\varphi\) be the homomorphism from \(\mathbb{Z}[x]\) to \(\mathbb{Z}[i]\) that sends \(x\) to \(i\). Then \(\varphi\) is surjective; the value \(a + bi\) is the image of the polynomial \(a + bx\). The kernel of \(\varphi\) is the ideal \(K\) generated by \(x^2 + 1\). The Correspondence Theorem therefore tells us that the ideals of \(\mathbb{Z}[x]\) containing \(x^2 + 1\) are precisely the inverse images of ideals of \(\mathbb{Z}[i]\) under \(\varphi\).

It turns out that \(\mathbb{Z}[i]\) is a principal ideal domain, although it's not clear whether Artin expects us to know this at this point in the text. So in fact each ideal \(I \subseteq \mathbb{Z}[i]\) can be identified by a generator \(a + bi\) and \(\varphi^{-1}(I)\) is then generated by \(x^2 + 1\) together with \(a + bx\), and every ideal of \(\mathbb{Z}[x]\) containing \(x^2 + 1\) takes this form.

Exercise 11.4.3

  1. It's not clear what the question is asking us to do when it asks to identify the ring, but here goes. Let \(I = (x^2 - 3, 2x + 4)\). Then \(2 \in I\) since \(2 = (2 - x)(2x + 4) + 2(x^2 - 3)\). Consider the homomorphism \(\varphi : \mathbb{Z}[x] \to \mathbb{F}_2[x]\) such that \(\varphi(x) = x\). Clearly \(\varphi\) is surjective and its kernel is \((2)\). Since \((2) \subseteq I\), the Correspondence Theorem tells us that \(\mathbb{Z}[x]/I \cong \mathbb{F}_2[x]/\varphi(I)\). Now \(\varphi(I) = (\varphi(x^2 - 3), \varphi(2x + 4)) = (\varphi(x^2 - 3)) = (x^2 - 1) = ((x - 1)^2)\), so \(\mathbb{Z}[x]/I \cong \mathbb{F}_2[x]/((x - 1)^2) \cong \mathbb{F}_2[y + 1]/(y^2) \cong \mathbb{F}_2[y]/(y^2)\) (where \(y = x - 1\)). So this ring has 4 elements.
  2. This problem parallels Exercise 11.4.5. By the Correspondence Theorem, \(\mathbb{Z}[i]/(2+i) \cong \mathbb{Z}[x]/(x^2 + 1, 2 + x)\), and letting \(\varphi : \mathbb{Z}[x] \to \mathbb{Z}\) be defined by \(\varphi(f) = f(-2)\), we obtain \(\mathbb{Z}[x]/(x^2 + 1, 2 + x) \cong \mathbb{Z}/(\varphi(x^2 + 1)) = \mathbb{Z}/5\mathbb{Z}\). This should not be surprising since \(2 + i\) generates the same ideal of \(\mathbb{Z}[i]\) as \(i - 2\) does.
  3. Let \(I = (6, 2x - 1)\). Then \(x + 1 = x^2(6) - (3x+1)(2x - 1) \in I\). Define \(\varphi : \mathbb{Z}[x] \to \mathbb{Z}\) by \(\varphi(f) = f(-1)\). Then \(\varphi\) is surjective and its kernel \((x + 1)\) is contained in \(I\). Therefore \(\mathbb{Z}[x]/I \cong \mathbb{Z}/\varphi(I) = \mathbb{Z}/(\varphi(6), \varphi(2x - 1)) = \mathbb{Z}/(6, -3) = \mathbb{Z}/3\mathbb{Z}\).
  4. Let \(I = (2x^2 - 4, 4x - 5)\). Then \(x - 3 = 2(2x^2 - 4) - (x + 1)(4x - 5) \in I\). Define \(\varphi : \mathbb{Z}[x] \to \mathbb{Z}\) by \(\varphi(f) = f(3)\). Then \(\varphi\) is surjective and its kernel \((x - 3)\) is contained in \(I\). Therefore \(\mathbb{Z}[x]/I \cong \mathbb{Z}/\varphi(I) = \mathbb{Z}/(\varphi(2x^2 - 4), \varphi(4x - 5)) = \mathbb{Z}/(14, 7) = \mathbb{Z}/7\mathbb{Z}\).
  5. \(\mathbb{Z}[x]/(x^2 + 3, 5) \cong \mathbb{F}_5[x]/(x^2 + 3)\). Since \(-3\) is not a square modulo 5, \(x^2 + 3\) is irreducible over \(\mathbb{F}_5\), so this ring is a finite field of order 25 (cf. Example 11.5.7(b)).

Exercise 11.4.4 They are not isomorphic. Here is one way to see this: let \(\pi_1\) and \(\pi_2\) be the canonical maps from \(\mathbb{Z}[x]\) to \(\mathbb{Z}[x]/(x^2 + 7)\) and \(\mathbb{Z}[x]/(2x^2 + 7)\), respectively. We have \(\pi_1(x)^2 = \pi_1(x^2) = -7\). If the two quotient rings were isomorphic then \(\pi_1(x)\) would have to be mapped to an element of \(\mathbb{Z}[x]/(2x^2 + 7)\) that also squares to \(-7\), since any isomorphism between the two rings must preserve the integers. Let \(P(x)\) be a polynomial such that \(\pi_2(P(x))^2 = -7\). Now \(\pi_2(P(x))^2 = \pi_2(P(x)^2) = -7\) implies \(P(x)^2 = Q(x)(2x^2 + 7) -7\) where \(Q(x) \in \mathbb{Z}[x]\). If \(P\) is regarded as a polynomial over \(\mathbb{C}\), then substituting \(x = \sqrt{-7/2}\) yields \(P(\sqrt{-7/2})^2 = -7\) so \(P(\sqrt{-7/2}) = \sqrt{-7}\). But all even powers of \(\sqrt{-7/2}\) are rational and all odd powers are rational multiples of \(\sqrt{-7/2}\) so this implies \(r_1 + r_2\sqrt{-7/2} = \sqrt{-7}\), which is not possible as \(1, \sqrt{-7/2}, \sqrt{-7}\) are linearly independent over \(\mathbb{Q}\). So it is not possible to construct an isomorphism between the two quotient rings.

Some possibly nicer solutions can be found here.