# Is the ideal of a variety the annihilator of a subspace of the symmetric algebra?

Let $V$ be a vector space over an algebraically closed field $K$.

Let $\mathrm{Sym}(V^*)=\mathrm{Sym}(V)^*$ be the symmetric algebra on $V$, i.e. if we give a basis $e_1,...,e_n$ of $V$ and let $x_1,...,x_n$ be the dual basis, this is just $K[x_1,...,x_n]$.

Suppose we are given an algebraic variety with ideal $I=\langle f_1,...,f_n \rangle$.

These are the set of polynomials that vanish on the vanishing set of $f_1,...,f_n$.

Is $I \subset\mathrm{Sym}(V)^*$ the annihilator of a subspace $W \subset \mathrm{Sym}(V)$?

• What is the action of $Sym(V^*)$ on $Sym(V)$? – J. David Taylor May 4 '16 at 2:43
• It is via the natural isomorphism of $Sym(V^*) \cong Sym(V)^*$. Sorry for being unclear. Here is another way of formulating my question. Let $\phi: Sym(V^*) \to Sym(V)^*$ be the natural isomorphism. Then is there a subspace $W$ of $Sym(V)$ such that $\phi(I)$ is the set of functionals that are zero on $W$? – user062295 May 4 '16 at 2:55
• Ok. The action that I specified when asked was $\lambda (v_1\otimes ...v_k)=\sum_{\text{all permutations}} \lambda(v_{i_1} \otimes ...v_{i_r}) ...v_{i_k}$. Now the condition of module theoretic annihilation under this action is unreasonably strong. Thus in my question I was asking about the vector space definition of annihilation. An element in $Sym^k(V)^*$ evaluates on $Sym^l(V)$ as 0 if $l\neq 0$ and as usual otherwise. Under the vector space definition of annihilation, is my question still correct? – user062295 May 4 '16 at 11:25

## 2 Answers

Given any vector space $V$ and a subspace $W^* \subset V^*$, $W^*$ is always the annihilator of $\{v|\lambda(v)=0 \forall \lambda \in W^* \}$(one inclusion is clear, the other follows from dimension reasons). Now apply this statement to $V=Sym V$ and $W^*=I$.

This is a response to Prof. J. D. Taylor, that was too large to put in a comment.

This was not my identification of $Sym(V^*)$ with $Sym(V)^*$. $Sym(V^*)= \oplus Sym^i(V^*)$ of graded components. We have a pairing $Sym^i(V^*) \otimes Sym^i(V) \to K$ defined by $\langle \lambda_1 \cdot ...\lambda_i, v_1 \otimes... v_n \rangle=\sum_{\begin{pmatrix} j_1 & ... & j_i \\ 1&...&i \end{pmatrix} \in S_n } \lambda_{j_1}(v_1)...\lambda_{j_i}(v_i)$. Extend this pairing trivially to get a pairing $Sym^i(V^*) \otimes Sym(V) \to K$. Take the direct sum to get a pairing $Sym(V^*) \otimes Sym(V) \to K$.

In the context of your counterexample, when we fix $x_1,x_2$, the functional $\langle x_1 \cdot x_2, \cdot \rangle: Sym(V) \to K$ is the zero functional when restricted to $V \subset Sym(V)$.