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Let $k$ be an algebraically closed field. The ideal-variety correspondence says that the equations \begin{align} \mathbf{I}(X) &= \left\{f\in k[x_1,\dotsc,x_n]:p\in X\Rightarrow f(p)=0\right\} \\ \mathbf{Z}(S) &= \left\{p\in \mathbb{A}^n_k:f\in S\Rightarrow f(p)=0\right\} \end{align} define inverse functors \begin{align} \mathbf{I} &: \mathrm{cl}\left(\mathbb{A}^n_k\right)^{\mathrm{op}}\rightarrow\sqrt{k[x_1,\dotsc,x_n]} \\ \mathbf{Z} &: \sqrt{k[x_1,\dotsc,x_n]}\rightarrow\mathrm{cl}\left(\mathbb{A}^n_k\right)^{\mathrm{op}}. \end{align} Here, $\mathrm{cl}\left(\mathbb{A}^n_k\right)$ denotes the collection of algebraic subsets of $\mathbb{A}^n_k$ viewed as a full subcategory of the poset $\cal{P}\left(\mathbb{A}^n_k\right)$ (the power set of $\mathbb{A}^n_k$). Similarly, $\sqrt{k[x_1,\dotsc,x_n]}$ denotes the collection of radical ideals of $k[x_1,\dotsc,x_n]$ viewed as a full subcategory of the poset $\cal{P}(k[x_1,\dotsc,x_n])$.

Now, obviously $\mathbf{I}$ and $\mathbf{Z}$ extend to functors \begin{align} \mathbf{I}&:\cal{P}\left(\mathbb{A}^n_k\right)^{\mathrm{op}}\rightarrow\cal{P}(k[x_1,\dotsc,x_n]) \\ \mathbf{Z}&:\cal{P}(k[x_1,\dotsc,x_n])\rightarrow\cal{P}\left(\mathbb{A}^n_k\right)^{\mathrm{op}}. \end{align} Of course, these functors are no longer inverses of one another after extension. I'd like to know what we can say about these extended functors. Specifically:

  1. Are $\mathbf{I}$ and $\mathbf{Z}$ adjoint functors?
  2. If $\mathbf{I}$ and $\mathbf{Z}$ are adjoint, then how general can we make the situation? Do they remain adjoint if $k$ is only assumed to be a commutative ring?

Since $S\subseteq\mathbf{I}(X)$ if and only if $X\subseteq\mathbf{Z}(S)$ I'm inclined to say that $\mathbf{I}$ is left adjoint to $\mathbf{Z}$ and only assuming $k$ to be a commutative ring doesn't seem to alter the situation, but I want to make sure I'm not making a mistake.

Brian Fitzpatrick
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1 Answers1

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Yes, they are adjoint. Quite generally, if you have two sets $A$ and $B$ and a binary relation between them, say $R\subseteq A\times B$, then you get a pair of contravariant functors adjoint on the right between the partially ordered sets $\mathcal P(A)$ and $\mathcal P(B)$ by the sending $X\subseteq A$ to $\{b\in B:(\forall x\in X)\,(x,b)\in R\}$ and sending $Y\subseteq B$ to $\{a\in A:(\forall y\in Y)\,(a,y)\in R\}$. In your situation, $A$ is the affine space, $B$ is the polynomial ring, and $R=\{(p,f)\in A\times B:f(p)=0\}$.

Andreas Blass
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