Is $\mathbb{C}[[x]] \simeq \mathbb{C}[x]_{(x)}$? 
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 $(x^n)$. 
 A: If two integral domains are isomorphic, then their fraction fields are also isomorphic. In our case this means that $\mathbb C(X)\simeq\mathbb C((X))$, and this is not true since the transcendence degree of $\mathbb C(X)$ over $\mathbb C$ is $1$, while the transcendence degree of $\mathbb C((X))$ over $\mathbb C$ is infinite.
A: 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 $p$. 
A: Their quotient fields have different transcendence degrees over $\mathbb{C}$, which is the largest field contained in each, so they are not isomorphic rings.
$\mathbb{C}[x]_{(x)}$ is the ring of rational functions which are defined (ie, do not have a pole) at $0$.
$\mathbb{C}[[x]]$ is the ring of power series centered at $0$, and it is a strictly larger ring. This ring includes all meromorphic functions (and thus all rational functions) defined at $0$, as well as many more power series that do not converge. As Thomas indicated in the comments, this is the completion of $\mathbb{C}[x]_{(x)}$ at its unique maximal ideal--it is also the completion of $\mathbb{C}[x]$ at the same ideal.
The fact that they have the same $\operatorname{spec}$ is indicative of the fact that both are rings of functions on the same underlying space (defined at $0$, and not everywhere blowing up at the generic point $\mathbb{C}$), but the latter ring admits a wider class of functions on this space.
