Source: Section 1 of Gathmann

Problem Set 1

Prove that every affine variety \(X\subset {\mathbb{A}}^n/k\) consisting of only finitely many points can be written as the zero locus of \(n\) polynomials.

Hint: Use interpolation. It is useful to assume at first that all points in \(X\) have different \(x_1{\hbox{-}}\)coordinates.

Let \(X = \left\{{\mathbf{p}_1, \cdots, \mathbf{p}_d}\right\} =\left\{{\mathbf{p}_j}\right\}_{j=1}^d\), where each \(\mathbf{p}_j\in {\mathbb{A}}^n\) can be written in coordinates \begin{align*}\mathbf{p}_j \coloneqq{\left[ {p_j^1, p_j^2, \cdots, p_j^n} \right]}.\end{align*}

Proof idea: for some fixed \(k\) with \(2\leq k \leq n\), consider the pairs \((p_j^1, p_j^k) \in {\mathbb{A}}^2\). Letting \(j\) range over \(1\leq j \leq d\) yields \(d\) points of the form \((x, y) \in {\mathbb{A}}^2\), so construct an interpolating polynomial such that \(f(x) = y\) for each tuple. Then \(f(x) - y\) vanishes at every such tuple.

Doing this for each \(k\) (keeping the first coordinate always of the form \(p_j^1\) and letting the second coordinate vary) yields \(n-1\) polynomials in \(k[x_1, x_k] \subseteq k[x_1, \cdots, x_{n}]\), then adding in the polynomial \(p(x) = \prod_j (x-p_j^1)\) yields a system the vanishes precisely on \(\left\{{\mathbf{p}_j}\right\}\).

Without loss of generality, we can assume all of the first components \(\left\{{p_j^1}\right\}_{j=1}^d\) are distinct.

\todo[inline]{Todo: follows from "rotation of axes"?}

We will use the following fact:

Given a set of \(d\) points \(\left\{{(x_i, y_i)}\right\}_{i=1}^d\) with all \(x_i\) distinct, there exists a unique polynomial of degree \(d\) in \(f \in k[x]\) such that \(\tilde f(x_i) = y_i\) for every \(i\).

This can be explicitly given by \begin{align*} \tilde f(x) = \sum_{i=1}^d y_i \qty{\prod_{\substack{0\leq m \leq d \\ m\neq i}} \qty{x - x_m \over x_i - x_m }} .\end{align*}

Equivalently, there is a polynomial \(f\) defined by \(f(x_i) = \tilde f(x_i) - y_i\) of degree \(d\) whose roots are precisely the \(x_i\).

\vspace{2em}

Using this theorem, we define a system of \(n\) polynomials in the following way:

Define \(f_1 \in k[x_1] \subseteq k[x_1, \cdots, x_n]\) by \begin{align*}f_1(x) = \prod_{i=1}^d \qty{x - p_i^1}.\end{align*} Then the roots of \(f_1\) are precisely the first components of the points \(p\).

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Define \(f_2 \in k[x_1, x_2] \subseteq k[x_1, \cdots, x_n]\) by considering the ordered pairs \begin{align*}\left\{{(x_1, x_2) = (p_j^1, p_j^2)}\right\},\end{align*} then taking the unique Lagrange interpolating polynomial \(\tilde f_2\) satisfying \(\tilde f_2(p_j^1) = p_j^2\) for all \(1\leq j \leq d\). Then set \(f_2 \coloneqq\tilde f_2(x_1) - x_2 \in k[x_1, x_2]\).

Define \(f_3 \in k[x_1, x_3] \subseteq k[x_1, \cdots, x_n]\) by considering the ordered pairs \begin{align*}\left\{{(x_1, x_3) = (p_j^1, p_j^3)}\right\},\end{align*} then taking the unique Lagrange interpolating polynomial \(\tilde f_3\) satisfying \(\tilde f_2(p_j^1) = p_j^3\) for all \(1\leq j \leq d\). Then set \(f_3 \coloneqq\tilde f_3(x_1) - x_3 \in k[x_1, x_3]\).
\(\cdots\)

\vspace{2em}

Continuing in this way up to \(f_n \in k[x_1, x_n]\) yields a system of \(n\) polynomials.

\(V(f_1, \cdots, f_n) = X\).

\(X\subseteq V(f_i)\):

This is essentially by construction. Letting \(p_j\in X\) be arbitrary, we find that \begin{align*} f_1(p_j) = \prod_{i=1}^d \qty{p_j^1 - p_i^1} = (p_j^1 - p_j^1) \prod_{\substack{i\leq d \\ i\neq j}} \qty{p_j^1 - p_i^1} = 0 .\end{align*}

Similarly, for \(2\leq k \leq n\), \begin{align*} f_k(p_j) = \tilde f_k(p_j^1) - p_j^k = 0 ,\end{align*} which follows from the fact that \(\tilde f_k(p_j^1) = p_j^k\) for every \(k\) and every \(j\) by the construction of \(\tilde f_k\).

\(X^c \subseteq V(f_i)^c\):

This follows from the fact the polynomials \(f\) given by Lagrange interpolation are unique, and thus the roots of \(\tilde f\) are unique. But if some other point was in \(V(f_i)\), then one of its coordinates would be another root of some \(\tilde f\).

Determine \(\sqrt{I}\) for \begin{align*} I\coloneqq\left\langle{x_1^3 - x_2^6,\, x_1 x_2 - x_2^3}\right\rangle {~\trianglelefteq~}{\mathbb{C}}[x_1, x_2] .\end{align*}

For notational purposes, let \(\mathcal{I}, \mathcal{V}\) denote the maps in Hilbert’s Nullstellensatz, we then have \begin{align*}(\mathcal{I} \circ \mathcal{V})(I) = \sqrt{I}.\end{align*}

So we consider \(\mathcal{V}(I) \subseteq {\mathbb{A}}^2/{\mathbb{C}}\), the vanishing locus of these two polynomials, which yields the system \begin{align*} \begin{cases} x^3 - y^6 & = 0 \\ xy - y^3 & = 0. \end{cases} \end{align*} In the second equation, we have \((x- y^2)y = 0\), and since \({\mathbb{C}}[x, y]\) is an integral domain, one term must be zero.

If \(y=0\), then \(x^3 = 0 \implies x= 0\), and thus \((0, 0) \in \mathcal{V}(I)\), i.e. the origin is contained in this vanishing locus.
Otherwise, if \(x-y^2 = 0\), then \(x=y^2\), with no further conditions coming from the first equation.

Combining these conditions, \begin{align*}P\coloneqq\left\{{(t^2, t) {~\mathrel{\Big\vert}~}t\in {\mathbb{C}}}\right\} \subset \mathcal{V}(I).\end{align*}

where \(I = \left\langle{x^3 - y^6, xy-y^3}\right\rangle\).

We have \(P = \mathcal{V}(I)\), and so taking the ideal generated by \(P\) yields \begin{align*} \qty{\mathcal{I} \circ \mathcal{V}} (I) = \mathcal{I}(P) = \left\langle{y-x^2}\right\rangle \in {\mathbb{C}}[x ,y] \end{align*}

and thus \(\sqrt{I} = \left\langle{y-x^2}\right\rangle\).

Let \(X\subset {\mathbb{A}}^3/k\) be the union of the three coordinate axes. Compute generators for the ideal \(I(X)\) and show that it can not be generated by fewer than 3 elements.

Claim: \begin{align*}I(X) = \left\langle{x_2 x_3,\, x_1 x_3,\, x_1 x_2}\right\rangle.\end{align*}

We can write \(X = X_1 \cup X_2 \cup X_3\), where

The \(x_1{\hbox{-}}\)axis is given by \(X_1 \coloneqq V(x_2 x_3)\) \(\implies I(X_1) = \left\langle{x_2 x_3}\right\rangle\),
The \(x_2{\hbox{-}}\)axis is given by \(X_2 \coloneqq V(x_1 x_3)\) \(\implies I(X_2) = \left\langle{x_1 x_3}\right\rangle\),
The \(x_3{\hbox{-}}\)axis is given by \(X_3 \coloneqq V(x_1 x_2)\) \(\implies I(X_3) = \left\langle{x_1 x_2}\right\rangle\).

Here we’ve used, for example, that \begin{align*}I(V(x_2 x_3)) = \sqrt{\left\langle{x_2 x_3}\right\rangle} = \left\langle{x_2 x_3}\right\rangle\end{align*} by applying the Nullstellensatz and noting that \(\left\langle{x_2x_3}\right\rangle\) is radical since it is generated by a squarefree monomial.

We then have \begin{align*} I(X) &= I(X_1 \cup X_2 \cup X_3) \\ &= I(X_1) \cap I(X_2) \cap I(X_3) \\ &= \sqrt{I(X_1) + I(X_2) + I(X_3)} \\ &= \sqrt{\left\langle{x_2, x_3}\right\rangle + \left\langle{x_1 x_3}\right\rangle + \left\langle{x_1 x_2}\right\rangle} \\ &= \sqrt{\left\langle{x_2x_3,\, x_1 x_3,\, x_1 x_2}\right\rangle} \hspace{8em}\text{since } \left\langle{a}\right\rangle + \left\langle{b}\right\rangle = \left\langle{a, b}\right\rangle \\ &= {\left\langle{x_2x_3,\, x_1 x_3,\, x_1 x_2}\right\rangle} ,\end{align*} where in the last equality we’ve again used the fact that an ideal generated by squarefree monomials is radical.

\(I(X)\) can not be generated by 2 or fewer elements.

Let \(J\coloneqq I(X)\) and \(R\coloneqq k[x_1, x_2, x_3]\), and toward a contradiction, suppose \(J = \left\langle{r, s}\right\rangle\). Define \({\mathfrak{m}}\coloneqq\left\langle{x, y, z}\right\rangle\) and a quotient map \begin{align*}\pi: J \to J/{\mathfrak{m}}J\end{align*} and consider the images \(\pi(r), \pi(s)\).

Note that \(J/{\mathfrak{m}}J\) is an \(R/{\mathfrak{m}}{\hbox{-}}\)module, and since \(R/{\mathfrak{m}}\cong k\), \(J/{\mathfrak{m}}J\) is in fact a \(k{\hbox{-}}\)vector space. Since \(\pi(r), \pi(s)\) generate \(J/{\mathfrak{m}}J\) as a \(k{\hbox{-}}\)module, \begin{align*}\dim_k J/{\mathfrak{m}}J \leq 2.\end{align*}

But this is a contradiction, since we can produce 3 \(k{\hbox{-}}\)linearly independent elements in \(J/{\mathfrak{m}}J\): namely \(\pi(x_1 x_2), \pi(x_1 x_3), \pi(x_2 x_3)\). Suppose there exist \(\alpha_i\) such that \begin{align*} \alpha_1 \pi(x_1 x_2) + \alpha_2 \pi(x_1 x_3) + \alpha_3 \pi(x_2 x_3) = 0 \in J/{\mathfrak{m}}J \iff \alpha_1 x_1 x_2 + \alpha_2 x_1 x_3 + \alpha_3 x_2 x_3 \in {\mathfrak{m}}J ,\end{align*}

But we can then note that \begin{align*} {\mathfrak{m}}J = \left\langle{x_1, x_2. x_3}\right\rangle\left\langle{x_1 x_2, x_1 x_3, x_2 x_3}\right\rangle = \left\langle{x_1^2 x_2,\, x_1^2 x_3,\, x_1x_2 x_3, \cdots}\right\rangle .\end{align*} can’t contain any nonzero elements of degree \(d<3\), so no such \(\alpha_i\) can exist and these elements are \(k{\hbox{-}}\)linearly independent.

Let \(Y\subset {\mathbb{A}}^n/k\) be an affine variety and define \(A(Y)\) by the quotient \begin{align*} \pi: k[x_1,\cdots, x_n] \to A(Y) \coloneqq k[x_1, \cdots, x_n]/I(Y) .\end{align*}

Show that \(V_Y(J) = V(\pi^{-1}(J))\) for every \(J{~\trianglelefteq~}A(Y)\).
Show that \(\pi^{-1} (I_Y(X)) = I(X)\) for every affine subvariety \(X\subseteq Y\).

Using the fact that \(I(V(J)) \subset \sqrt{J}\) for every \(J{~\trianglelefteq~}k[x_1, \cdots, x_n]\), deduce that \(I_Y(V_Y(J)) \subset \sqrt{J}\) for every \(J{~\trianglelefteq~}A(Y)\).

Conclude that there is an inclusion-reversing bijection \begin{align*} \left\{{\substack{\text{Affine subvarieties}\\ \text{of } Y}}\right\} \iff \left\{{\substack{\text{Radical ideals} \\ \text{in } A(Y)}}\right\} .\end{align*}

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Let \(J {~\trianglelefteq~}k[x_1, \cdots, x_n]\) be an ideal, and find a counterexample to \(I(V(J)) =\sqrt{J}\) when \(k\) is not algebraically closed.

Take \(J = \left\langle{x^2+1}\right\rangle {~\trianglelefteq~}{\mathbb{R}}[x]\), noting that \(J\) is nontrivial and proper but \({\mathbb{R}}\) is not algebraically closed. Then \(V(J) \subseteq {\mathbb{R}}\) is empty, and thus \(I(V(J)) = I(\emptyset)\).

\(I(V(J)) = {\mathbb{R}}[x]\).

Checking definitions, for any set \(X \subset {\mathbb{A}}^n/k\) we have \begin{align*} I(X) = \left\{{f\in {\mathbb{R}}[x] {~\mathrel{\Big\vert}~}\forall x\in X,\, f(x)=0}\right\} \\ \end{align*} and so we vacuously have \begin{align*} I(\emptyset) = \left\{{f\in {\mathbb{R}}[x] {~\mathrel{\Big\vert}~}\forall x\in \emptyset,\, f(x)=0}\right\} = \left\{{f\in {\mathbb{R}}[x]}\right\} = {\mathbb{R}}[x] .\end{align*}

\(\sqrt{J} \neq {\mathbb{R}}[x]\).

This follows from the fact that maximal ideals are radical, and \({\mathbb{R}}[x]/ J \cong {\mathbb{C}}\) being a field implies that \(J\) is maximal. In this case \(\sqrt{J} = J \neq {\mathbb{R}}[x]\).

That maximal ideals are radical follows from the fact that if \(J{~\trianglelefteq~}R\) is maximal, we have \(J \subset \sqrt{J} \subset R\) which forces \(\sqrt{J} = J\) or \(\sqrt{J}=R\).

But if \(\sqrt{J}=R\), then \begin{align*} 1\in \sqrt{J} \implies 1^n \in J \text{ for some }n \implies 1 \in J \implies J=R ,\end{align*} contradicting the assumption that \(J\) is maximal and thus proper by definition.