Brian Bi

## 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.