Why is a vector space $V$ equal to the spectrum of the polynomial ring generated by its dual basis?

Question

I am recently reading this paper, "On the Chow Ring of a Geometric Quotient" by [Elingsrud, Stromme, 89]. There is an identity in section 3 that is a bit confusing:

$$V = Spec (k[x_1, \dots, x_n]).$$

The original text reads

Let $V$ be a vector space of dimension $n$. Choose a basis $\{v_1, \dots, v_n\}$ of $V$. Let $\{x_1, \dots, x_n\}$ be the dual basis such that $V = Spec (k[x_1, \dots, x_n])$.

A similar statement also appears in the introduction in this paper.

Our setting is a reductive group $G$ acting linearly on a vector space $V$, which we identify with the affine variety $Spec(Sym(V^{\vee}))$.

According to my understanding of the spectrum of a ring, the right-hand side should contain more points, unless this only means all the closed points of $Spec (k[x_1, \dots, x_n])$. This ring also looks a bit weird to me. Only those polynomials of homogeneous degree 1 are linear functions on $V$, while all others are not.

My question is, is this really a valid identity? Or it is just a convention that people use a lot. This identification actually has nothing to do with anything else in this paper. So I also wonder what this identification can be used for.

Please add the name (and possibly, the link) of the paper, and the full context around that equation. Also, $V^*$ never appears in your question, so why mentioning it? — Martin Brandenburg, Oct 25 '24 at 21:20
@MartinBrandenburg Thank you. I just added it. I mention the dual space also because it appears in another similar sentence in the paper, which I also added to the question. — P.Zhang, Oct 25 '24 at 21:39
I think you would want to say $V$ is equivalent to the $k$-valued points of the spectrum (which is only the same as the closed points in case $k$ is algebraically complete). — Daniel Schepler, Oct 25 '24 at 21:39
@KReiser "On the Chow Ring of a Geometric Quotient". The link is just added. — P.Zhang, Oct 25 '24 at 21:41
@DanielSchepler Yes. In this paper the field $k$ is assumed to be algebraically closed. I agree with you. It should just be the $k$-valued points of the spectrum. — P.Zhang, Oct 25 '24 at 21:46

Martin Brandenburg · Accepted Answer · 2024-10-25T22:59:24.777

There are three ways to fix this wrong identity $V = \mathbb{A}^n_k$, where $V$ is an $n$-dimensional vector space over $k$ and $\mathbb{A}^n_k = \mathrm{Spec} \, k[x_1,\dotsc,x_n]$.

Replace it with $V = \mathbb{A}^n_k(k)$, as mentioned in the comments. To be precise, it must be $|V| = \mathbb{A}^n_k(k)$, where $|-|$ is the underlying set.
Use the equivalence of categories between affine schemes of finite type and (classical) affine varieties to switch back and forth between these without writing down the equivalence explicitly (so this is the same as point 1 but without changing the notation). In other words, we "identify" a variety with its scheme. This is done a lot in the literature. See here for my opinion on "identifications", your question is a good example why they are problematic.
Use the equivalence of (symmetric monoidal, $k$-linear) categories between vector bundles over $\mathrm{Spec}(k)$ and vector spaces over $k$ to switch back and forth between these without writing down the equivalence explicitly. (Again, see here for my opinion on this kind of habit.) The vector bundle over a scheme $S$ associated to a locally free sheaf $\mathcal{F}$ on $S$ has as its underlying scheme $\mathrm{Spec}\,\mathrm{Sym}(\mathcal{F}^*)$, the vector bundle structure comes from adjunctions. See Equivalence of categories of vector bundles and locally free sheaves for details. So the formal equation here should be $\mathrm{Spec} \,\mathrm{Sym}(V^*) \cong \mathbb{A}^n_k$, which is true because $x_1,\dotsc,x_n$ is a basis of $V^*$ and then $\mathrm{Sym}(V^*) = k[x_1,\dotsc,x_n]$.

Also, associating $V$ with $\operatorname{Spec} \operatorname{Sym}(V^)$ gives a "coordinate-free" functorial correspondence. Here, the $k$-valued points of $\operatorname{Spec} \operatorname{Sym}(V^)$ correspond to $k$-algebra homomorphisms $\operatorname{Sym}(V^) \to k$, which correspond to $k$-linear maps $V^ \to k$, which by definition is just $V^{*}$, and since $V$ is finite-dimensional that's isomorphic to $V$. (And they even become isomorphic as vector spaces, given that you use the canonical co-$k$-module structure of $\operatorname{Sym}(V^)$ as an object of CommRing.) — Daniel Schepler, Oct 25 '24 at 22:56
Also, as far as I know some sources will use $\mathbf{Spec}$ for the operation to turn a sheaf of $\mathscr{O}_X$-algebras into a scheme with an affine morphism to $X$, whereas the non-bold $\operatorname{Spec}$ will be reserved for the functor from rings to schemes. — Daniel Schepler, Oct 25 '24 at 23:03
The symmetric algebra is a co-($k$-module object) in $\mathbf{CAlg}_k$, not in $\mathbf{CRing}$. And yes people who haven't understood that $\mathrm{Spec}$ for algebras or sheaves of algebras is just the same construction from a categorical point of view will use different letters for it. :P Same for $\mathrm{Sym}$. — Martin Brandenburg, Oct 25 '24 at 23:03

score 1 · Answer 2 · answered Oct 26 '24 at 13:31

Copied from Milne, Algebraic Groups, 2.6. For a $k$-vector space $V$, we let $V_{\mathfrak{a}}$ denote the functor $R\rightsquigarrow V\otimes R$. Recall that for a $k$-vector space $W$, the symmetric algebra $Sym(W)$ on $W$ has the following universal property: every $k$-linear map $W\rightarrow A$ from $W$ to a $k$-algebra $A$ extends uniquely to a $k$-algebra homomorphism $Sym(W)\rightarrow A$. Assume that $V$ is finite-dimensional, and let $V^{\vee}$ be its dual. Then, for a $k$-algebra $R$, \begin{align*} V\otimes R & \simeq Hom_{k}(V^{\vee},R)&&\text{(homomorphisms of }k\text{-vector spaces)}\\ & \simeq Hom_{k}(Sym(V^{\vee}),R)&&\text{(homomorphisms of } k\text{-algebras).} \end{align*} Therefore, $V_{\mathfrak{a}{}}$ is an algebraic group with $\mathcal{O}% {}(V_{\mathfrak{a}{}})=$ $Sym(V^{\vee})$.

Why is a vector space $V$ equal to the spectrum of the polynomial ring generated by its dual basis?

2 Answers2