--- title: "Problem Set 1" notoc: true --- > Source: [Section 1 of Gathmann](https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2019/alggeom-2019-c1.pdf) # Problem Set 1 :::{.exercise title="Gathmann 1.19"} Prove that every affine variety $X\subset \AA^n/k$ 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 $x_1\dash$coordinates. ::: :::{.solution} Let $X = \ts{\vector p_1, \cdots, \vector p_d} =\ts{\vector p_j}_{j=1}^d$, where each $\vector p_j\in \AA^n$ can be written in coordinates $$\vector p_j \da \thevector{p_j^1, p_j^2, \cdots, p_j^n}.$$ :::{.remark} Proof idea: for some fixed $k$ with $2\leq k \leq n$, consider the pairs $(p_j^1, p_j^k) \in \AA^2$. Letting $j$ range over $1\leq j \leq d$ yields $d$ points of the form $(x, y) \in \AA^2$, 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 $k[x_1, x_k] \subseteq \kx{n}$, then adding in the polynomial $p(x) = \prod_j (x-p_j^1)$ yields a system the vanishes precisely on $\ts{\vector p_j}$. ::: :::{.claim} Without loss of generality, we can assume all of the first components $\ts{p_j^1}_{j=1}^d$ are distinct. ::: \todo[inline]{Todo: follows from "rotation of axes"?} We will use the following fact: :::{.theorem title="Lagrange"} Given a set of $d$ points $\ts{(x_i, y_i)}_{i=1}^d$ 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 \[ \tilde f(x) = \sum_{i=1}^d y_i \qty{\prod_{\substack{0\leq m \leq d \\ m\neq i}} \qty{x - x_m \over x_i - x_m }} .\] 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 $$f_1(x) = \prod_{i=1}^d \qty{x - p_i^1}.$$ 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 $$\ts{(x_1, x_2) = (p_j^1, p_j^2)},$$ 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 $f_2 \da \tilde f_2(x_1) - x_2 \in k[x_1, x_2]$. \ - Define $f_3 \in k[x_1, x_3] \subseteq k[x_1, \cdots, x_n]$ by considering the ordered pairs $$\ts{(x_1, x_3) = (p_j^1, p_j^3)},$$ 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 $f_3 \da \tilde f_3(x_1) - x_3 \in k[x_1, x_3]$. - $\cdots$ \vspace{2em} Continuing in this way up to $f_n \in k[x_1, x_n]$ yields a system of $n$ polynomials. \ :::{.proposition} $V(f_1, \cdots, f_n) = X$. ::: :::{.proof} :::{.claim} $X\subseteq V(f_i)$: ::: This is essentially by construction. Letting $p_j\in X$ be arbitrary, we find that \[ f_1(p_j) = \prod_{i=1}^d \qty{p_j^1 - p_i^1} = (p_j^1 - p_j^1) \prod_{\substack{i\leq d \\ i\neq j}} \qty{p_j^1 - p_i^1} = 0 .\] Similarly, for $2\leq k \leq n$, \[ f_k(p_j) = \tilde f_k(p_j^1) - p_j^k = 0 ,\] 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} $X^c \subseteq V(f_i)^c$: ::: 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 title="Gathmann 1.21"} Determine $\sqrt{I}$ for \[ I\da \gens{x_1^3 - x_2^6,\, x_1 x_2 - x_2^3} \normal \CC[x_1, x_2] .\] ::: :::{.solution} For notational purposes, let $\mathcal{I}, \mathcal{V}$ denote the maps in Hilbert's Nullstellensatz, we then have $$(\mathcal{I} \circ \mathcal{V})(I) = \sqrt{I}.$$ So we consider $\mathcal{V}(I) \subseteq \AA^2/\CC$, the vanishing locus of these two polynomials, which yields the system \[ \begin{cases} x^3 - y^6 & = 0 \\ xy - y^3 & = 0. \end{cases} \] In the second equation, we have $(x- y^2)y = 0$, and since $\CC[x, y]$ 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, $$P\da \ts{(t^2, t) \suchthat t\in \CC} \subset \mathcal{V}(I).$$ where $I = \gens{x^3 - y^6, xy-y^3}$. We have $P = \mathcal{V}(I)$, and so taking the ideal generated by $P$ yields \[ \qty{\mathcal{I} \circ \mathcal{V}} (I) = \mathcal{I}(P) = \gens{y-x^2} \in \CC[x ,y] \] and thus $\sqrt{I} = \gens{y-x^2}$. ::: :::{.exercise title="Gathmann 1.22"} Let $X\subset \AA^3/k$ 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} **Claim**: \[I(X) = \gens{x_2 x_3,\, x_1 x_3,\, x_1 x_2}.\] We can write $X = X_1 \union X_2 \union X_3$, where - The $x_1\dash$axis is given by $X_1 \da V(x_2 x_3)$ $\implies I(X_1) = \gens{x_2 x_3}$, - The $x_2\dash$axis is given by $X_2 \da V(x_1 x_3)$ $\implies I(X_2) = \gens{x_1 x_3}$, - The $x_3\dash$axis is given by $X_3 \da V(x_1 x_2)$ $\implies I(X_3) = \gens{x_1 x_2}$. Here we've used, for example, that $$I(V(x_2 x_3)) = \sqrt{\gens{x_2 x_3}} = \gens{x_2 x_3}$$ by applying the Nullstellensatz and noting that $\gens{x_2x_3}$ is radical since it is generated by a squarefree monomial. We then have \[ I(X) &= I(X_1 \union X_2 \union X_3) \\ &= I(X_1) \intersect I(X_2) \intersect I(X_3) \\ &= \sqrt{I(X_1) + I(X_2) + I(X_3)} \\ &= \sqrt{\gens{x_2, x_3} + \gens{x_1 x_3} + \gens{x_1 x_2}} \\ &= \sqrt{\gens{x_2x_3,\, x_1 x_3,\, x_1 x_2}} \hspace{8em}\text{since } \gens{a} + \gens{b} = \gens{a, b} \\ &= {\gens{x_2x_3,\, x_1 x_3,\, x_1 x_2}} ,\] where in the last equality we've again used the fact that an ideal generated by squarefree monomials is radical. :::{.claim} $I(X)$ can not be generated by 2 or fewer elements. Let $J\da I(X)$ and $R\da k[x_1, x_2, x_3]$, and toward a contradiction, suppose $J = \gens{r, s}$. Define $\mfm \da \gens{x, y, z}$ and a quotient map $$\pi: J \to J/\mfm J$$ and consider the images $\pi(r), \pi(s)$. \ Note that $J/\mfm J$ is an $R/\mfm\dash$module, and since $R/\mfm \cong k$, $J/\mfm J$ is in fact a $k\dash$vector space. Since $\pi(r), \pi(s)$ generate $J/\mfm J$ as a $k\dash$module, $$\dim_k J/\mfm J \leq 2.$$ \ But this is a contradiction, since we can produce 3 $k\dash$linearly independent elements in $J/\mfm J$: namely $\pi(x_1 x_2), \pi(x_1 x_3), \pi(x_2 x_3)$. Suppose there exist $\alpha_i$ such that \[ \alpha_1 \pi(x_1 x_2) + \alpha_2 \pi(x_1 x_3) + \alpha_3 \pi(x_2 x_3) = 0 \in J/\mfm J \iff \alpha_1 x_1 x_2 + \alpha_2 x_1 x_3 + \alpha_3 x_2 x_3 \in \mfm J ,\] But we can then note that \[ \mfm J = \gens{x_1, x_2. x_3}\gens{x_1 x_2, x_1 x_3, x_2 x_3} = \gens{x_1^2 x_2,\, x_1^2 x_3,\, x_1x_2 x_3, \cdots} .\] can't contain any nonzero elements of degree $d<3$, so no such $\alpha_i$ can exist and these elements are $k\dash$linearly independent. ::: ::: :::{.exercise title="Gathmann 1.23: Relative Nullstellensatz"} Let $Y\subset \AA^n/k$ be an affine variety and define $A(Y)$ by the quotient \[ \pi: k[x_1,\cdots, x_n] \to A(Y) \da k[x_1, \cdots, x_n]/I(Y) .\] a. Show that $V_Y(J) = V(\pi^{-1}(J))$ for every $J\normal A(Y)$. 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 $J\normal k[x_1, \cdots, x_n]$, deduce that $I_Y(V_Y(J)) \subset \sqrt{J}$ for every $J\normal A(Y)$. Conclude that there is an inclusion-reversing bijection \[ \correspond{\text{Affine subvarieties}\\ \text{of } Y} \iff \correspond{\text{Radical ideals} \\ \text{in } A(Y)} .\] ::: \newpage :::{.exercise title="Extra"} Let $J \normal k[x_1, \cdots, x_n]$ be an ideal, and find a counterexample to $I(V(J)) =\sqrt{J}$ when $k$ is not algebraically closed. ::: :::{.solution} Take $J = \gens{x^2+1} \normal \RR[x]$, noting that $J$ is nontrivial and proper but $\RR$ is not algebraically closed. Then $V(J) \subseteq \RR$ is empty, and thus $I(V(J)) = I(\emptyset)$. :::{.claim} $I(V(J)) = \RR[x]$. Checking definitions, for any set $X \subset \AA^n/k$ we have \[ I(X) = \ts{f\in \RR[x] \st \forall x\in X,\, f(x)=0} \\ \] and so we vacuously have \[ I(\emptyset) = \ts{f\in \RR[x] \st \forall x\in \emptyset,\, f(x)=0} = \ts{f\in \RR[x]} = \RR[x] .\] ::: :::{.claim} $\sqrt{J} \neq \RR[x]$. This follows from the fact that maximal ideals are radical, and $\RR[x]/ J \cong \CC$ being a field implies that $J$ is maximal. In this case $\sqrt{J} = J \neq \RR[x]$. \ That maximal ideals are radical follows from the fact that if $J\normal R$ 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 \[ 1\in \sqrt{J} \implies 1^n \in J \text{ for some }n \implies 1 \in J \implies J=R ,\] contradicting the assumption that $J$ is maximal and thus proper by definition. ::: :::