--- notoc: true title: Problem Set 1 --- > Source: [Section 1 of Gathmann](https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2019/alggeom-2019-c1.pdf) # Problem Set 1 ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 1.19"} 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} ::: {.remark .proofenv .proofenv .proofenv .proofenv .proofenv} 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 $\left\{{\mathbf{p}_j}\right\}$. ::: ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} Without loss of generality, we can assume all of the first components $\left\{{p_j^1}\right\}_{j=1}^d$ are distinct. ::: ```{=tex} \todo[inline]{Todo: follows from "rotation of axes"?} ``` We will use the following fact: ::: {.theorem .proofenv .proofenv .proofenv .proofenv .proofenv title="Lagrange"} 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*} `{=html} 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 $x_i$. ::: ```{=tex} \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*} `{=html} Then the roots of $f_1$ are precisely the first components of the points $p$. \ - 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*} `{=html} 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*} `{=html} 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 $f_3 \coloneqq\tilde f_3(x_1) - x_3 \in k[x_1, x_3]$. - $\cdots$ ```{=tex} \vspace{2em} ``` Continuing in this way up to $f_n \in k[x_1, x_n]$ yields a system of $n$ polynomials. \ ::: {.proposition .proofenv .proofenv .proofenv .proofenv .proofenv} $V(f_1, \cdots, f_n) = X$. ::: ::: {.proof .proofenv .proofenv .proofenv .proofenv .proofenv} ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} $X\subseteq V(f_i)$: ::: 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*} `{=html} 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*} `{=html} 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 $\tilde f_k$. ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} $X^c \subseteq V(f_i)^c$: ::: 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 $\tilde f$. ::: ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 1.21"} 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*} `{=html} ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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*} `{=html} 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*} `{=html} 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. 1. 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. 2. 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*} `{=html} 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*} `{=html} and thus $\sqrt{I} = \left\langle{y-x^2}\right\rangle$. ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 1.22"} 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} **Claim**: ` \begin{align*}I(X) = \left\langle{x_2 x_3,\, x_1 x_3,\, x_1 x_2}\right\rangle.\end{align*} `{=html} 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*} `{=html} 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*} `{=html} where in the last equality we've again used the fact that an ideal generated by squarefree monomials is radical. ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} $I(X)$ 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*} `{=html} 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*} `{=html} \ 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*} `{=html} 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*} `{=html} 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. ::: ::: ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Gathmann 1.23: Relative Nullstellensatz"} 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*} `{=html} a. Show that $V_Y(J) = V(\pi^{-1}(J))$ for every $J{~\trianglelefteq~}A(Y)$. b. Show that $\pi^{-1} (I_Y(X)) = I(X)$ for every affine subvariety $X\subseteq Y$. c. 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*} `{=html} ::: ```{=tex} \newpage ``` ::: {.exercise .proofenv .proofenv .proofenv .proofenv .proofenv title="Extra"} 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. ::: ::: {.solution .proofenv .proofenv .proofenv .proofenv .proofenv} 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 $I(V(J)) = I(\emptyset)$. ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} $I(V(J)) = {\mathbb{R}}[x]$. 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*} `{=html} 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*} `{=html} ::: ::: {.claim .proofenv .proofenv .proofenv .proofenv .proofenv} $\sqrt{J} \neq {\mathbb{R}}[x]$. 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*} `{=html} contradicting the assumption that $J$ is maximal and thus proper by definition. ::: :::