Let \(S \subset k[X_1,\ldots , X_n]\). We let the affine variety defined by \(S\) to be \[ \mathcal {V}(S) = \{ (x_1,\ldots , x_n) \in k^n : f(x_1,\ldots , x_n) =0 \text { for all } f\in S \} \]
Note that if we set \(I = (S)\) the ideal generated by \(S\) and apply the Hilbert basis theorem to write \(I=(f_1,\ldots , f_h)\) for some finite number of polynomials, then \[ \mathcal {V}(S) = \mathcal {V}(I) = \mathcal {V}( \{f_1,\ldots , f_h \}) \] hence every affine variety is defined as the vanishing locus of a finite set of polynomials.
Given a subset \(U \subset k^n\), we define an ideal \[ \mathcal {I}(U) = \{ f \in k[X_1,\ldots , X_n] : f(x_1,x_2,\ldots , x_n) = 0 \text { for all } (x_1,x_2,\ldots , x_n) \in U \} \]
Thus, we have maps between subsets of \(k^n\) and subsets of \(k[X_1,\ldots , X_n]\). \[ \{ S \subset k[X_1,\ldots , X_n] \} \overset {\mathcal {V}}{\underset {\mathcal {I}}\rightleftarrows } \{ U \subset k^n \} \]
\(\sphericalangle \) Show that \(S_1 \subset S_2\) implies \(\mathcal {V}(S_2) \subset \mathcal {V}(S_1)\), and \(U_1 \subset U_2\) implies \(\mathcal {I}(U_2) \subset \mathcal {I}(U_1)\). Use the set \(U = \{ (0,i) : i \in \mathbb {Z} \} \subset \mathbb {C}^2\) to show that it can happen that \(U_1 \subsetneq U_2\) but \(\mathcal {I}(U_1) = \mathcal {I}(U_2)\).
\(\sphericalangle \) Note that \(U \subset \mathcal {V}(\mathcal {I}(U))\) for any subset \(U \subset k^n\). Prove that if \(U\) is an affine variety, then \(U= \mathcal {V}(\mathcal {I}(U))\).
In general, for a set \(U \subset k^n\), \(\mathcal {V}(\mathcal {I}(U))\) is the smallest affine variety containing \(U\), and is called the Zariski closure of \(U\) and denoted by \(\overline {U}\).
\(\blacktriangleright \) \(\mathcal {I}\) forgets multiplicities: in \(k[X]\) we have \(\mathcal {V}(X^2) = \{0\}\) and \(\mathcal {I}(\{0\}) = (X)\), so \(\mathcal {I}(\mathcal {V}(I)) \neq I\) in general. In fact \(\mathcal {I}(U)\) is always a radical ideal, since if \(f^m\) vanishes on \(U\) then so does \(f\).
\(\blacktriangleright \) \(\mathcal {V}(I+J) = \mathcal {V}(I) \cap \mathcal {V}(J)\).
\(\blacktriangleright \) \(\mathcal {V}(IJ) = \mathcal {V}(I \cap J) = \mathcal {V}(I) \cup \mathcal {V}(J)\).
\(\blacktriangleright \) Hilbert’s Nullstellensatz which we will see later in the course implies that if \(k\) is algebraically closed, \(\mathcal {V}(I : J^\infty ) = \overline {\mathcal {V}(I) \setminus \mathcal {V}(J)}\).
Basic questions:
We will answer all these questions in the next few sections but first we have to develop some tools.
We work in the ring \(k[X_1,\ldots , X_n]\). Let us introduce some notation.
\(\mathbb {N}^n = \{ \alpha = (\alpha _1,\ldots , \alpha _n) : \alpha _i \in \mathbb {N} \}\).
\(X^\alpha = X_1^{\alpha _1} X_2^{\alpha _2} \ldots X_n^{\alpha _n}\). We write \(|\alpha |= \alpha _1 + \alpha _2 + \ldots + \alpha _n\) for the total degree of \(X^\alpha \).
We write an element \(f \in k[X_1,\ldots , X_n]\) as \(f = \sum _\alpha a_\alpha X^\alpha \) with \(a_\alpha \in k\) such that \(\{ \alpha : a_\alpha \neq 0 \}\) is a finite set.
Unlike the case of \(k[X]\), when \(n>1\) it is unclear which monomial has the highest degree. For example, what is the highest degree monomial in \(f(X,Y) = X^2 Y + 42 X Y^2\) ?
To address this we introduce the following notion.
Definition 4.3. Let \( \leq \) be a total ordering on \(\mathbb {N}^n\). We say that \(\leq \) is a monomial ordering if
Since \(\alpha \mapsto X^\alpha \) identifies \(\mathbb {N}^n\) with the set of monomials, a monomial ordering is the same thing as an ordering of the monomials of \(k[X_1,\ldots ,X_n]\). Being a total ordering means that any two monomials are comparable, so every non-zero polynomial has a well-defined largest term.
\(\sphericalangle \) Show that \(X^\alpha | X^\beta \) implies that \(\alpha \leq \beta \) for any monomial ordering \(\leq \).
\(\blacktriangleright \) Lexicographic Order \(\leq _{lex}\): Let \(\alpha = (\alpha _1,\alpha _2, \ldots , \alpha _n)\), \(\beta = (\beta _1,\beta _2, \ldots , \beta _n)\). Define \(\alpha <_{lex} \beta \) if the first non-zero entry of \(\beta - \alpha \in \mathbb {Z}^n\) is positive, and \(\alpha \leq _{lex} \beta \) if \(\alpha <_{lex} \beta \) or \(\alpha = \beta \). Thus \(\alpha <_{lex} \beta \) means that \(\alpha _i < \beta _i\) for the first index \(i\) with \(\alpha _i \neq \beta _i\).
\(\blacktriangleright \) By contrast, the ordering of \(\mathbb {N}^n\) in which \(\alpha \leq \beta \) if \(\alpha _i \leq \beta _i\) for all \(i\), that is, \(X^\alpha | X^\beta \), satisfies both conditions of Definition 4.3 but is not total when \(n>1\): for example, \((1,0)\) and \((0,1)\) are not comparable. The lexicographic order is a total ordering refining it.
\(\blacktriangleright \) Graded Lexicographic Order \(\leq _{grlex}\): Define \(\alpha \leq _{grlex} \beta \) if either \(|\alpha | < |\beta |\) or if \(|\alpha |= |\beta |\) and \(\alpha \leq _{lex} \beta \).
\(\sphericalangle \) Order the terms of the polynomial \(f(X,Y,Z)= XY^2Z + Z^2 + X^3 + X^2 Z^2\) with respect to \(\leq _{lex}\) and \(\leq _{grlex}\).
\(\blacktriangleright \) Let \(A \in GL(n,\mathbb {R})\) be a matrix such that the first non-zero entry of each column is positive. Define \(\alpha <_A \beta \) if the first non-zero entry of \(A(\beta - \alpha ) \in \mathbb {R}^n\) is positive. Each row of \(A\) is a weight vector: monomials are compared by their first weight, ties are broken by the second weight, and so on. Since \(A\) is invertible, any two distinct \(\alpha , \beta \) are comparable, and the condition on the columns \(Ae_j\) says exactly that \(0 <_A e_j\) for all \(j\). For example, the matrices \[ \begin {pmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end {pmatrix}, \qquad \begin {pmatrix} 1 & 1 & \cdots & 1 & 1 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end {pmatrix}, \qquad \begin {pmatrix} 1 & 1 & \cdots & 1 & 1 \\ 0 & 0 & \cdots & 0 & -1 \\ 0 & 0 & \cdots & -1 & 0 \\ \vdots & \vdots & & \vdots & \vdots \\ 0 & -1 & \cdots & 0 & 0 \end {pmatrix} \] define \(\leq _{lex}\), \(\leq _{grlex}\) and the graded reverse lexicographic order \(\leq _{grevlex}\), respectively. In the last one, monomials of the same total degree are compared by their exponent of \(X_n\), the smaller exponent being the larger monomial, with ties broken by the exponent of \(X_{n-1}\), and so on. Note that this matrix has negative entries, but the first non-zero entry of each column is still positive. In fact, any monomial ordering on \(\mathbb {N}^n\) arises this way for some matrix \(A\).5
\(\sphericalangle \) Show that \(\leq _A\) is a monomial ordering.
\(\blacksquare \) In what follows, we fix a monomial ordering \(\leq \).
\(\blacksquare \) For \(\alpha ,\beta \in \mathbb {N}^n\), we often write \(X^\alpha \leq X^\beta \) to mean that \(\alpha \leq \beta \).
Given a non-zero polynomial \[ f = a_{\alpha ^{(1)}} X^{\alpha ^{(1)}} + a_{\alpha ^{(2)}} X^{\alpha ^{(2)}} + \ldots + a_{\alpha ^{(s)}} X^{\alpha ^{(s)}} \] where \(\alpha ^{(i)}\in \mathbb {N}^n\) with \(\alpha ^{(1)} > \alpha ^{(2)} > \ldots > \alpha ^{(s)}\) and all the coefficients \(a_{\alpha ^{(i)}}\) are non-zero, we define \begin {align*} LT(f) &\coloneqq a_{\alpha ^{(1)}} X^{\alpha ^{(1)}} \text {\ \ \ \ \ \bf leading term }\\ LM(f) &\coloneqq X^{\alpha ^{(1)}} \text {\ \ \ \ \ \ \ \ \ \ \ \bf leading monomial }\\ LC(f) &\coloneqq a_{\alpha ^{(1)}} \text {\ \ \ \ \ \ \ \ \ \ \ \ \bf leading coefficient } \end {align*}
The ideal membership problem is easy to solve for monomial ideals.
Lemma 4.4. Let \(I \subset k[X_1,\ldots , X_n]\) be a monomial ideal, that is, an ideal generated by a (possibly infinite) set \(G\) of monomials, and let \(f \in k[X_1,\ldots , X_n]\). Then \(f \in I\) if and only if every term of \(f\) is divisible by a monomial in \(G\).
Proof. Suppose \(f \in I\), then \(f = \sum _{i=1}^s h_i g_i\) for some polynomials \(h_i \in k[X_1,\ldots , X_n]\) and monomials \(g_i \in G\). Write \(h_i = \sum a_{i,\alpha } X^\alpha \), then \(f = \sum _{i,\alpha } a_{i,\alpha } X^\alpha g_i\). Hence, all the terms of \(f\) are divisible by monomials in \(G\). The other direction is clear. □
Lemma 4.5. Let \((g_i)_{i \geq 1}\) be a sequence of monomials such that \(g_1 \geq g_2 \geq g_3 \geq \ldots \). Then, there exists \(r \geq 1\) such that \(g_r=g_{r+1}=g_{r+2}=\ldots \).
Proof. Let \(I = (g_i : i \geq 1)\). By the Hilbert basis theorem (Theorem 4.1), the ascending chain of ideals \((g_1) \subset (g_1,g_2) \subset \ldots \) stabilises, so there exists \(r\) such that \(I = (g_1,\ldots , g_r)\). Now, for \(i > r\), we have \(g_i \leq g_r\). On the other hand, since \(g_i \in I\), by the previous lemma, there exists \(j \in \{ 1,\ldots , r\}\) such that \(g_j | g_i\), and so \(g_j \leq g_i\). Since \(j \leq r\), we get \(g_r \leq g_j \leq g_i\), hence \(g_i=g_r\). □
5L Robbiano - Term orderings on the polynomial ring.