TITLE: Is $\mathbb{C}[[x]] \simeq \mathbb{C}[x]_{(x)}$?
QUESTION [6 upvotes]: Let $\mathbb{C}[x]$ be the ring of polynomials and $\mathbb{C}[[x]]$ the formal power series. Is $\mathbb{C}[[x]] \simeq \mathbb{C}[x]_{(x)}$? Is it true? Is there a geometric interpretation of this isomorphism?

We have that $\operatorname{Spec}(\mathbb{C}[x]_{(x)})=\{(0),(x)\}=\operatorname{Spec}(\mathbb{C}[[x]])$. In fact $\mathbb{C}[[x]]$ is an integral domain and its ideals are $(x^n)$.

REPLY [3 votes]: If one is only interested at whether they are isomorphic as $\mathbb{C}$-algebras, then it can be solved by the transcendental degree over $\mathbb{C}$, since the transcendental degree of $\mathbb{C}((X))$ over $\mathbb{C}$ is uncountably infinite. 
If furthermore one is interested at whether the two rings are isomorphic as ring, the answer is still no. One way to show it is to use the valuation map. Suppose otherwise there is an isomorphism of rings $\alpha:\mathbb{C}[X]_{(X)}\rightarrow \mathbb{C}[[X]]$. Then $\alpha(X)$ is of the form $XF$ where $F\in \mathbb{C}[[X]]$ and $F(0)\neq 0$ by considering the valuation. Then after composed with another ring-isomorphism $\beta:\mathbb{C}[[X]]\rightarrow \mathbb{C}[[X]]$ which maps $XF$ to $X$, we may get an isomorphism of rings $\beta \alpha:\mathbb{C}[X]_{(X)} \rightarrow \mathbb{C}[[X]]$ which maps $X$ to $X$. Hence this isomorphism maps $X+1$ to $X+1$, which is an contradiction since that $X+1$ is not a square in the ring $\mathbb{C}[X]_{(X)}$ but it is a square in the ring $\mathbb{C}[[X]]$ ! 
More generally, these two rings are not isomorphic for any field $k$. The above arguement is still valid in case of characteristic $0$. In case of characteristc $p$, we just need to change "the square" to the "$q$-power" where $q$ is a prime number different from $p$.