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. □

A polynomial \(f \in k[X,Y]\) lies in \(R\) if and only if \(f(0,Y) \in k\), that is, if and only if \(f(0,Y) = f(0,0)\). Since \(f \mapsto f(0,Y)\) is a ring homomorphism \(k[X,Y] \to k[Y]\), we have \((f \pm g)(0,Y) = f(0,Y) \pm g(0,Y)\) and \((fg)(0,Y) = f(0,Y) g(0,Y)\), so \(R\) is closed under addition, subtraction and multiplication, and it contains \(0\) and \(1\). Hence \(R\) is a subring. Note that \(R = k + X k[X,Y]\), and that \(R\) is generated as a \(k\)-algebra by \(X, XY, XY^2, \ldots \), since \(X^i Y^j = X^{i-1} \cdot XY^j\) for \(i > 0\).

To see that \(R\) is not Noetherian, consider the ideals \[ I_N = (X, XY, XY^2, \ldots , XY^N) \subset R, \qquad N \geq 0 . \] Clearly \(I_0 \subset I_1 \subset I_2 \subset \ldots \), and we claim that each inclusion is strict, so that \(R\) does not satisfy the ascending chain condition. Suppose that \(XY^{N+1} \in I_N\), say \[ XY^{N+1} = \sum _{j=0}^{N} g_j \, XY^j \] with \(g_j \in R\). Cancelling \(X\), which is allowed since \(k[X,Y]\) is an integral domain, gives \(Y^{N+1} = \sum _{j=0}^N g_j Y^j\), and setting \(X = 0\) gives \[ Y^{N+1} = \sum _{j=0}^N g_j(0,Y) \, Y^j = \sum _{j=0}^N g_j(0,0) \, Y^j , \] which is impossible since the right hand side has degree at most \(N\). Hence \(XY^{N+1} \notin I_N\).

Thus a subring of a Noetherian ring need not be Noetherian. Note that the union \(\bigcup _N I_N = X k[X,Y]\) is the kernel of the homomorphism \(R \to k\), \(f \mapsto f(0,0)\), so it is a maximal ideal of \(R\) which is not finitely generated. □

With respect to \(\leq _{grlex}\) we have \(LT(f_1) = X^3\) and \(LT(f_2) = X^2Y\), the terms of highest degree. Since \(f_1, f_2 \in I\), we have \((X^3, X^2Y) \subset \ini (I)\). On the other hand, \[ X f_2 - Y f_1 = (X^3Y - 2XY^2 + X^2) - (X^3Y - 2XY^2) = X^2 \] lies in \(I\), so \(X^2 = LT(X^2) \in \ini (I)\). But \(X^2 \notin (X^3, X^2Y)\): by Lemma 4.4 of the notes, a polynomial lies in a monomial ideal if and only if each of its terms is divisible by one of the generating monomials, and \(X^2\) is divisible by neither \(X^3\) nor \(X^2Y\). Hence \((LT(f_1), LT(f_2)) \subsetneq \ini (I)\); in other words, \(\{f_1, f_2\}\) is not a Gröbner basis of \(I\).

In fact, since \(X^2 \in I\) we also get \(XY = \tfrac 12 (X \cdot X^2 - f_1) \in I\) and \(Y^2 - \tfrac 12 X = \tfrac 12 (Y \cdot X^2 - f_2) \in I\), and conversely \(f_1 = X \cdot X^2 - 2 \cdot XY\) and \(f_2 = Y \cdot X^2 - 2 (Y^2 - \tfrac 12 X)\), so \[ I = (X^2, \; XY, \; Y^2 - \tfrac 12 X) . \] The \(S\)-polynomials of these three generators are \(0\), \(\tfrac 12 X^3\) and \(\tfrac 12 X^2\), each of which has remainder \(0\) on division by \(X^2\), so by Buchberger’s criterion this is a Gröbner basis of \(I\). Hence \(\ini (I) = (X^2, XY, Y^2)\), \(C(I) = \{1, X, Y\}\), and \(\dim _k k[X,Y]/I = 3\) by Theorem 4.8. □

The leading terms are \(LT(f) = X^2Y\), \(LT(f_1) = XY\) and \(LT(f_2) = Y^2\). We follow the proof of Proposition 4.6: at each step we look at the leading term of the current polynomial, subtract a multiple of the first divisor in the list whose leading term divides it, and if there is no such divisor we move the leading term to the remainder.

Division by \(f_1, f_2\). Since \(XY \mid X^2Y\), we subtract \(X f_1\): \[ f - X f_1 = XY^2 + X + Y^2 . \] The leading term \(XY^2\) is again divisible by \(XY\), so we subtract \(Y f_1\): \[ XY^2 + X + Y^2 - Y f_1 = X + Y^2 + Y . \] The leading term \(X\) is divisible by neither \(XY\) nor \(Y^2\), so \(X\) goes to the remainder, leaving \(Y^2 + Y\). Its leading term is divisible by \(Y^2\), and \(Y^2 + Y - f_2 = Y + 1\), neither of whose terms is divisible by \(XY\) or \(Y^2\). Altogether \[ f = (X + Y) f_1 + f_2 + (X + Y + 1) , \] with remainder \(r = X + Y + 1\).

Division by \(f_2, f_1\). Now \(Y^2 \nmid X^2Y\) but \(XY \mid X^2Y\), so the first step is the same: \(f - X f_1 = XY^2 + X + Y^2\). This time \(f_2\) is the first divisor whose leading term divides \(XY^2\), so we subtract \(X f_2\): \[ XY^2 + X + Y^2 - X f_2 = 2X + Y^2 . \] The term \(2X\) goes to the remainder, and \(Y^2 - f_2 = 1\) goes to the remainder as well. Altogether \[ f = (X + 1) f_2 + X f_1 + (2X + 1) , \] with remainder \(r = 2X + 1\).

The two remainders are different, so the remainder depends on the order in which the divisors are listed. By Theorem 4.8 this cannot happen for a Gröbner basis, and indeed \(\{f_1, f_2\}\) is not a Gröbner basis of \(I = (f_1, f_2)\): the difference of the two remainders, \((X + Y + 1) - (2X + 1) = Y - X = X f_2 - Y f_1\), lies in \(I\), but its leading term \(-X\) is divisible by neither \(XY\) nor \(Y^2\). □

(a) For the lexicographic order with \(Z > Y > X\), the generators \(g_1 = Y - X^2\) and \(g_2 = Z - X^3\) have leading terms \(Y\) and \(Z\). By Buchberger’s criterion (Theorem 4.11) it suffices to check that \(S(g_1, g_2)\) has remainder \(0\) on division by \(g_1, g_2\). We have \[ S(g_1, g_2) = Z g_1 - Y g_2 = X^3 Y - X^2 Z , \] whose leading term is \(-X^2 Z\), since \(Z > Y\). Subtracting \(-X^2 g_2\) gives \(X^3 Y - X^5\), and subtracting \(X^3 g_1\) gives \(0\). Hence \(\{Y - X^2, Z - X^3\}\) is a Gröbner basis of \(I\), and it is reduced.

(b) For the lexicographic order with \(X > Y > Z\) we write the generators as \(g_1 = X^2 - Y\) and \(g_2 = X^3 - Z\), with leading terms \(X^2\) and \(X^3\), and run Buchberger’s algorithm. First, \(S(g_1, g_2) = X g_1 - g_2 = -XY + Z\) is its own remainder on division by \(g_1, g_2\), so we add \(g_3 = XY - Z\) to the generators. Since \(g_2 = X g_1 + g_3\), we may discard \(g_2\). Next, \[ S(g_1, g_3) = Y g_1 - X g_3 = XZ - Y^2 \] is its own remainder on division by \(g_1, g_3\), so we add \(g_4 = XZ - Y^2\). Now \(S(g_1, g_4) = Z g_1 - X g_4 = XY^2 - YZ = Y g_3\) has remainder \(0\), but \[ S(g_3, g_4) = Z g_3 - Y g_4 = Y^3 - Z^2 \] is its own remainder on division by \(g_1, g_3, g_4\), so we add \(g_5 = Y^3 - Z^2\). Finally, division with remainder gives \begin {align*} S(g_1, g_5) &= X^2 Z^2 - Y^4 = Z^2 g_1 - Y g_5, \\ S(g_3, g_5) &= XZ^2 - Y^2 Z = Z g_4, \\ S(g_4, g_5) &= XZ^3 - Y^5 = Z^2 g_4 - Y^2 g_5 , \end {align*}

all with remainder \(0\), and \(S(g_1,g_3) = g_4\) and \(S(g_3,g_4) = g_5\) now have remainder \(0\) as well. By Buchberger’s criterion, \[ G = \{ X^2 - Y, \; XY - Z, \; XZ - Y^2, \; Y^3 - Z^2 \} \] is a Gröbner basis of \(I\) for the lexicographic order with \(X > Y > Z\); it is in fact the reduced Gröbner basis. The ideal \(I\) is the ideal of the twisted cubic curve \(t \mapsto (t, t^2, t^3)\); compare with the example in Lecture 8, where the same four polynomials form a Gröbner basis for the graded lexicographic order. Note that \(G \cap k[Y,Z] = \{Y^3 - Z^2\}\), which by Problem 21 generates \(I \cap k[Y,Z]\): the projection \((t,t^2,t^3) \mapsto (t^2,t^3)\) of the twisted cubic to the \((Y,Z)\)-plane lies on the cuspidal cubic \(Y^3 = Z^2\). □

Write \(h_1 = 3X - 6Y - 2Z\), \(h_2 = 2X - 4Y + 4W\), \(h_3 = X - 2Y - Z - W\) and \(I = (h_1, h_2, h_3)\). All three leading terms are multiples of \(X\), so the \(S\)-polynomials are \begin {align*} S(h_1, h_3) &= h_1 - 3 h_3 = Z + 3W, \\ S(h_2, h_3) &= h_2 - 2 h_3 = 2Z + 6W, \\ S(h_1, h_2) &= \tfrac 13 h_1 - \tfrac 12 h_2 = -\tfrac 23 Z - 2W . \end {align*}

None of these is divisible by \(X\), so by Buchberger’s criterion \(\{h_1, h_2, h_3\}\) is not a Gröbner basis, and we add \(g_2 = Z + 3W\) to the generators. Since \(h_1 = 3 h_3 + g_2\) and \(h_2 = 2 h_3 + 2 g_2\), we have \(I = (h_3, g_2)\), and replacing \(h_3\) by \(h_3 + g_2\) we get \(I = (g_1, g_2)\) with \[ g_1 = X - 2Y + 2W, \qquad g_2 = Z + 3W . \] It remains to check that \(S(g_1, g_2) = Z g_1 - X g_2 = -3XW - 2YZ + 2ZW\) has remainder \(0\) on division by \(g_1, g_2\). Its leading term \(-3XW\) is divisible by \(X\), and \(S(g_1,g_2) + 3W g_1 = -2YZ - 6YW + 2ZW + 6W^2\). Continuing, \(-2YZ - 6YW + 2ZW + 6W^2 + 2Y g_2 = 2ZW + 6W^2 = 2W g_2\), so the remainder is \(0\). Hence \(\{X - 2Y + 2W, \; Z + 3W\}\) is a Gröbner basis of \(I\), and it is reduced.

The coefficient matrix of the linear system and its reduced row echelon form are \[ \begin {pmatrix} 3 & -6 & -2 & 0 \\ 2 & -4 & 0 & 4 \\ 1 & -2 & -1 & -1 \end {pmatrix} \quad \rightsquigarrow \quad \begin {pmatrix} 1 & -2 & 0 & 2 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end {pmatrix} , \] and the non-zero rows of the latter are exactly the coefficient vectors of \(g_1\) and \(g_2\). This is not a coincidence: for linear forms, the steps of Buchberger’s algorithm are row operations. The \(S\)-polynomial of two linear forms with the same leading variable is a linear combination of the two in which that variable has been eliminated, and reducing a linear form by linear forms with other leading variables is back substitution. In general, the non-zero rows of the reduced row echelon form of a system of linear forms give the reduced Gröbner basis of the ideal they generate, for the lexicographic order in which the variables are ordered as the columns. □

Let \(A\) be the \(m \times n\) matrix whose \(j\)-th row is the coefficient vector \((a_{1j}, \ldots , a_{nj})\) of \(f_j\). Then \(V = \{ x \in k^n : Ax = 0 \}\) is the kernel of \(A\), so \(\dim _k V = n - r\) by the rank–nullity theorem, where \(r = \mathrm {rank}(A)\).

Row operations on \(A\) replace \(f_1, \ldots , f_m\) by linear forms which span the same \(k\)-subspace of \(k[X_1,\ldots ,X_n]\), and hence generate the same ideal \(I\). Bringing \(A\) into reduced row echelon form, we therefore get \(I = (g_1, \ldots , g_r)\) with \[ g_i = X_{p_i} + \sum _{l \in F} b_{il} X_l , \qquad 1 \leq i \leq r, \] where \(p_1 < \ldots < p_r\) are the pivot columns and \(F \subset \{1,\ldots ,n\}\) is the set of the remaining \(n - r\) indices.

Let \(\phi : k[X_1,\ldots ,X_n] \to k[X_1,\ldots ,X_n]\) be the \(k\)-algebra homomorphism with \(\phi (X_{p_i}) = g_i\) for \(1 \leq i \leq r\) and \(\phi (X_l) = X_l\) for \(l \in F\). It is an automorphism: its inverse is the \(k\)-algebra homomorphism \(\psi \) with \(\psi (X_{p_i}) = X_{p_i} - \sum _{l \in F} b_{il} X_l\) and \(\psi (X_l) = X_l\) for \(l \in F\), as one checks on the variables. Both \(\phi \) and \(\psi \) send each variable to a polynomial of degree \(\leq 1\), so both map \(k[X_1,\ldots ,X_n]_{\leq s}\) into itself, and being mutually inverse they restrict to bijections of \(k[X_1,\ldots ,X_n]_{\leq s}\). Let \(J = (X_{p_1}, \ldots , X_{p_r})\). Since \(\phi \) is an automorphism sending the generators of \(J\) to the generators of \(I\), we have \(\phi (J) = I\), and then \(\phi (J_{\leq s}) \subset I_{\leq s}\) and \(\psi (I_{\leq s}) \subset J_{\leq s}\), so \(\phi (J_{\leq s}) = I_{\leq s}\). Hence \(\phi \) induces a \(k\)-linear isomorphism \[ k[X_1,\ldots ,X_n]_{\leq s} / J_{\leq s} \; \xrightarrow {\ \sim \ } \; k[X_1,\ldots ,X_n]_{\leq s} / I_{\leq s} , \] and \(HF_I(s) = HF_J(s)\) for all \(s\).

Now \(J\) is a monomial ideal, so by Lemma 4.4 a polynomial lies in \(J\) if and only if each of its terms is divisible by one of \(X_{p_1}, \ldots , X_{p_r}\). Hence \(J_{\leq s}\) is spanned by the monomials of degree \(\leq s\) which involve a pivot variable, and \(k[X_1,\ldots ,X_n]_{\leq s} / J_{\leq s}\) has a basis given by the classes of the monomials of degree \(\leq s\) in the \(n - r\) variables \(X_l\), \(l \in F\). As in Example (b) of Lecture 8, there are \(\binom {n - r + s}{n - r}\) of these, so \[ HF_I(s) = HF_J(s) = \binom {s + n - r}{n - r} = \frac {(s+1)(s+2) \cdots (s + n - r)}{(n-r)!} , \] which is a polynomial in \(s\) of degree \(n - r\). Therefore \(HP_I(t) = \binom {t + n - r}{n - r}\) and \(\deg HP_I(t) = n - r = \dim _k V\).

Geometrically, \(\phi \) corresponds to a linear change of coordinates on \(k^n\) in which \(V\) becomes the coordinate subspace \(\{ X_{p_1} = \ldots = X_{p_r} = 0 \}\). The argument used that \(\phi \) preserves degrees. An automorphism of \(k[X_1,\ldots ,X_n]\) which does not, such as \(X_1 \mapsto X_1 + X_2^2\), \(X_2 \mapsto X_2\), can change the Hilbert function: it sends \((X_1)\), whose Hilbert function is \(s + 1\), to \((X_1 + X_2^2)\), whose Hilbert function is \(2s + 1\). Alternatively, one can show that \(g_1, \ldots , g_r\) form a Gröbner basis of \(I\) for the graded lexicographic order (compare Problem 18), so that \(\ini (I) = J\), and deduce \(HF_I = HF_J\) from Macaulay’s lemma. □

Write \(f_1 = X^2 + Y + Z - 1\), \(f_2 = X + Y^2 + Z - 1\) and \(f_3 = X + Y + Z^2 - 1\), with leading terms \(X^2\), \(X\) and \(X\), and run Buchberger’s algorithm. First, \(S(f_2, f_3) = f_2 - f_3\) is its own remainder on division by \(f_1, f_2, f_3\), so we add \[ g_2 = Y^2 - Y - Z^2 + Z \] to the generators. Next, reducing \(S(f_1, f_3) = f_1 - X f_3 = -XY - XZ^2 + X + Y + Z - 1\) modulo \(f_2, f_3, g_2\) gives the remainder \[ g_3 = 2YZ^2 + Z^4 - Z^2 , \] and reducing \(S(g_2, g_3) = Z^2 g_2 - \tfrac 12 Y g_3\) modulo \(g_3\) gives \(\tfrac 14 g_4\), where \[ g_4 = Z^6 - 4Z^4 + 4Z^3 - Z^2 . \] Since \(f_2 = f_3 + g_2\) and \(f_1 = X f_3 + S(f_1,f_3)\), the ideal \(I\) is generated by \(g_1 = f_3\), \(g_2\), \(g_3\) and \(g_4\). One checks that all remaining \(S\)-polynomials of \(g_1, g_2, g_3, g_4\) have remainder \(0\); this is tedious by hand and best left to a computer, for example

R = QQ[x,y,z, MonomialOrder => Lex];
I = ideal (x^2+y+z-1, x+y^2+z-1, x+y+z^2-1);
gens gb I

in Macaulay2. Hence \[ G = \{ X + Y + Z^2 - 1, \; Y^2 - Y - Z^2 + Z, \; 2YZ^2 + Z^4 - Z^2, \; Z^6 - 4Z^4 + 4Z^3 - Z^2 \} \] is a Gröbner basis of \(I\) for the lexicographic order with \(X > Y > Z\).

The point of the lexicographic order is that \(G\) is triangular: \(g_4\) involves only \(Z\), \(g_2\) and \(g_3\) involve only \(Y\) and \(Z\), and \(g_1\) is linear in \(X\). Since \[ g_4 = Z^2 (Z - 1)^2 (Z^2 + 2Z - 1), \qquad g_2 = (Y - Z)(Y + Z - 1), \qquad g_3 = Z^2 (2Y + Z^2 - 1) , \] every point of \(\mathcal {V}(I)\) has \(Z \in \{0, 1, -1 + \sqrt {2}, -1 - \sqrt {2} \}\), and for each such \(Z\) we find \(Y\) from \(g_2 = g_3 = 0\) and then \(X = 1 - Y - Z^2\) from \(g_1 = 0\). Since \(G\) generates \(I\), the points obtained lie in \(\mathcal {V}(I)\).

• \(Z = 0\): \(g_3 = 0\) holds automatically, \(g_2 = 0\) gives \(Y \in \{0, 1\}\), and \(X = 1 - Y\). This gives the points \((1,0,0)\) and \((0,1,0)\).
• \(Z = 1\): \(g_3 = 0\) gives \(Y = 0\), which also satisfies \(g_2 = 0\), and \(X = 0\). This gives the point \((0,0,1)\).
• \(Z = \zeta \) with \(\zeta ^2 + 2\zeta - 1 = 0\): \(g_3 = 0\) gives \(Y = \tfrac 12 (1 - \zeta ^2) = \zeta \), which also satisfies \(g_2 = 0\), and \(X = 1 - \zeta - \zeta ^2 = \zeta \). This gives the points \((\zeta ,\zeta ,\zeta )\) with \(\zeta = -1 \pm \sqrt {2}\).

Hence \(\mathcal {V}(I)\) consists of the five points \[ (1,0,0), \quad (0,1,0), \quad (0,0,1), \quad (-1 + \sqrt {2}, -1 + \sqrt {2}, -1 + \sqrt {2}), \quad (-1 - \sqrt {2}, -1 - \sqrt {2}, -1 - \sqrt {2}) . \]

It is instructive to compare \(|\mathcal {V}(I)| = 5\) with \(\dim _{\mathbb {C}} \mathbb {C}[X,Y,Z]/I\). From \(G\) we read off \(\ini (I) = (X, Y^2, YZ^2, Z^6)\) for the lexicographic order, so \(C(I)\) consists of the \(8\) monomials \[ 1, \; Y, \; Z, \; YZ, \; Z^2, \; Z^3, \; Z^4, \; Z^5 , \] and \(\dim _{\mathbb {C}} \mathbb {C}[X,Y,Z]/I = 8\) by Theorem 4.8. Since this number does not depend on the monomial order, we can use it to find the initial ideal for the graded lexicographic order without further computation. The leading terms of \(f_1, f_2, f_3\) are now \(X^2, Y^2, Z^2\), and the complement of \((X^2, Y^2, Z^2)\) consists of the \(8\) monomials \[ 1, \; X, \; Y, \; Z, \; XY, \; XZ, \; YZ, \; XYZ . \] As \((X^2, Y^2, Z^2) \subset \ini _{grlex}(I)\) and a strictly larger monomial ideal would have a strictly smaller complement, we get \(\ini _{grlex}(I) = (X^2, Y^2, Z^2)\), so \(\{f_1, f_2, f_3\}\) is a Gröbner basis for \(\leq _{grlex}\). Macaulay’s lemma (Lecture 8) now gives the Hilbert function: \(HF_I(0) = 1\), \(HF_I(1) = 4\), \(HF_I(2) = 7\) and \(HF_I(s) = 8\) for \(s \geq 3\), so \(HP_I(t) = 8\). Counting instead the elements of degree \(\leq s\) in the lexicographic complement would give \(1, 3, 5, 6, 7, 8, 8, \ldots \), which is wrong: Macaulay’s lemma needs a graded order.

Finally, here is why \(5 \neq 8\). The polynomial \(h = XY - YZ = Y(X - Z)\) vanishes at all five points of \(\mathcal {V}(I)\), and \[ h^2 = (X - Z) \big ( (X - Z) f_2 + f_3 - f_1 \big ) \in I , \] but \(h \notin I\): no term of \(h\) is divisible by \(X^2\), \(Y^2\) or \(Z^2\), so \(h\) is its own remainder on division by the Gröbner basis \(\{f_1, f_2, f_3\}\), and by Theorem 4.8 the remainder of an element of \(I\) is \(0\). Thus \(I\) is not a radical ideal, and \(\mathcal {I}(\mathcal {V}(I)) \neq I\). One can show, using the Nullstellensatz (Lecture 9), that if \(J\) is a radical ideal with \(\mathcal {V}(J)\) finite then \(|\mathcal {V}(J)| = \dim _{\mathbb {C}} \mathbb {C}[X_1,\ldots ,X_n]/J\). Indeed, Macaulay2 tells us that

I == radical I
gens gb radical I

return false and the Gröbner basis \[ \sqrt {I} = (Z^4 + Z^3 - 3Z^2 + Z, \; 2YZ + Z^3 - Z, \; Y^2 - Y - Z^2 + Z, \; X + Y + Z^2 - 1) , \] with initial ideal \((X, Y^2, YZ, Z^4)\) and \(C(\sqrt {I}) = \{1, Y, Z, Z^2, Z^3\}\), so that \(\dim _{\mathbb {C}} \mathbb {C}[X,Y,Z]/\sqrt {I} = 5 = |\mathcal {V}(I)|\). □

Fix \(l\) and write \(G_l = G \cap k[X_{l+1},\ldots ,X_n]\) and \(I_l = I \cap k[X_{l+1},\ldots ,X_n]\). The intersection \(I_l\) of the ideal \(I\) with the subring \(k[X_{l+1},\ldots ,X_n]\) is an ideal of that subring, and \(G_l \subset I_l\), so the ideal \((G_l)\) generated by \(G_l\) in \(k[X_{l+1},\ldots ,X_n]\) is contained in \(I_l\).

The key observation is that a non-zero \(g \in k[X_1,\ldots ,X_n]\) whose leading monomial lies in \(k[X_{l+1},\ldots ,X_n]\) lies in \(k[X_{l+1},\ldots ,X_n]\) itself. Indeed, let \(X^\alpha \) be a monomial in which some \(X_i\) with \(i \leq l\) occurs, and let \(X^\beta \in k[X_{l+1},\ldots ,X_n]\), so that \(\beta _1 = \ldots = \beta _l = 0\). Then the first non-zero entry of \(\alpha - \beta \) occurs at an index \(\leq l\) and is positive, so \(X^\alpha >_{lex} X^\beta \). Hence a term of \(g\) involving one of \(X_1, \ldots , X_l\) would be larger than \(LM(g)\), which is impossible.

Now let \(f \in I_l\) be non-zero. We show that \(f \in (G_l)\) by induction on \(LM(f)\), which is legitimate by Lemma 4.5. Since \(G\) is a Gröbner basis, \(LT(f) \in \ini (I) = (LT(g) : g \in G)\), so by Lemma 4.4 there is \(g \in G\) with \(LM(g) \mid LM(f)\). As \(LM(f) \in k[X_{l+1},\ldots ,X_n]\), so is its divisor \(LM(g)\), hence \(g \in G_l\) by the key observation. Put \[ f' = f - \frac {LT(f)}{LT(g)} \, g . \] Then \(f' \in I\) and \(f' \in k[X_{l+1},\ldots ,X_n]\), so \(f' \in I_l\), and either \(f' = 0\) or \(LM(f') < LM(f)\). By induction \(f' \in (G_l)\), and therefore \(f = f' + \frac {LT(f)}{LT(g)} g \in (G_l)\). This proves \(I_l = (G_l)\).

In fact the argument shows more: for every non-zero \(f \in I_l\) there is \(g \in G_l\) with \(LM(g) \mid LM(f)\), so \(G_l\) is a Gröbner basis of \(I_l\) for the lexicographic order on \(k[X_{l+1},\ldots ,X_n]\). This is what makes the lexicographic order useful for solving systems of polynomial equations: in Problem 20, the element \(g_4 \in G \cap \mathbb {C}[Z]\) generates \(I \cap \mathbb {C}[Z]\), and the elements of \(G \cap \mathbb {C}[Y,Z]\) generate \(I \cap \mathbb {C}[Y,Z]\), which is why we could solve for \(Z\) first, then for \(Y\) and finally for \(X\). □