Minimal field containing two field of formal Laurent series

Question

Let $L/K$ be a field extension and $X,Y$ be two $K$-algebraically independent elements in $L$.

What is the minimal subfield of $L$ that contains both $K((X))$ and $K((Y))$?

The field $K((X,Y))=\operatorname{Frac}(K[[X,Y]])$ seems to be a good candidate, but I'm not sure how to prove that.

Maybe another (somehow equivalent) question is that if some subfield of $L$ contains both $K[[X]]$ and $K[[Y]]$, can we deduce that it contains $K[[X,Y]]$?

Answer 1

Finally, I found the following results:

Let $K$ be a field, R be the 2-variables formal power series ring $K[[X,Y]]$, and $A$ be the minimal subring of $R$ containing both $K[[X]]$ and $K[[Y]]$.

Theorem 1 ([AHW91, Corollary 1.3]). $A\neq R$. In particular $u_1=\sum X^i Y^i\notin A$.

Theorem 2 ([AHW91, Proposition 1.6]). $R$ has uncountable transcendence degree over $\operatorname{Frac}A$. In particular $\operatorname{Frac}A\neq\operatorname{Frac}R$.

[AHW91] S. S. Abhyankar, W. Heinzer, and S. Wiegand. “On the compositum of two power series rings”. In: Proceedings of the American Mathematical Society 112.3 (1991), pp. 629–636. doi: 10/bn9jt9

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Thus the answer to both of my original questions is NO.

Answer 2

I answered yesterday but I mistook formal series with polynomials, my bad. It makes the problem a lot harder ! I think the answer is actually no but I can only answer it for the second question (where I assume you talk about a subring and not a subfield of $L$). If $R$ is the smallest ring containing $K[\![X]\!] \cup K[\![Y]\!]$ i.e. the intersection of all rings containing this union, $R \subset K[\![X,Y]\!]$ because $K[\![X]\!] \cup K[\![Y]\!] \subset K[\![X,Y]\!]$ and it is a ring.

In fact, you have by the same argument that $$ R \subset R' = \{f(g_1(X),\ldots,g_n(X),h_1(Y),\ldots,h_m(Y))|(n,m) \in \mathbb{N}^2, \forall i, g_i \in K[\![X]\!], \forall j, h_j \in K[\![Y]\!], f \in K[Z_1,\ldots,Z_n,T_1,\ldots,T_m]\} $$ It is easy to check that $R'$ is a ring containing $K[\![X]\!] \cup K[\![Y]\!]$ and since any element of $R'$ can be constructed with finite sums and products of elements of $K[\![X]\!] \cup K[\![Y]\!]$, it contains $R$ hence $R = R'$.

However, $R \neq K[\![X,Y]\!]$. To prove it, let us show that any element of $R$ can be written as $f(g_1(X),\ldots,g_n(X),h_1(Y),\ldots,h_n(Y))$ where $f : K^n \times K^n \rightarrow K$ is a bilinear map.

Indeed, using $R = R'$, any element $r$ of $R$ can be written as $f(g_1(X),\ldots,g_n(X),h_1(Y),\ldots,h_m(Y))$ with $f$ a polynomial with $n + m$ variables. Write $$ f(Z_1,\ldots,Z_n,T_1,\ldots,T_m) = \sum_{\underset{\max(i) \leqslant N}{i \in \mathbb{N}^n}}\sum_{\underset{\max(j) \leqslant N}{j \in \mathbb{N}^m}} a_{ij}Z^iT^j, $$ where $N \in \mathbb{N}$, $\max(i) = \max\{i_1,\ldots,i_n\}$, the $a_{ij}$ are scalars in $K$ and $Z^i$ means $Z_1^{i_1}\cdots Z_n^{i_n}$. Then, \begin{align*} r(X,Y) & = f(g_1(X),\ldots,g_n(X),h_1(Y),\ldots,h_m(Y))\\ & = \sum_{\underset{\max(i) \leqslant N}{i \in \mathbb{N}^n}}\sum_{\underset{\max(j) \leqslant N}{j \in \mathbb{N}^m}} a_{ij}g(X)^ih(Y)^j\\ & = \sum_{\underset{\max(i) \leqslant N}{i \in \mathbb{N}^n}}\sum_{\underset{\max(j) \leqslant N}{j \in \mathbb{N}^m}} a_{ij}\tilde{g}_i(X)\tilde{h}_j(Y), \end{align*} where for all $i$ and $j$, $\tilde{g}_i(X) = g(X)^i$ and $\tilde{h}_j(Y) = h(Y)^j$. $$ \tilde{f}\left(\left(\tilde{Z}_i\right)_{i \in \mathbb{N}^n, \max(i) \leqslant N}\left(\tilde{T}_j\right)_{j \in \mathbb{N}^n, \max(j) \leqslant N}\right) = \sum_{i \in \mathbb{N}^n}\sum_{j \in \mathbb{N}^m} a_{ij}\tilde{Z}_i\tilde{T}_j : K^{(N + 1)^n} \times K^{(N + 1)^n} \rightarrow K $$ is the wanted bilinear map and we have $r(X,Y) = \tilde{f}\left(\left(\tilde{g}_i(X)\right),\left(\tilde{h}_j(Y)\right)\right)$.

Now, we we build an element of $K[\![X,Y]\!]\backslash R$. For all integer $m$, let $n_m$ be the largest integer such that $(2m + 1)^2 > 2n_m(2m + 1) + n_m^2$. As we have a square on the left hand side, $n_m \rightarrow +\infty$ when $m \rightarrow +\infty$. Let for all $m$, $$ p_m : \left\{\begin{array}{rcl} K^{2n_m(2m + 1) + n_m^2} & \rightarrow & K^{(2m + 1)^2} \\ \begin{pmatrix} (g_{ip})_{1 \leqslant i \leqslant n_m,m^2 \leqslant p < (m + 1)^2} \\ (h_{j,q})_{1 \leqslant j \leqslant n_m,m^2 \leqslant q < (m + 1)^2} \\ (a_{ij})_{1 \leqslant i,j \leqslant n_m}\end{pmatrix} & \mapsto & \displaystyle \left(\sum_{i,j = 1}^{n_m} a_{ij}g_{ip}h_{jq}\right)_{m^2 \leqslant p,q < (m + 1)^2}\end{array}\right. $$ $p_m$ is a polynomial map hence a morphism of algebraic varieties over $K$. Since $(2m + 1)^2 > 2n_m(2m + 1) + n_m^2$, we can use a dimensional argument if $K$ is infinite, else a cardinality argument if $K$ is finite, to deduce that $p_m$ is not surjective. Let $(s_{pq})_{m^2 \leqslant p,q < (m + 1)^2}$ be an element of $K^{(2m + 1)^2}$ that is not in the image of $p_m$.

Let $s(X,Y) = \sum_{p,q \in \mathbb{N}^2} s_{pq}X^pY^q \in K[\![X,Y]\!]$ and let us assume that $s \in R$. In this case, we showed that we can write $$ s(X,Y) = \sum_{i,j = 1}^n a_{ij}g_i(X)h_j(Y). $$ Let $g_i(X) = \sum_{p \in \mathbb{N}} g_{ip}X^p$ and $h_j(Y) = \sum_{q \in \mathbb{N}} h_{jq}Y^q$. We deduce that, \begin{align*} \sum_{p,q \in \mathbb{N}^2} s_{pq}X^pY^q & = s(X,Y)\\ & = \sum_{i,j = 1}^n a_{ij}g_i(X)h_j(Y)\\ & = \sum_{i,j = 1}^n a_{ij}\sum_{p,q \in \mathbb{N}^2} g_{ip}h_{jq}X^pY^q\\ & = \sum_{p,q \in \mathbb{N}^2} \left(\sum_{i,j = 1}^n a_{ij}g_{ip}h_{jq}\right)X^pY^q. \end{align*} Therefore, for all $p,q$, $s_{pq} = \sum_{i,j = 1}^n a_{ij}g_{ip}h_{jq}$. As $n_m \rightarrow +\infty$, there is a $m$ such that $n_m \geqslant n$. In this case, set $a_{ij} = 0$ if $i > n$ or $j > n$ and we have $p_m((g_{ip}),(h_{jq}),(a_{ij})) = (s_{pq})_{m^2 \leqslant p,q < (m + 1)^2}$, which contradicts the definition of $(s_{pq})$. By contradiction, $s \notin R$.

I think that $F = (K(\!(X)\!),K(\!(Y)\!))$ (which is by the way the field of fractions of $R$) is not $K(\!(X,Y)\!)$ with the same kind or argument, but I can't get rid of the denominator. I hope you will find how to solve this problem !