# 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 $(x^n)$.

• $\mathbb{C}[[x]]$ is the $(x)$-adic completion of $\mathbb{C}[x]_{(x)}$, so they are not isomorphic. – Matt Sep 22 '13 at 16:34
• @Matt but they have the same spec.... – ArthurStuart Sep 22 '13 at 16:36
• @ArthurStuart The fact that their spectrums are isomorphic means what to you? $\mathbb Z_{(p)}$ and $\mathbb Z_p$ - the $p$-adic integers - have the same spectrum, too, but they are not isomorphic - the first is isomorphic to a subring of $\mathbb Q$ while the second is not. – Thomas Andrews Sep 22 '13 at 16:44
• I didn't write that comment, but if $I$ is an ideal in a commutative ring $R$, in general, you can define the $I$-adic completion that is very similar to the $p$-adic completion. See: en.wikipedia.org/wiki/Completion_(ring_theory) , specifically the section on Krull topology. – Thomas Andrews Sep 22 '13 at 16:53
• @ArthurStuart: any two fields have isomorphic spectra. This indicates that the spectrum is not a sufficiently fine object to capture the whole structure of a ring in a geometric way, and that some extra data (namely the structure sheaf) is needed. – Nils Matthes Sep 23 '13 at 12:11

## 3 Answers

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.

• I have to accept only an answer... you underline the geometric interpretation of this problem... thanks – ArthurStuart Sep 22 '13 at 18:02

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.

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 $p$.

• Dear Ding, The question is not about $\mathbb C[X]_X$ (polynomials with $X$ inverted) but about $\mathbb C[X]_{(X)}$ (polynomials with all $f(X)$ having non-zero constant term inverted). Regards, – Matt E Sep 23 '13 at 1:56
• @Matt E: Thanks a lot, I make a mistake in a hurry. I have rewritten it. – Joy-Joy Sep 23 '13 at 2:24
• Dear YACP, you are right, thank you very much to point out the two mistakes. But I think the two rings are still not isomorphic. By considering the valuation we can show that $\alpha(X)$ is of the form $XF$ where $F\in \mathbb{C}[[X]]$ and $F(0)\neq 0$. – Joy-Joy Sep 23 '13 at 10:42
• Dear YACP, do you mean that you would like to know why the two rings are not isomorphic? 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 $F(0)\neq 0$ as you know. 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 $X+1$,contradiction! – Joy-Joy Sep 23 '13 at 16:07
• You are right. I am sorry to make such a stupid mistake. But $X+1$ is a square on one side but not on the other side. This arrives an contradiction anyway. – Joy-Joy Sep 23 '13 at 17:14