*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

- 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. - 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.
- 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 \(\mathbb{Z}[x]/I \cong \mathbb{Z}/\varphi(I) = \mathbb{Z}/(\varphi(6), \varphi(2x - 1)) = \mathbb{Z}/(6, -3) = \mathbb{Z}/3\mathbb{Z}\).
- 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 \(\mathbb{Z}[x]/I \cong \mathbb{Z}/\varphi(I) = \mathbb{Z}/(\varphi(2x^2 - 4), \varphi(4x - 5)) = \mathbb{Z}/(14, 7) = \mathbb{Z}/7\mathbb{Z}\).
- \(\mathbb{Z}[x]/(x^2 + 3, 5) \cong \mathbb{F}_5[x]/(x^2 + 3)\). 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.