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Suppose $k$ is an algebraically closed field of characteristic zero and $A$ is a finitely generated commutative associative reduced $k$-algebra.

Suppose the group $\mathbb{Z}_2$ acts on $A$ in such a way that the induced action on maxSpec $A$ is free. (Side note: This is equivalent to the induced grading $A = A_0 \oplus A_1$ being a strong grading, i.e., $A_1^2 = A_0$.)

Let $S^2 A = (A \otimes A)^{S_2}$, where $S_2$ acts by swapping the factors in the tensor product. The action of $\mathbb{Z}_2$ on $A$ induces a diagonal action of $\mathbb{Z}_2$ on $A \otimes A$. Since this action commutes with the $S_2$ action, we have an induced action of $\mathbb{Z}_2$ on $S^2 A$.

Now let $(S^2 A)_{\mathbb{Z}_2}$ be the corresponding algebra of coinvariants. By definition, this is the quotient of $S^2 A$ by the ideal generated by elements of the form $(b - g \cdot b)$ for $g \in \mathbb{Z}_2$ and $b \in S^2A$.

My question is: Is $(S^2 A)_{\mathbb{Z}_2}$ isomorphic to $A^{\mathbb{Z}_2}$ (as an algebra)?

My reason for thinking this is true is the following intuition:

  • $S^2 A$ is the coordinate ring of the set of unordered pairs of points of $A$.

  • Thus $(S^2 A)_{\mathbb{Z}_2}$ is the coordinate ring of set of unordered pairs of points of $A$ that are invariant under the $\mathbb{Z}_2$-action.

  • Since the action of $\mathbb{Z}_2$ is free, unordered pairs of points of $A$ that are invariant under the $\mathbb{Z}_2$ action are just $\mathbb{Z}_2$-orbits.

  • Thus $(S^2 A)_{\mathbb{Z}_2}$ should be isomorphic to $A^{\mathbb{Z}_2}$, which is precisely the coordinate ring of the variety of $\mathbb{Z}_2$-orbits.

I think the above reasoning is rigorous on the level of maximal ideals. But I'd like to know that the rings are actually isomorphic.

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1 Answer 1

You are right, they are isomorphic.

Denote by $g$ the generator of $\Bbb Z_2$. The map given by $x\otimes y\mapsto x^gy$ is obviously a homomorphism $h:A\otimes A\to A$ of algebras. Since $S^2A=S^2A_0+S^2A_1+K$, where $K$ is spanned by $a_0\otimes a_1+a_1\otimes a_0$, $a_i\in A_i$, and $A_1^2=A_0$, we conclude that $h(S^2A)=A_0$ (even $h(S^2A_1)=A_0$) and that $K$ generates in $S^2A$ the ideal $I$ such that $S^2A/I$ is the algebra of coinvariants.

In order to show that the kernel of $h$ on $S^2A$ equals $I$, we observe that, providing $\sum_ia_{0i}a_{1i}b_{1i}=\sum_jc_{1j}d_{1j}$, we have $$\sum_i(a_{0i}\otimes a_{1i}b_{1i}+a_{1i}b_{1i}\otimes a_{0i})+\sum_j(c_{1j}\otimes d_{1j}+d_{1j}\otimes c_{1j})=$$ $$=\sum_i(a_{0i}\otimes a_{1i}+a_{1i}\otimes a_{0i})(1\otimes b_{1i}+b_{1i}\otimes1)-$$ $$-\sum_i(1\otimes a_{0i}b_{1i}+a_{0i}b_{1i}\otimes1)(1\otimes a_{1i}+a_{1i}\otimes1)+$$ $$+\sum_j(1\otimes c_{1j}+c_{1j}\otimes1)(1\otimes d_{1j}+d_{1j}\otimes1)\in I,$$ where $a_{0i}\in A_0$ and $a_{1i},b_{1i},c_{1j},d_{1j}\in A_1$.

Let $s_i\in S^2A_i$ be such that $h(s_0+s_1)=0$. We can write $s_0=\sum_i(a_{0i}\otimes a_{1i}b_{1i}+a_{1i}b_{1i}\otimes a_{0i})$ and $s_1=\sum_j(c_{1j}\otimes d_{1j}+d_{1j}\otimes c_{1j})$ for suitable $a_{0i}\in A_0$ and $a_{1i},b_{1i},c_{1j},d_{1j}\in A_1$ because $A_0=A_1^2$. From $h(s_0+s_1)=0$, we obtain $\sum_ia_{0i}a_{1i}b_{1i}=\sum_jc_{1j}d_{1j}$ (the characteristic of $k$ is different from $2$). By the above idenity, $s_0+s_1\in I$.

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