Section 11.8. Maximal Ideals
Exercise 11.8.1 No principal ideal of \(\Z[x]\) is maximal:
- The ideals \((-1), (1)\) are not proper, and therefore, are not maximal.
- The ideals \((n)\), where \(n\) is an integer other than \(\pm 1\), are not maximal since \((n)\) is contained in the larger ideal \((n, x)\), which is still proper.
- Suppose \(P \in \Z[x]\) is not a constant. There exists some \(n \in \Z\) such that \(P(n) \notin \{-1, 0, 1\}\). Consider the ideal \(I = (P, P(n))\), which is strictly larger than \((P)\). If \(Q \in I\), then \(P(n) \divides Q(n)\), so \(1 \notin I\), so \(I\) is still a proper ideal. Therefore \((P)\) isn't maximal.
Exercise 11.8.2
Let \(I_1 = \R \times \{0\}\). It is clear that \(I_1\) is an ideal. If \(I_1\) is contained in a larger ideal \(I\), then \((a, b) \in I\) for some \(b \ne 0\), so \((b, b) = (b - a, 0) + (a, b) \in I\), so \((1, 1) = (b^{-1}, b^{-1})(b, b) \in I\), therefore \(I = \R \times \R\). Therefore, we conclude that \(I_1\) is a maximal ideal. Analogously, \(I_2 = \{0\} \times \R\) is a maximal ideal.
Conversely, let \(I\) be a maximal ideal. \(I\) can't be the zero ideal, since the zero ideal is contained in \(I_1\) and \(I_2\). So \(I\) contains some element \((a, b)\) where either \(a\) or \(b\) is nonzero, but not both (if it were both, then \((1, 1) = (a^{-1}, b^{-1})(a, b) \in I\), a contradiction). If \(a \ne 0\) and \(b = 0\), then \(I\) contains \(I_1\), so \(I = I_1\). If \(a = 0\) and \(b \ne 0\), then \(I\) contains \(I_2\), so \(I = I_2\). We conclude that \(I_1\) and \(I_2\) are the only maximal ideals of \(\R \times \R\).
Let \(R = \R[x]/(x^2)\). Use \(\alpha\) to denote the residue of \(x\) in \(R\). Suppose \(I \subseteq R\) is a maximal ideal. Let \(\pi : R \to R/I\) be the canonical projection. Since \(R/I\) is a field, and \(\pi(\alpha)\pi(\alpha) = \pi(\alpha^2) = \pi(0) = 0\), it must be that \(\pi(\alpha) = 0\), that is, \(\alpha \in I\).
Obviously \(I\) can't contain any nonzero constant without being the entire ring. If \(I\) is not just \((\alpha)\), then \(I\) must contain an element of the form \(a\alpha + b\) where \(a, b\) are both nonzero. In that case \(b = (a\alpha + b) - a\alpha \in I\), so again \(I = R\). We conclude that \((\alpha)\) is the sole maximal ideal of \(R\). (It is obvious enough that \((\alpha)\) is not the entire ring, but we can also check by taking \(R/(\alpha)\), which is isomorphic to \(\R\), which is a field.)
- Let \(R = \R[x]/(x^2 - 3x + 2)\). Let \(\alpha\) denote the residue of \(x\) in \(R\). Since \(x^2 - 3x + 2\) factors as \((x - 1)(x - 2)\), it follows that \((\alpha - 1)(\alpha - 2) = 0\). If \(I\) is a maximal ideal of \(R\), then \(R/I\) is a field, so at least one of \(\alpha - 1\) or \(\alpha - 2\) has to lie in \(I\). Obviously it can't be both, because then \(1 \in I\), a contradiction. Now \(R/(\alpha - 1) \cong \R[x]/(x - 1, x^2 - 3x + 2) = \R[x]/(x - 1) \cong \R\), a field, so \((\alpha - 1)\) is a maximal ideal already. Likewise \((\alpha - 2)\) is a maximal ideal. So \((\alpha - 1)\) and \((\alpha - 2)\) are the maximal ideals of \(R\).
- The polynomial \(x^2 + x + 1\) is irreducible over \(\R\), so \((x^2 + x + 1)\) is already a maximal ideal of \(\R[x]\); so the quotient ring \(\R[x]/(x^2 + x + 1)\) is in fact a field, and its only maximal ideal is the zero ideal.
Exercise 11.8.3 If \(x^3 + x + 1\) splits over \(\mathbb{F}_2\), then it has to have at least one linear factor. But for all \(x \in Z\), \(x^3\) and \(x\) have the same parity, so \(x^3 + x + 1\) is odd; therefore \(x^3 + x + 1 = 0\) has no solutions over \(\mathbb{F}_2\). We conclude that \(x^3 + x + 1\) is irreducible over \(\mathbb{F}_2\). By Proposition 11.8.4(a), \((x^3 + x + 1)\) is a maximal ideal of \(\mathbb{F}_2[x]\), so \(\mathbb{F}_2[x]/(x^3 + x + 1)\) is a field.
Over \(\mathbb{F}_3\), the polynomial \(x^3 + x + 1\) has the divisor \(x - 1\) since \(x^3 + x + 1 = 0\) when \(x = 1\). By Proposition 11.8.4(a), \((x^3 + x + 1)\) is not a maximal ideal in \(\mathbb{F}_3[x]\). Therefore \(\mathbb{F}_3[x]/(x^3 + x + 1)\) is not a field.
Exercise 11.8.4 By Proposition 11.8.4(a), the maximal ideals of \(\R[x]\) are the principal ideals generated by the monic irreducible polynomials. It is well-known that the only irreducibles in \(\R[x]\) are the linear polynomials and the quadratic polynomials with negative discriminant. So the maximal ideals of \(\R[x]\) are the principal ideals generated by polynomials of the form \(x + a\) and \(x^2 + bx + c\) where \(b^2 < 4c\). It is clear that two different polynomials from this set generate different ideals. So we want to find a bijection between these polynomials and the upper half-plane. In principle it's easy to show that one exists since these two sets both have the cardinality of the continuum, but it is also easy to exhibit the bijection explicitly. Define \begin{align*} f(x + a) &= (a, 0) \\ f(x^2 + bx + c) &= (b, 4c - b^2) \\ \end{align*} The following facts imply that \(f\) is a bijection from the monic irreducible polynomials in \(\R[x]\) to the upper half-plane (including the x-axis):
- By construction, the image of \(f\) is contained in the upper half-plane.
- Each quadratic polynomial maps to a point above the x-axis, while every linear polynomial maps onto the x-axis. Two different linear polynomials obviously map to different points. It isn't hard to see that this is also the case for two different quadratic polynomials. So \(f\) is one-to-one.
- Let \((p, q) \in \R^2\) be given, with \(q \ge 0\). If \(q = 0\), then \((p, q) = f(x + p)\). If \(q > 0\), then \((p, q) = f(x^2 + px + (q + p^2)/4)\) and it's easy to verify that this quadratic is irreducible. So \(f\) is onto.