\[ \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal} \]

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Section 11.5. Adjoining Elements

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.