What is $k[x,y]_{(x,y)}$?

Question

After intensive internet research, I couldn't find any source that gave me a solution, probably the question is too easy. It would be great, if someone can tell me if I'm correct or not.

Let $k$ be any field (I'm usually working with $\mathbb{C}$, but every other field should be okay as well), is the polynomial ring localized at $(x,y)$ isomorphic to the formal power series ring, i.e. $$ k[x,y]_{(x,y)} \simeq k[[x,y]]? $$ Here $(x,y)$ is seen as an ideal, so the localization at $(x,y)$ could be interpreted as dividing by $k[x,y]\setminus (x,y)$.

Why should this be true? As a heuristic: On the LHS, only $x$ and $y$ (and multiples of these) have no inverse, the same is true for the RHS.

Does this hold for an arbitrary finite amount of variables? I would say yes.

Am I correct and it really is this easy or am I just standing on the hose?

Answer 1

No, they are not necessarily isomorphic. For one thing, if you choose $k$ to be countable, then $k[x,y]_{(x,y)}$ is countable, but $k[[x,y]]$ is uncountable.

For another thing, if I recall correctly, the former one is not complete with respect to its maximal ideal, whereas the second one is.

But they do superficially look similar: two local algebras whose maximal ideal is generated by $(x,y)$.

Answer 2

To complement Rschwieb’s answer, the formal power series ring $k[[x,y]]$ is the completion of the localization $k[x,y]_{(x,y)}$ at the unique maximal ideal. Geometrically, the latter corresponds to germs of functions at the origin in the affine plane, whereas the former corresponds to functions on a formal disc centered at the origin.

As for why the two are not isomorphic, suppose there was an isomorphism $$f:k[x,y]_{(x,y)} \longrightarrow k[[x,y]].$$ Then $f$ would necessarily send the unique maximal ideal in one ring to the unique maximal ideal in the other. Hence we can match up the completions of both sides. But the canonical map from $k[[x,y]]$ to its completion is an isomorphism, which is not true of $k[x,y]_{(x,y)}$.