Is $R[[x]][[y]]$ the same as $R[[x,y]]?$ Let $x,y$ be two central indeterminates in $R$, is $R[[x]][[y]]=R[[x,y]]$?
My take on this is:
Let $f(x,y) \in R[[x]][[y]]$, then $f(x,y) = \sum_{i=0}^{\infty}g_i(x)y^i = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}r_{i,j}x^jy^i \in R[[x,y]]$. So $R[[x]][[y]] \subset R[[x,y]]$.
Then let $f(x,y) \in R[[x,y]]$ we also have $f(x,y) = \sum_{i=0}^{\infty}(\sum_{j=0}^{\infty}r_{i,j}x^i)y^j \in R[[x]][[y]]$.
We conclude $R[[x]][[y]]=R[[x,y]]$.
I suspect there is something wrong on this proof, but I don't know. Is this proof wrong? Is this proposition wrong? Can someone provide a counter-example for me and perhaps give me some explanation?
 A: I’ve been thinking about this question, and I’ve decided that it merits a more complete treatment than I gave in my comment.
I believe it’ll clear the mind if we discuss four different rings, $\mathscr R_0=R[x][y]$, $\mathscr R_1=R[[x]][y]$, $\mathscr R_2=R[x][[y]]$, and $\mathscr R_3=R[[x]][[y]]$.
When we discuss one of these rings, let’s consider, for any element of the ring, the configuration of monomials that may have nonzero coefficients. And if the monomial $x^my^n$ has a nonzero coefficient, let’s put a star, $\star$, at the lattice point $(m,n)$ in the Cartesian plane. For example, if we expand $\frac1{1-xy}$ as a power series, we get $\sum_{i=0}^\infty x^iy^i$, which tells us to put a star at every point on the diagonal line $Y=X$, nowhere else. (I use capital letters for coordinates in the Cartesian plane here, and you understand that they have nothing to do with our indeterminates $x$ and $y$.)
Now let’s look at our first ring $\mathscr R_0=R[x][y]$. Elements here are polynomials in $y$, and the coefficients are polynomials in $x$. This ring is a polynomial ring, and an element $f$ there has a degree $N$, so that we may write $f=\sum_{n=0}^N\lambda_n(x)y^n$. All the stars are to be found in the rows numbered $0,1,\cdots,N$, no other rows. Each row denotes a polynomial $\lambda_n(x)$, and each has its own degree, but since there are only finitely many of them, there is a maximal $x$-degree $D$. Thus all nonzero stars appear in the rectangle $0\le m\le D$, $0\le n\le N$. There are only finitely many stars, therefore, and we have only finitely many nonzero terms: a polynomial in $x$ and $y$. So we have seen that $\mathscr R_0=R[x][y]=R[x,y]=R[y][x]$.
In the same way, $\mathscr R_1=R[[x]][y]$ is a polynomial ring, in the letter $y$, so any element $f$ has a $y$-degree, say $N$ again. Again $f$ has the shape $\sum_{n=0}^N\lambda_n(x)y^n$, but each $\lambda_n$ is a power series, and may have infinitely many nonzero monomials. Still only finitely many rows, but each may extend infinitely. For an example, take $\sum_{n=0}^N\frac1{1-x}y^n$. In general, the finitely many rows each may have infinitely many stars; in the example, each row is completely filled with stars.
It’s completely different for $\mathscr R_2=R[x][[y]]$. Each element $f$ is a power series in $y$, doesn’t have a degree. Rather $f$ will have the shape $\sum_{n=0}^\infty\lambda_n(x)y^n$, where now the $\lambda_n$ are polynomials in $x$, each with its own degree $d_n$. So we may have infinitely many rows, but each row’s stars can extend only finitely far to the right, to a distance of $d_n$. But these numbers may be unbounded, there may be infinitely many of them. It’s hard to say what the constellation of stars will generally look like here, but an example is $\sum_{n=0}^\infty\frac{x^{n+1}-1}{x-1}y^n$, where the coefficient of $y^n$ will be the polynomial $1+x+\cdots+x^n$. In this example, the constellation consists of all stars above the line $Y=X$.
Finally your case, $\mathscr R_3=R[[x]][[y]]$. Here, as you saw, you have a $y$-power series, and each coefficient is a power series in $x$. Infinitely many rows are possible, and each row may extend infinitely to the right. Thus no limitation on what the constellation of stars may look like, and indeed $\sum_{(m,n)\in\Bbb N\times\Bbb N}x^my^n$ is in your ring, giving a first quadrant filled with stars.
And now I think your correct conclusion becomes plainly justified: that $R[[x]][[y]]=R[[x,y]]$, and thus of course the same as, or at least isomorphic to $R[[y]][[x]]$.
