Commutative Algebra: Problems and Solutions

Since \(R \neq \{0\}\), the zero ideal \((0)\) is an ideal other than \(R\), hence it is prime by hypothesis. Therefore \(R\) is an integral domain: if \(xy = 0\) then \(xy \in (0)\), so \(x \in (0)\) or \(y \in (0)\).

Now let \(a \in R\) be a non-zero element. We show that \(a\) is a unit. Consider the ideal \((a^2)\). There are two cases.

If \((a^2) = R\), then \(1 \in (a^2)\), so \(1 = r a^2 = (ra) a\) for some \(r \in R\), and \(a\) is a unit.

If \((a^2) \neq R\), then \((a^2)\) is prime by hypothesis. Since \(a \cdot a = a^2 \in (a^2)\), primality gives \(a \in (a^2)\), so \(a = r a^2\) for some \(r \in R\). Hence \(a(1 - ra) = 0\). As \(R\) is an integral domain and \(a \neq 0\), we get \(1 - ra = 0\), that is, \(ra = 1\), and again \(a\) is a unit.

Thus every non-zero element of \(R\) is invertible, and \(R \neq \{0\}\), so \(R\) is a field. □

We use the fact that every proper ideal of \(R\) is contained in a maximal ideal; this is where Zorn’s lemma enters.

Suppose first that \(x \in \mathcal {J}(R)\) and let \(y \in R\). If \(1-xy\) were not a unit, then \((1-xy)\) would be a proper ideal, hence contained in some maximal ideal \(\mathfrak {m}\). Since \(x \in \mathcal {J}(R) \subset \mathfrak {m}\), we have \(xy \in \mathfrak {m}\), and therefore \(1 = (1-xy) + xy \in \mathfrak {m}\), contradicting the fact that \(\mathfrak {m}\) is proper. Hence \(1-xy\) is a unit.

Conversely, suppose that \(1-xy\) is a unit for all \(y \in R\), and suppose for contradiction that \(x \notin \mathcal {J}(R)\). Then there is a maximal ideal \(\mathfrak {m}\) with \(x \notin \mathfrak {m}\). By maximality, \(\mathfrak {m} + (x) = R\), so we can write \(1 = m + xy\) with \(m \in \mathfrak {m}\) and \(y \in R\). But then \(1-xy = m \in \mathfrak {m}\) is a unit, which forces \(\mathfrak {m} = R\), a contradiction. Hence \(x \in \mathcal {J}(R)\). □

Since \(\mathfrak {p} = I \cap J\), we have \(\mathfrak {p} \subset I\) and \(\mathfrak {p} \subset J\). Suppose for contradiction that neither inclusion is an equality, and choose \(a \in I \setminus \mathfrak {p}\) and \(b \in J \setminus \mathfrak {p}\). Then \(ab \in I\) because \(a \in I\), and \(ab \in J\) because \(b \in J\), so \(ab \in I \cap J = \mathfrak {p}\). As \(\mathfrak {p}\) is prime, \(a \in \mathfrak {p}\) or \(b \in \mathfrak {p}\), contradicting the choice of \(a\) and \(b\). Hence \(\mathfrak {p} = I\) or \(\mathfrak {p} = J\). □

Suppose first that \(R \setminus S = \bigcup _{\lambda } \mathfrak {p}_\lambda \) for some family of prime ideals \(\mathfrak {p}_\lambda \). Since \(1\) lies in no proper ideal, \(1 \in S\). If \(x, y \in S\) and \(xy \notin S\), then \(xy \in \mathfrak {p}_\lambda \) for some \(\lambda \), so \(x \in \mathfrak {p}_\lambda \) or \(y \in \mathfrak {p}_\lambda \) by primality; but \(\mathfrak {p}_\lambda \subset R \setminus S\), a contradiction. Hence \(S\) is a multiplicative set. It is saturated: if \(xy \in S\) and, say, \(x \notin S\), then \(x \in \mathfrak {p}_\lambda \) for some \(\lambda \), hence \(xy \in \mathfrak {p}_\lambda \subset R \setminus S\), a contradiction.

Conversely, suppose that \(S\) is a saturated multiplicative set. We claim that every \(x \in R \setminus S\) lies in a prime ideal disjoint from \(S\); since every such prime ideal is contained in \(R \setminus S\), this shows that \(R \setminus S\) is the union of the prime ideals disjoint from \(S\). So let \(x \notin S\). The ideal \((x)\) is disjoint from \(S\): if \(rx \in S\) for some \(r \in R\), then \(x \in S\) by saturation, which is not the case. By Krull’s lemma from the lectures (whose proof uses Zorn’s lemma), there exists a prime ideal \(\mathfrak {p}\) containing \((x)\) and disjoint from \(S\). Thus \(x \in \mathfrak {p} \subset R \setminus S\), as claimed. □

For \(z = a + bi \in \mathbb {Z}[i]\) we have \(N(z) = a^2 + b^2 = |z|^2\), where \(|z|\) denotes the usual absolute value of the complex number \(z\). In particular \(N\) takes values in \(\mathbb {N}\), \(N(z) = 0\) only for \(z = 0\), and \(N(zw) = N(z)N(w)\) since \(|zw| = |z||w|\). Also, \(\mathbb {Z}[i]\) is an integral domain, being a subring of the field \(\mathbb {C}\).

Let \(\alpha , \beta \in \mathbb {Z}[i]\) with \(\beta \neq 0\). We must find \(q, r \in \mathbb {Z}[i]\) with \(\alpha = q\beta + r\) and \(N(r) < N(\beta )\). Consider the complex number \(\alpha /\beta = u + vi\) with \(u, v \in \mathbb {Q}\). Choose integers \(m, n\) with \(|u - m| \leq \frac {1}{2}\) and \(|v - n| \leq \frac {1}{2}\), and put \(q = m + ni \in \mathbb {Z}[i]\). Then \[ \left | \frac {\alpha }{\beta } - q \right |^2 = (u-m)^2 + (v-n)^2 \leq \frac {1}{4} + \frac {1}{4} = \frac {1}{2} < 1 . \] This is the geometric fact in the hint: every complex number lies at distance less than \(1\) from some Gaussian integer. Now set \(r = \alpha - q\beta \in \mathbb {Z}[i]\). Then \[ N(r) = |\alpha - q\beta |^2 = |\beta |^2 \left | \frac {\alpha }{\beta } - q \right |^2 \leq \frac {1}{2} N(\beta ) < N(\beta ), \] as required. □

If \(f = 0\) then clearly \(f\) vanishes at every point of \(k^n\). For the converse we argue by induction on \(n\).

For \(n = 1\), a non-zero polynomial \(f \in k[X_1]\) of degree \(d\) has at most \(d\) roots in \(k\). Since \(k\) is infinite, a polynomial vanishing at every point of \(k\) must therefore be zero.

Now let \(n > 1\) and suppose that \(f\) vanishes at every point of \(k^n\). Write \(f\) as a polynomial in \(X_n\) with coefficients in \(k[X_1,\ldots , X_{n-1}]\): \[ f = \sum _{j=0}^{d} g_j(X_1,\ldots , X_{n-1}) X_n^j . \] Fix a point \((x_1,\ldots , x_{n-1}) \in k^{n-1}\). The one-variable polynomial \[ f(x_1,\ldots , x_{n-1}, X_n) = \sum _{j=0}^{d} g_j(x_1,\ldots , x_{n-1}) X_n^j \in k[X_n] \] vanishes at every \(x_n \in k\), so by the case \(n=1\) it is the zero polynomial, that is, \(g_j(x_1,\ldots , x_{n-1}) = 0\) for all \(j\). As this holds for every point of \(k^{n-1}\), the induction hypothesis gives \(g_j = 0\) for all \(j\), hence \(f = 0\).

For \(k = \mathbb {F}_2\), the polynomial \(f = X_1^2 - X_1 \in \mathbb {F}_2[X_1,\ldots , X_n]\) is non-zero, but \(f(x_1,\ldots ,x_n) = x_1^2 - x_1 = 0\) for every point of \(\mathbb {F}_2^n\), since \(0^2 = 0\) and \(1^2 = 1\). (More generally, \(X_1^q - X_1\) vanishes on \(\mathbb {F}_q^n\) for any finite field \(\mathbb {F}_q\).) □

The inclusion \(\iota : \mathbb {Z} \to \mathbb {Z}[X]\) is a ring homomorphism and \(\mathfrak {p} \cap \mathbb {Z} = \iota ^{-1}(\mathfrak {p})\). The preimage of a prime ideal under a ring homomorphism is prime (Lecture 2), so \(\mathfrak {p} \cap \mathbb {Z}\) is a prime ideal of \(\mathbb {Z}\). The prime ideals of \(\mathbb {Z}\) are \((0)\) and \((p)\) for \(p\) a prime number, which gives the second claim.

(i) Suppose \(\mathfrak {p} \cap \mathbb {Z} = \{0\}\) and \(\mathfrak {p} \neq (0)\). Recall that the content of a non-zero polynomial in \(\mathbb {Z}[X]\) is the greatest common divisor of its coefficients, that a polynomial is primitive if its content is \(1\), and Gauss’s lemma: the content of a product is the product of the contents. In particular, if \(f\) is primitive and \(f\) divides \(dg\) in \(\mathbb {Z}[X]\) for some non-zero \(d \in \mathbb {Z}\), then \(f\) divides \(g\): writing \(dg = qf\), the content of \(q\) equals the content of \(qf = dg\), which is divisible by \(d\), so \(q = d q'\) with \(q' \in \mathbb {Z}[X]\) and \(g = q' f\).

Choose a non-zero \(f \in \mathfrak {p}\) of smallest possible degree. Writing \(f = c f_0\) with \(c \in \mathbb {Z}\) the content of \(f\) and \(f_0\) primitive, we have \(c \notin \mathfrak {p}\) (as \(\mathfrak {p} \cap \mathbb {Z} = \{0\}\) and \(c \neq 0\)), so \(f_0 \in \mathfrak {p}\) by primality. Replacing \(f\) by \(f_0\), we may assume that \(f\) is primitive. We claim that \(\mathfrak {p} = (f)\). Let \(g \in \mathfrak {p}\). Dividing \(g\) by \(f\) in \(\mathbb {Q}[X]\) and clearing denominators, we find a non-zero \(d \in \mathbb {Z}\) and \(q, r \in \mathbb {Z}[X]\) with \[ d g = q f + r, \qquad r = 0 \text { or } \deg r < \deg f. \] Since \(r = dg - qf \in \mathfrak {p}\), the minimality of \(\deg f\) forces \(r = 0\). Thus \(f\) divides \(dg\), and by the consequence of Gauss’s lemma above, \(f\) divides \(g\) in \(\mathbb {Z}[X]\). Hence \(\mathfrak {p} = (f)\). Finally, \(f\) is irreducible: it is non-zero and not a unit, and if \(f = gh\) then \(g \in \mathfrak {p}\) or \(h \in \mathfrak {p}\), say \(g = f u\) with \(u \in \mathbb {Z}[X]\); then \(f = fuh\), so \(uh = 1\) and \(h\) is a unit.

(ii) Suppose \(\mathfrak {p} \cap \mathbb {Z} = (p)\) with \(p\) a prime number. Then \((p) \subset \mathfrak {p}\). Let \(\pi : \mathbb {Z}[X] \to \mathbb {Z}[X]/(p) \simeq \mathbb {F}_p[X]\) be the reduction mod \(p\), and write \(\overline {g} = \pi (g)\). The ideals of \(\mathbb {F}_p[X]\) correspond to the ideals of \(\mathbb {Z}[X]\) containing \((p)\), and under this correspondence \(\mathfrak {p}\) corresponds to \(\overline {\mathfrak {p}} = \pi (\mathfrak {p})\), which is a prime ideal since \(\mathbb {F}_p[X]/\overline {\mathfrak {p}} \simeq \mathbb {Z}[X]/\mathfrak {p}\) is an integral domain. As \(\mathbb {F}_p[X]\) is a PID, \(\overline {\mathfrak {p}} = (\overline {f})\) where \(\overline {f} = 0\) or \(\overline {f}\) is an irreducible polynomial, which we may take to be monic after multiplying by a unit. Let \(f \in \mathbb {Z}[X]\) be \(0\) if \(\overline {f} = 0\), and otherwise a monic polynomial lifting \(\overline {f}\) (lift each coefficient to an integer, taking the leading coefficient to be \(1\)). Then for \(g \in \mathbb {Z}[X]\) we have \begin {align*} g \in \mathfrak {p} &\iff \overline {g} \in (\overline {f}) \\ &\iff \overline {g} = \overline {h}\,\overline {f} \text { for some } h \in \mathbb {Z}[X] \\ &\iff g - hf \in (p) \text { for some } h \in \mathbb {Z}[X] \\ &\iff g \in (p,f), \end {align*}

so \(\mathfrak {p} = (p, f)\) as required. □

First, \(I\) is a proper ideal, since \(R = (1)\) is principal. Suppose for contradiction that \(I\) is not prime, so that there exist \(a, b \in R \setminus I\) with \(ab \in I\).

The ideal \(I + (a)\) strictly contains \(I\) because \(a \notin I\), so by maximality it is principal: \(I + (a) = (c)\) for some \(c \in R\). Consider the ideal \[ J \coloneqq (I : c) = \{ r \in R : rc \in I \}. \] We have \(I \subset J\), since \(I\) is an ideal. Moreover \(b \in J\): writing \(c = x + ta\) with \(x \in I\) and \(t \in R\), we get \(bc = bx + t(ab) \in I\). As \(b \notin I\), the ideal \(J\) strictly contains \(I\), so it is principal: \(J = (d)\) for some \(d \in R\).

We claim that \(I = (cd)\). On the one hand, \(d \in J\) means \(cd \in I\), so \((cd) \subset I\). On the other hand, let \(y \in I\). Since \(I \subset I + (a) = (c)\), we can write \(y = rc\) with \(r \in R\); then \(rc \in I\) says \(r \in J = (d)\), so \(r = sd\) for some \(s \in R\) and \(y = s\,cd \in (cd)\). Hence \(I = (cd)\) is principal, contradicting the hypothesis. Therefore \(I\) is prime. □

Let \(R\) be a local ring with unique maximal ideal \(\mathfrak {m}\). A unit lies in no proper ideal, since an ideal containing a unit contains \(1\); hence every unit lies in \(R \setminus \mathfrak {m}\). Conversely, let \(x \in R \setminus \mathfrak {m}\) and suppose that \(x\) is not a unit. Then \((x)\) is a proper ideal, so it is contained in some maximal ideal of \(R\) (this uses Zorn’s lemma, through the proposition of Lecture 2 that every proper ideal lies in a maximal ideal). Since \(\mathfrak {m}\) is the only maximal ideal, \(x \in (x) \subset \mathfrak {m}\), a contradiction. Thus the units are exactly the elements of \(R \setminus \mathfrak {m}\).

Now let \(R\) be a ring whose set \(N\) of non-units is an ideal. Then \(N\) is a proper ideal, as \(1 \notin N\). Every proper ideal of \(R\) consists of non-units, by the remark above, and is therefore contained in \(N\). It follows that \(N\) is a maximal ideal: an ideal strictly containing \(N\) is not contained in \(N\), so it is not proper, that is, it equals \(R\). It also follows that \(N\) is the only maximal ideal: a maximal ideal is proper, hence contained in \(N\), and then equal to \(N\) by maximality. Therefore \(R\) is a local ring with maximal ideal \(N\). □

Take \(R = \mathbb {Z}\), \(I = (2)\) and \(J = (3)\). Then \(2 \in I \cup J\) and \(3 \in I \cup J\), but \(2 + 3 = 5\) is divisible by neither \(2\) nor \(3\), so \(5 \notin I \cup J\). Hence \(I \cup J\) is not closed under addition and is not an ideal.

The smallest ideal containing \(I \cup J\) is \(I + J = (2) + (3) = (2,3)\). Since \(1 = 3 - 2 \in (2,3)\), this is the whole ring \(\mathbb {Z}\). More generally, for any ideals \(I, J\) of a ring \(R\) the smallest ideal containing \(I \cup J\) is \(I + J\): it is an ideal containing \(I\) and \(J\), and any ideal containing \(I \cup J\) is closed under addition, so it contains every sum \(a + b\) with \(a \in I\), \(b \in J\).

In fact, for any ring the union \(I \cup J\) is an ideal only when \(I \subset J\) or \(J \subset I\), in which case \(I \cup J\) is the larger of the two. Indeed, if \(a \in I \setminus J\) and \(b \in J \setminus I\), then \(a + b \notin I\) (otherwise \(b = (a+b) - a \in I\)) and similarly \(a + b \notin J\), so \(a + b \notin I \cup J\). □

Write \(\mathcal {N}\) and \(\mathcal {J}\) for the nilradical and the Jacobson radical of \(R[X]\). Since every maximal ideal is prime, \(\mathcal {N}\), the intersection of all prime ideals of \(R[X]\), is contained in \(\mathcal {J}\), the intersection of all maximal ideals. It remains to show that \(\mathcal {J} \subset \mathcal {N}\).

Let \(f = a_0 + a_1 X + \cdots + a_n X^n \in \mathcal {J}\). By Problem 2 applied with \(y = X\), the polynomial \(1 - fX\) is a unit of \(R[X]\). Let \(\mathfrak {p}\) be any prime ideal of \(R\) and reduce modulo \(\mathfrak {p}\): the image of \(1 - fX\) in \((R/\mathfrak {p})[X]\) is a unit, being the image of a unit under the ring homomorphism \(R[X] \to (R/\mathfrak {p})[X]\). Now \(R/\mathfrak {p}\) is an integral domain, so degrees add under multiplication in \((R/\mathfrak {p})[X]\) and a unit must have degree \(0\). Hence the image of \(fX\) in \((R/\mathfrak {p})[X]\) is \(0\), that is, \(a_0, a_1, \ldots , a_n \in \mathfrak {p}\). As \(\mathfrak {p}\) was arbitrary, each \(a_i\) lies in the intersection of all prime ideals of \(R\), which is the nilradical \(\mathcal {N}(R)\). So each \(a_i\) is nilpotent in \(R\), hence each \(a_i X^i\) is nilpotent in \(R[X]\), and \(f\) is a sum of nilpotent elements, hence nilpotent, since the nilradical of \(R[X]\) is an ideal. Therefore \(f \in \mathcal {N}\).

In fact the argument shows more: \(\mathcal {J} = \mathcal {N} = \mathcal {N}(R)[X]\), the polynomials all of whose coefficients are nilpotent in \(R\). □

We adapt the proof of the Hilbert basis theorem (Theorem 4.1 of the notes), replacing leading coefficients by lowest-order coefficients. For a non-zero \(f = \sum _{i \geq 0} c_i T^i \in R[[T]]\) write \(\mathrm {ord}(f)\) for the smallest \(i\) with \(c_i \neq 0\), and set \(\mathrm {ord}(0) = \infty \).

Let \(I \subset R[[T]]\) be an ideal. For \(n \geq 0\) define \[ J_n = \{ a \in R : \exists f \in I \text { of the form } f = aT^n + c_{n+1}T^{n+1} + c_{n+2}T^{n+2} + \ldots \}. \] As in the proof of Theorem 4.1, \(J_n\) is an ideal of \(R\) (if \(a, a' \in J_n\) come from \(f, f' \in I\) and \(r \in R\), then \(a+a'\) and \(ra\) come from \(f + f'\) and \(rf\)), and \(J_n \subset J_{n+1}\) since \(f \in I\) implies \(Tf \in I\). Since \(R\) is Noetherian, the chain \(J_0 \subset J_1 \subset \ldots \) stabilises: there is \(s\) with \(J_s = J_{s+i}\) for all \(i \geq 0\). For each \(m \leq s\) write \(J_m = (a_{m,1}, \ldots , a_{m,r_m})\) and choose \(f_{m,j} \in I\) of the form \(a_{m,j} T^m + (\text {terms of order} > m)\). Let \(I' \subset I\) be the ideal generated by the finitely many \(f_{m,j}\). We claim that \(I = I'\).

Let \(f \in I\). First, we show that there is \(g \in I'\) with \(\mathrm {ord}(f - g) \geq s\). If \(\mathrm {ord}(f) = m < s\) with lowest coefficient \(a\), then \(a \in J_m\), so \(a = \sum _i b_i a_{m,i}\) with \(b_i \in R\), and \(f - \sum _i b_i f_{m,i} \in I\) has order \(> m\). Repeating this at most \(s\) times gives the required \(g\).

So it remains to show that every \(h \in I\) with \(\mathrm {ord}(h) \geq s\) lies in \(I'\). In the polynomial case the analogous step was an induction on the degree, which terminates. Here the order can only be pushed up indefinitely, so we construct the coefficients of a solution step by step. Put \(h_0 = h\). Suppose \(h_n \in I\) with \(\mathrm {ord}(h_n) \geq s + n\) has been constructed, and let \(c_n\) be its coefficient of \(T^{s+n}\). Then \(c_n \in J_{s+n} = J_s\), so we can write \(c_n = \sum _i b_i^{(n)} a_{s,i}\) with \(b_i^{(n)} \in R\), and we set \[ h_{n+1} = h_n - \sum _{i=1}^{r_s} b_i^{(n)} T^n f_{s,i} \in I, \] which has order \(\geq s + n + 1\). Now define the power series \(g_i = \sum _{n \geq 0} b_i^{(n)} T^n \in R[[T]]\) for \(1 \leq i \leq r_s\). For every \(N \geq 0\) we have \[ h - \sum _{i=1}^{r_s} g_i f_{s,i} = h_{N+1} - \sum _{i=1}^{r_s} \Big ( \sum _{n > N} b_i^{(n)} T^n \Big ) f_{s,i}, \] and both terms on the right have order \(\geq s + N + 1\). Hence every coefficient of \(h - \sum _i g_i f_{s,i}\) vanishes, that is, \(h = \sum _i g_i f_{s,i} \in I'\). Therefore \(f = g + (f - g) \in I'\), and \(I = I'\) is finitely generated. □

Let \(x \in \mathcal {V}(S_2)\), so \(f(x) = 0\) for all \(f \in S_2\). In particular \(f(x) = 0\) for all \(f \in S_1 \subset S_2\), that is, \(x \in \mathcal {V}(S_1)\). Similarly, let \(f \in \mathcal {I}(U_2)\), so \(f\) vanishes at every point of \(U_2\). In particular \(f\) vanishes at every point of \(U_1 \subset U_2\), that is, \(f \in \mathcal {I}(U_1)\).

Now let \(U_1 = U = \{ (0,i) : i \in \mathbb {Z} \}\) and let \(U_2 = \mathcal {V}(X) = \{ (0,y) : y \in \mathbb {C} \}\) be the \(Y\)-axis in \(\mathbb {C}^2\). Then \(U_1 \subsetneq U_2\), since for example \((0, \tfrac 12) \in U_2 \setminus U_1\). We claim that \(\mathcal {I}(U_1) = \mathcal {I}(U_2) = (X)\).

First, \(X\) vanishes on \(U_2\), so \((X) \subset \mathcal {I}(U_2)\), and \(\mathcal {I}(U_2) \subset \mathcal {I}(U_1)\) by the first part. It remains to show that \(\mathcal {I}(U_1) \subset (X)\). Let \(f \in \mathcal {I}(U_1)\). Collecting the terms of \(f\) that involve \(X\), we can write \(f = X g(X,Y) + h(Y)\) with \(g \in \mathbb {C}[X,Y]\) and \(h(Y) = f(0,Y) \in \mathbb {C}[Y]\). For every \(i \in \mathbb {Z}\) we have \(0 = f(0,i) = h(i)\), so the one-variable polynomial \(h\) has infinitely many roots and must be \(0\). Hence \(f = Xg \in (X)\), as required.

Note that \(\mathcal {V}(\mathcal {I}(U_1)) = \mathcal {V}(X) = U_2\): the \(Y\)-axis is the Zariski closure of the set \(U_1\) of integer points on it. □