Brian Bi

Exercise 11.5.1 This is just an application of Proposition 11.5.5(c), although we will want to reduce $$\alpha^5 + 1$$ first before the multiplication. Since $$x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$$, it follows that $$x^5 + 1$$ leaves the remainder 2 upon division by $$x^4 + x^3 + x^2 + x + 1$$. So $$(x^5 + 1)(x^3 + x^2 + x)$$ leaves the remainder $$2x^3 + 2x^2 + 2x$$. So the element in question is $$2\alpha^3 + 2\alpha^2 + 2\alpha$$.

Exercise 11.5.2 We are being asked to show that the ring $$R[\alpha]$$, which is defined to be $$R[x]/(x - a)$$, is isomorphic to $$R$$. Define the homomorphism $$\varphi : R[x] \to R$$ by $$\varphi(f) = f(a)$$. Then $$\varphi$$ is surjective and its kernel is $$(x - a)$$. The desired result follows from the first isomorphism theorem.

Exercise 11.5.3 This is the ring $$(\mathbb{Z}/12\mathbb{Z})[x]/(2x - 1)$$, which is isomorphic to $$\mathbb{Z}[x]/(12, 2x - 1)$$ according to the Correspondence Theorem. Notice that $$x(12) - 6(2x - 1) = 6$$, so the ideal $$(12, 2x - 1)$$ is the same as the ideal $$(6, 2x - 1)$$. We determined in Exercise 11.4.3(c) that the ring obtained by quotienting out this ideal is isomorphic to $$\mathbb{Z}/3\mathbb{Z}$$.

Exercise 11.5.4

1. $$R' \cong \Z[x]/I$$ where $$I = (2x - 6, 6x - 15)$$. Evidently $$3 = (6x - 15) - 3(2x - 6) \in I$$. Define $$\varphi : \Z[x] \to \Z_3[x]$$ to be a homomorphism with $$\varphi(x) = x$$. Then $$\varphi$$ is surjective and its kernel contains $$(3)$$, so $$\Z[x]/I \cong \Z_3[x]/\varphi(I) = \Z_3[x]/(2x) = \Z_3[x]/(x) \cong \Z_3$$.
2. $$R' \cong \Z[x]/I$$ where $$I = (2x - 6, x - 10)$$. We kill $$x - 10$$ first to obtain a ring isomorphic to $$\Z$$, and then kill the residue of $$2x - 6$$ in the resulting ring, which is $$2\cdot 10 - 6 = 14$$, yielding $$\Z_{14}$$.
3. $$R' \cong \Z[x]/I$$ where $$I = (x^3 + x^2 + 1, x^2 + x)$$. Now $$1 = (x^3 + x^2 + 1) - x(x^2 + x) \in I$$, so $$I = (1)$$. Therefore $$R'$$ is the zero ring.

Exercise 11.5.5 Yes; it's true when $$F = \Z/2\Z$$, since $$(\Z/2\Z)[x]/(x^2) \cong (\Z/2\Z)[x+1]/((x + 1)^2) = (\Z/2\Z)[x]/(x^2 + 2x + 1) = (\Z/2\Z)[x]/(x^2 - 1)$$. More generally this will be true whenever $$F$$ has characteristic 2.

Exercise 11.5.6

1. Every element $$\beta \in R'$$ can be written in the form $$b_0 + b_1 \alpha + \ldots + b_k \alpha^k$$ since it's the image of some element of $$R[x]$$. But for each $$i \in \{0, 1, \ldots, k\}$$, we have $$b_i \alpha^i = \alpha^k (b_i a^{k - i})$$. So $$\beta = \alpha^k b$$ where $$b = \sum_{i=0}^k b_i a^{k - i}$$ and $$b \in R$$.
2. Let $$\varphi$$ denote the map from $$R$$ to $$R'$$. Since $$\varphi$$ is the composition of the natural inclusion map from $$R$$ to $$R[x]$$ and the quotient map from $$R[x]$$ to $$R'$$, the kernel of $$\varphi$$ is the set of constant polynomials $$b$$ such that $$b = P(x)(ax - 1)$$ where $$P \in R[x]$$. Write $$P(x) = c_0 + c_1 x + \ldots + c_n x$$; then we have $$b = (ax - 1) (c_0 + c_1 x + \ldots + c_n x)$$. Expanding and equating terms of like powers, we obtain: \begin{align*} -c_0 &= b \\ ac_0 - c_1 &= 0 \\ \vdots \\ ac_{n-1} - c_n &= 0 \\ ac_n &= 0 \end{align*} This set of equations has a solution iff $$a^{n+1}b = 0$$, from which the desired result follows.
3. If $$a$$ is nilpotent, then part (b) implies that $$\ker \varphi$$ contains all of $$R$$, so all constant polynomials in $$R[x]$$ are mapped to zero, so all of $$R[x]$$ is mapped to zero. Conversely, if $$a$$ isn't nilpotent, then there is no $$n$$ such that $$a^n 1 = 0$$, so $$1 \notin \ker \varphi$$, so $$R'$$ is not the zero ring.

Exercise 11.5.7 It is clear that $$R'$$ is spanned by the residues of the elements $$t^i x^j$$ where $$i, j \in \N$$ (with coefficients in $$F$$). It's also clear that $$tx = 1$$, so the set $$B = \{\ldots, t^2, t, 1, x, x^2, \ldots\}$$ spans $$R'$$. We now show that they're linearly independent in $$R'$$. If $$P(x, t) = 0$$ in $$R'$$, then $$P(x, t) = Q(x, t)(tx - 1)$$ for some $$Q \in F[x, t]$$, but every nonzero polynomial of the form $$Q(x, t)(tx - 1)$$ contains some term with both $$x$$ and $$t$$, so it can't be any nonzero linear combination of the elements of $$B$$. So $$B$$ is a basis for $$R'$$, that is, every element of $$R'$$ can be uniquely written in the form $$a_n t^n + a_{n-1} t^{n-1} + \ldots + a_1 t + a_0 + a_{-1}x + \ldots + a_{-m} x^m$$, so $$R'$$ is in bijection with the Laurent polynomials if we identify $$x$$ with $$t^{-1}$$. Finally, since $$xt = 1$$, the multiplication rule for $$R'$$ is the same as that of the Laurent polynomials under this identification.