Dual space of polynomial algebra

Question

Let $k$ be an infinite field and let's consider the ring $R=k[x_1,\dots,x_n]$. This ring has a structure of $k$-vector space (or a $k$-algebra).

I am interested to know about the structure of the vector space $R^*$.

It would be enough to know about the structure of the vector space $k[x]^*$ since $R\cong k[x]^{\otimes n}=\stackrel{n\text{ times}}{k[x]\otimes\cdots\otimes k[x]}$ (this, as a $k$-algebra) and $(V\otimes W)^*\cong V^*\otimes W^*$ as here (as a vector space).

So these are my questions:

1. Is there any known vector space which is isomorphic to $k[x]^*$?
2. If $A$ is a $k$-algebra, does this extra structure induce a $k$-algebra structure over the dual space $A^*$?
3. If 2. is true, does this structure preserve the property $(V\otimes W)^*\cong V^*\otimes W^*$?

While I was writing I found this. This answers 1: $k[x]^*\cong k[[x]]$. About 2, the homomorphism defined there doesn't seem to be an algebra homomorphism (if I am not mistaken $$\sum_{i=0}^\infty f(x^i)g(x^i)x^i\neq \sum_{i=0}^\infty \sum_{n=0}^i f(x^n)g(x^{i-n})x^i$$ in general) so it just preserves the structure as a vector space.

So

4. Is there any way to save 2. and/or 3.?

And last:

5. What can we say about the structure of $k[[x]]^{\otimes n}$?

Note: I am not sure if the notation $R^{\otimes n}$ is correct, I just thought it would be shorter that way.

Answer 1

I am fairly certain there are some inaccuracies on this page, and since an answer has already been accepted and upvoted I would like to try to address them. This post only addresses the infinite dimensional case, since everything is fine over finite dimensional spaces.

$V^*\otimes W^* \cong (V\otimes W)^*$

OK this is true, they have the same dimension, but I want to point out that the natural map $V^*\otimes W^* \hookrightarrow (V\otimes W)^*$ is never an isomorphism. Neil Strickland has written an extremely pleasant proof over on mathoverflow.

Especially in the case we care about, $A^*\otimes A^*$ vs $(A\otimes A)^*$ one can check that $\dim(A^*\otimes A^*) = \dim (A\otimes A)^* = \dim A^*$, but this doesn't give us any meaningful way identifying the two spaces.

the multiplication $m: A\otimes A \to A$ and unit $k\to A$ maps on an algebra A give rise to comuliplication $A^*\to A^*\otimes A^*$.

This is false. The multiplication dualizes to a map $A^* \to (A\otimes A)^*$ which in general does not factor through $A^*\otimes A^* \subseteq (A\otimes A)^*$. Sometimes it does, such as if $m$ is the zero map (allowable in a non-unital algebra). For $A = k[x]$ it does not. In general, for any $A$ there is a coalgebra structure on an appropriate subspace $A^\circ \leq A^*$, which for finite dimensional $A$ is just $A^\circ = A^*$. A good reference for this is Section 2.5 of Radford's Hopf Algebras - the definition is actually just $A^\circ = (m^*)^{-1}(A^*\otimes A^*)$. In the case of $k[x]$ see for example Exercise 2.5.17 of Radford.

$k[[x_1]]\otimes\cdots\otimes k[[x_n]] \cong k[[x_1,\cdots,x_n]]$ just as for polynomial algebras.

For polynomials, the isomorphism $k[x]\otimes k[y] \to k[x,y]$ is multiplication. For power series, the map of multiplication is not an isomorphism. In other words, completion does not commute with tensor products (see e.g. this mse question).

Upon identifying $A^* = k[[x]]$, The injection $A^*\otimes A^* \to (A\otimes A)^*$ is identified with the map of multiplication $k[[x_1]]\otimes k[[x_2]] \to k[[x_1,x_2]].$ As an example of something not in the image of this map we borrow from Strickland's overflow post the power series $$1 + x_1x_2 + x_1^2x_2^2 + x_1^3x_2^3 + \cdots = \sum_{i=0}^\infty (x_1x_2)^i.$$ Strickland has already provided a beautiful proof, but perhaps I can offer an explanation (not a proof!) why this is not in the image. It is easier to see this is not in the image of the map $A^*\times A^* \to (A\otimes A)^*$, namely to have $$(a_0 + a_1x_1 + a_2x_1^2 + \cdots )(b_0 + b_1x_2 + b_2x_2^2 + \cdots ) = \sum_{i=0}^\infty (x_1x_2)^i$$ we must solve equations $a_ib_j = \delta_{ij}$ for all $i,j\geq 0$. But $a_1b_1 = 1, a_2b_2 = 1$ and $a_1b_2 = 0$ are already contradictory, and in fact no matter what you do infinitely many pairs $(i,j)$ will have $a_ib_j \neq \delta_{ij}$. Now, in $A^*\otimes A^*$ you actually get to have finite sums of elements from $A^*\times A^*$, which allows you to try a little harder to solve more of the equations $a_ib_j = \delta_{ij}$. But the moral is that just having finite sums won't let you fix the infinitely many problems you have.

However, $k[[x_1,x_2]]$ is a completed tensor product, i.e. $k[[x_1,x_2]] = k[[x_1]]\hat\otimes k[[x_2]]$. The meaning of this is that $k[[x_1]]\otimes k[[x_2]]$ is a local ring (mathoverflow) with maximal ideal $\mathfrak m = (x_1\otimes 1, 1\otimes x_2)$; and the $\mathfrak m$-adic completion is isomorphic to $k[[x_1,x_2]]$.

Answer 2

(1) If $V\cong\bigoplus_I k$ is isomorphic to a direct sum of $I$-indexed copies of $k$, then $V^*\cong\prod_I k$ is isomorphic to a direct product of $I$-indexed copies of $k$. The duality map $V\otimes V^*\to k$ can be reinterpreted as $\big(\bigoplus_I k\big)\otimes\big(\prod_I k\big)\to k$ via the obvious dot product. In particular, since $k[x]$ has the canonical basis $\{1,x,x^2,\cdots\}$ it is isomorphic to $\bigoplus_{\mathbb{N}}k$, and so its dual space $k[x]^*$ is isomorphic to $\prod_{\mathbb{N}}k$, which is the function space $\hom(\mathbb{N},k)$ with pointwise addition. This is of course itself isomorphic (as a vector space) to $k[[x]]$.

(2) Any linear map $f:A\to B$ induces a dual map $f^*:B^*\to A^*$ (it is in this way that taking dual spaces is a contravariant endofunctor of the category $\mathsf{Vect}_k$ of $k$-vector spaces). The map is given by precomposition, so $f^*$ sends $g\in B^*$ to $g\circ f\in A^*$.

In particular, this means the multiplication $A\otimes A\to A$ and unit $k\to A$ maps on an algebra $A$ give rise to comuliplication $A^*\to A^*\otimes A^*$ and counit $A^*\to k$ on the dual vector space $A^*$. The map $A^*\to (A\otimes A)^*$ takes a linear functional $f:A\to k$ and sends it to a bilinear functional $(a,b)\mapsto f(ab)$ (then note the the isomorphism $(A\otimes A)^*\cong A^*\otimes A^*$), and the counit $A^*\to k$ takes a linear functional $f:A\to k$ and evaluates it at $1\in A$. This makes $A^*$ a coalgebra.

One can reverse this all too: given a coalgebra $A$, the dual space $A^*$ gets an algebra structure.

(3) Just as one may define tensor products of algebras, one may also define tensor products of coalgebras. Then for any algebras $A$ and $B$, $(A\otimes B)^*$ is coalgebra isomorphic to $A^*\otimes B^*$, and similarly for any two coalgebras $A$ and $B$, $(A\otimes B)^*$ and $A^*\otimes B^*$ are algebra isomorphic.

Indeed, the underlying isomorphism is just the usual one for showing $(A\otimes B)^*$ and $A^*\otimes B^*$ are naturally isomorphic as vector spaces (or more accurately, as bifunctors).

(4) As I said, the fix is to realize dual-taking exchanges algebras and coalgebras.

(5) Well, $k[[x_1]]\otimes\cdots\otimes k[[x_n]]\cong k[[x_1,\cdots,x_n]]$ just as for polynomial algebras. As vector spaces over $k$, the tensor product has dimension $(\dim k[[x]])^n=\dim k[[x]]$, so it's isomorphic to $k[[x]]$ alone as a vector space. Not sure what more you want to have said about it.