Section 1.1 Classical Affine Varieties
Throughout this class, \(k\) will always denote a field and we will often assume that \(k\) is an algebraically closed field. It will rarely cause problems to assume \(k = \mathbb{C}\text{,}\) in fact.
Definition 1.1.1.
For an arbitrary field \(k\text{,}\) affine \(n\)-space over \(k\text{,}\) written \(\mathbb{A}^n_k\text{,}\) is the set of \(n\)-tuples of elements of \(k\text{:}\)
\begin{equation*}
\mathbb{A}^n_k = \{(a_1, \dots, a_n) \mid a_i \in k\}.
\end{equation*}
So, \(\mathbb{A}^n_k\) is just \(k^n\text{,}\) but we avoid using the latter since \(k^n\) is the usual notation of the standard \(n\)-dimensional \(k\)-vector space. Affine space is not thought of as a vector space typically.
Definition 1.1.2.
Recall \(k[x_1, \dots, x_n]\) is the ring of all polynomials in the \(n\) variables \(x_1, \dots, x_n\) having \(k\) coefficients. For any subset \(S\) of \(k[x_1, \dots, x_n]\text{,}\) the zero locus of \(S\text{,}\) written \(V(S)\text{,}\) is the subset of \(\mathbb{A}^n_k\) giving the common zeroes of all the members of \(S\text{;}\) that is,
\begin{equation*}
V(S) := \{(a_1, \dots, a_n) \in \mathbb{A}^n_k \mid f(a_1, \dots, a_n) = 0, \text{ for all $f \in S$}\} \subseteq \mathbb{A}^n_k.
\end{equation*}
Definition 1.1.4.
A classical affine variety is a subset of \(\mathbb{A}^n_k\text{,}\) for some \(n\text{,}\) that is equal to \(V(S)\) for some subset \(S\) of \(k[x_1, \dots, x_n]\text{.}\)
Exploration 1.1.5.
Draw \(V_{\mathbb{R}}(x^2 + y^2 - z, x+y+z+1)\) (when \(k = {\mathbb{R}}\)). (Yes, I realize \({\mathbb{R}}\) is not algebraically closed; one instance in which we drop this assumption is when we wish to draw a variety.)
Exploration 1.1.6.
Let \(V = \{(t, t^2, t^3) \mid t \in k\} \subseteq \mathbb{A}^3_k\text{.}\) Prove \(V\) is an affine variety.
Exploration 1.1.7.
Let \(V = \{(t, e^t) \mid t \in \mathbb{C}\} \subseteq \mathbb{A}^2_\mathbb{C}\text{.}\) Prove \(V\) is not an affine variety.
Example 1.1.8.
Any singleton \(\{(a_1, \dots, a_n)\}\) is an affine variety, since it is the zero locus of the set \(\{x_1 - a_1, \dots, x_n - a_n\}\text{.}\)
Exploration 1.1.9.
Prove that any finite set of points of \(\mathbb{A}^n_k\) for any \(n \geq 1\) is an affine variety.
Exploration 1.1.10.
Is a countably infinite set of points in \(\mathbb{A}^n_\mathbb{C}\) an affine variety?
Exploration 1.1.11.
Thinking of \(\mathbb{A}^n_k\) as a vector space for once, prove that every linear subspace is an affine variety. More generally, show every translate of a linear subspace is an affine variety.
Exploration 1.1.12.
Show \(S = \mathbb{A}^2_k \setminus \{(0,0)\} = \{(a,b) \in \mathbb{A}^2_k \mid (a,b) \ne (0,0)\}\) is not an affine variety whenever \(k\) is an infinite field. (We will eventually define the more general notion of a “variety”, and we will see that this is a (non-affine) variety.) Is this true when \(k\) is a finite field?
For a subset \(S\) of \(k[x_1, \dots, x_n]\text{,}\) let \(\langle S \rangle\) denote the ideal of this ring generated by \(S\text{;}\) that is,
\begin{equation*}
\langle S \rangle= \{f_1 g_1 + \cdots + f_m g_m \mid m \geq 0, f_i \in k[x_1, \dots, x_n], g_i \in S, \text{ for all $i$}\}.
\end{equation*}
For any ideal \(I\text{,}\) we let \(\sqrt{I} = \{f \in k[x_1, \dots, x_n] \mid f^m \in I, \text{ for some $m \geq 1$}\}\text{,}\) the radical of \(I\). The radical of an ideal is again an ideal.
Proposition 1.1.13.
For any subset \(S\) of \(k[x_1, \dots, x_n]\) we have
\begin{equation*}
V(S) = V(\langle S \rangle) = V(\sqrt{\langle S \rangle }).
\end{equation*}
In particular, \(V(I) = V(\sqrt{I})\) for any ideal \(I\text{.}\)
Proof.
The containments \(\supseteq\) hold since \(S \subseteq \langle S \rangle \subseteq \sqrt{(S)}\text{.}\) Say \(P \in V(S)\text{.}\) An arbitrary element of \(\langle S \rangle\) has the form \(f = f_1 g_1 + \cdots + f_m g_m\) with \(f_i \in k[x_1, \dots, x_n]\) and \(g_i \in S\) for all \(i\text{.}\) We have \(f(P) = \sum_i f_i(P) g_i(P) = \sum_i f_i(P) \cdot 0 = 0\text{.}\) This proves \(V(S) = V(\langle S \rangle)\text{.}\) Given any ideal \(I\text{,}\) if \(P \in V(I)\) and \(f\) is such that \(f^n \in I\text{,}\) then \(f(P)^n = (f^n)(P) = 0\) and hence \(f(P) = 0\text{.}\) This proves \(V(I) = V(\sqrt{I})\text{.}\)
Thanks to this proposition, we can redifine an affine variety to be a subset of the form \(V(J)\) where \(J\) is an ideal (or even a radical ideal).
Here are some easy-to-prove properties of the rule \(I \mapsto V(I)\) sending ideals to varieties.
Proposition 1.1.14.
For any fixed \(n\text{:}\)
If \(I \subseteq J \subseteq k[x_1, \dots, x_n]\) then \(V(J) \subseteq V(I)\text{.}\)
If \(I, J \subseteq k[x_1, \dots, x_n]\) then \(V(I) \cup V(J)= V(I \cdot J)\) where \(I \cdot J\) is the ideal generated by all products of elements of the form \(ab\) with \(a \in I\) and \(b \in J\text{.}\) In particular, the union of a pair of affine varieties in \(\mathbb{A}^n_k\) is again an affine variety.
If \(\{I_\alpha\}_{\alpha \in A}\) is any collection of ideals of \(k[x_1, \dots, x_n]\text{,}\) then
\begin{equation*}
\bigcap_{\alpha \in A} V(I_\alpha) = V(\sum_{\alpha \in A} I_\alpha).
\end{equation*}
where \(\sum_{\alpha\in A} I_\alpha\) is the smallest ideal containing \(I_\alpha\) for all \(\alpha\text{.}\) In particular, an arbitrary intersection of affine varieties in \(\mathbb{A}^n\) is again an affine variety.
\(V(0) = \mathbb{A}^n_k\text{.}\) In particular, \(\mathbb{A}^n_k\) is an affine variety.
\(V(1) = \emptyset\text{.}\) In particular, the empty set is an affine variety.
Proof.
I’ll just prove the second one: If
\(P \in V(I) \cup V(J)\) then for all
\(f \in I\text{,}\) \(g \in J\) we have
\((fg)(P) = f(P) g(P) = 0\) since either
\(f(P) = 0\) or
\(g(P) = 0\text{.}\) Thus
\(P \in V(\{f g \mid f \in I, g \in J\}) = V(IJ)\text{,}\) where the last equality uses
Proposition 1.1.13.
Say \(P \notin V(I) \cup V(J)\text{.}\) Then for any \(f \in I, g \in J'\text{,}\) we have \(f(P) \ne 0\) and \(g(P) \ne 0\) and thus \((fg)(P) = f(P) g(P) \ne 0\text{.}\) This proves \(P \notin V(\{f g \mid f \in I, g \in J\}) = V(IJ)\text{.}\)
Recall:
Theorem 1.1.15.
The ring \(k[x_1, \dots, x_n]\) is noetherian. That is, every ideal of it is finitely generated or, equivalently, the collection of all ideals of this ring has the acc.
We won’t prove this Theorem in this class; I assume you’ve seen it before.
Corollary 1.1.16.
For any field \(k\text{,}\) every affine variety \(V\) in \(\mathbb{A}^n_k\) is the zero locus of a finite list of polynomials.
Proof.
By
Proposition 1.1.13 we may assume
\(V = V(I)\) for a (radical) ideal
\(I\text{.}\) By the HBT we have
\(I = \langle g_1, \dots, g_m \rangle\) for some
\(g_i\)’s and hence (using the Proposition again)
\(V = V(g_1, \dots, g_m)\text{.}\)
So far we have started with ideals (or even just sets) of polynomials and used them to produce ideals. There is a “backwards” function too:
Definition 1.1.17.
For and \(n\) and any subset \(W\) of \(\mathbb{A}^n_k\text{,}\) let \(I(W)\) denote the set of all polynomials that vanish at every point of \(W\text{:}\)
\begin{equation*}
I(W) = \{f \in k[x_1, \dots, x_n] \mid \, \text{$f(a_1, \dots, a_n) = 0$ for all $(a_1, \dots, a_n) \in W$} \}.
\end{equation*}
Proposition 1.1.18.
For any subset \(W\) of \(\mathbb{A}^n_k\text{,}\) \(I(W)\) is a radical ideal in \(k[x_1, \dots, x_n]\text{.}\)
Proof.
If \(g \in I(W)\) and \(f \in k[x_1, \dots, x_n]\) then for all \(P \in I(W)\) we have \((fg)(P) = f(P) g(P) = f(P) 0 = 0\) and hence \(fg \in I(W)\text{.}\) Similarly one shows \(I(W)\) is closed under addition. Since clearly \(0 \in I(W)\text{,}\) \(I(W)\) is an ideal. If \(f^n \in I(W)\) for some \(n \geq 1\text{,}\) then for all \(P \in W\) we have \(0 = (f^n)(P) = (f(P))^n\) and hence \(f(P) = 0\text{.}\) So \(f \in I(W)\text{,}\) and this proves \(I(W)\) is radical.
The next result is immediate from the definitions:
Proposition 1.1.19.
If \(W\) and \(W'\) are two subsets of \(\mathbb{A}^n_k\text{,}\) then:
\(I(W \cup W') = I(W) \cap I(W')\text{.}\)
If \(W \subseteq W'\) then \(I(W) \supseteq I(W')\text{.}\)
What would you guess is true about \(I(W \cap W')\text{?}\)
Exploration 1.1.20.
Find a counter-example to the “obvious guess” that \(I(W \cap W') = \sqrt{I(W) + I(W')}\) for \(k = {\mathbb{R}}\text{.}\)
The function \(V\) takes radical ideals to affine varieties and the function \(I\) takes affine varieties to radical ideals. Are they mutually inverse?
Proposition 1.1.21.
For any field \(k\) the following hold:
Given a radical ideal \(J\) of \(k[x_1, \dots, x_n]\) we have
\begin{equation*}
I(V(J)) \supseteq J.
\end{equation*}
(This holds in fact for any subset \(J\text{.}\))
Given an affine variety \(W\) in \(\mathbb{A}^n_k\) we have
\begin{equation*}
V(I(W)) = W
\end{equation*}
(For any subset W, we have \(V(I(W)) \supseteq W\))
Proof.
If \(f \in J\text{,}\) then for each \(P \in V(J)\) we have, by definition, that \(f(P) = 0\) and hence, again by definition, \(f \in I(V(J))\text{.}\) Note that this remains valid if \(J\) is any subset.
For the second one, pick \(P \in W\text{.}\) For any \(f \in I(W)\) we have, by definition, \(f(P) = 0\text{.}\) Again by definition this means \(P \in V(I(W))\text{.}\) Thus \(W \subseteq V(I(W))\text{.}\) Note that this containment holds if \(W\) is any subset. Now assume \(W\) is an affine variety. Then \(W = V(J)\) for some radical ideal \(J\text{.}\) By the first part we have \(I(V(J)) \supseteq J\) and hence \(V(I(V(J))) \subseteq V(J)\text{.}\) Substituting \(W\) for \(V(J)\) yields \(V(I(W)) \subseteq W\) and hence \(V(I(W)) = W\text{.}\)
Can you find an example where the containment \(I(V(J)) \supseteq J\) is strict, say for the ring \({\mathbb{R}}\text{?}\)
Here is one:
Example 1.1.22.
Take \(J = \langle x^2 + y^2 \rangle\) and \(k = {\mathbb{R}}\text{,}\) and note that \(J\) is a radical ideal since \(x^2+y^2\) is an irreducible polynomial (why?). We have \(V(J) = \{(0,0)\}\) and so \(I(V(J)) = \langle x,y \rangle \supsetneq J\text{.}\)
