Let $R$ and $S$ be non-zero rings with identity. Is it possible to have $R[x] \cong S[[x]]$ ?
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1$\begingroup$ What about $R=\mathbb{Z}[[x]]$, $S=\mathbb{Z}[x]$? $\endgroup$– Piotr AchingerDec 29, 2012 at 9:17
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8$\begingroup$ @Piotr: that doesn't work, or at least the natural map isn't an isomorphism. $\mathbb{Z}[[x]] [y]$ is the subring of $\mathbb{Z}[y][[x]]$ in which the coefficients of the $x^n$, as a polynomial in $y$, have uniformly bounded degrees. $\endgroup$– Qiaochu YuanDec 29, 2012 at 9:28
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1$\begingroup$ @Qiaochu: Wow, thanks, I never realized that! $\endgroup$– Piotr AchingerDec 29, 2012 at 10:10
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1$\begingroup$ $S[[x]]$ can be given the structure of a complete metric space, by defining $d(a,b)=2^{-v_x(a-b))}$, with $v_x$ the $x$-adic valuation. On the other hand $R[x]$ is naturally a countable union of quite small subsets (polynomials of degree at most $n$ for $n=0,1,2,3,\ldots$. Can one now try to argue topologically using some kind of Baire Category Theorem argument? Not that I can get it to work... $\endgroup$– user30035Dec 29, 2012 at 11:47
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1$\begingroup$ I made a similar comment on your other question, but please ammend your title style to include complete questions. For example, "Can $R[x] \cong S[[x]]$?" is a better title than the current one. Actually, the entire body of your question would fit in the title — titles on MO may be longer than tweets. $\endgroup$– Theo Johnson-FreydJan 1, 2013 at 4:03
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3 Answers
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Here's a proof that no such commutative rings $R$, $S$ exist. (See the edit for an extension to noncommutative rings.)
Suppose we have an isomorphism $\phi: R[x] \to S[[x]]$; let $a = \phi^{-1}(x)$. First we claim that for all $b \in R[x]$, the element $1 + ab$ is invertible in $R[x]$. Indeed, the element $\phi(1 + ab) = 1 + (\phi(b))x$ has an inverse given by a formal geometric series, so $\phi^{-1}$ applied to this element must also be invertible.
In particular, $1 + ax$ must be invertible in $R[x]$. But it is well-known (in the commutative case) that any invertible polynomial $a_0 + a_1 x + \ldots + a_n x^n$ has $a_0$ invertible and the other $a_i$ nilpotent. It follows quickly that $a$ must be nilpotent. But then $\phi(a) = x$ is nilpotent in $S[[x]]$. We have reached an absurdity.
Edit: Aided by Martin's excellent suggestion in his comment, we may easily extend the argument to noncommutative rings. Indeed, since $x$ is central in $S[[x]]$, we have that $a$ is central in $R[x]$. In particular, $a$ commutes with scalars; writing $a$ as a polynomial, it follows that each coefficient of $a$ is central in $R$. This is true also of the polynomial $1 + ax$, and now the proof that all but the unit coefficient of $1 + ax$ is nilpotent goes through as in the commutative case (see for example the nice argument given here). Thus $a$ is nilpotent.
answered Dec 29, 2012 at 16:02
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4$\begingroup$ In order to reduce to the commutative case, one may try to compute the centers of $R[x]$ and $S[[x]]$. $\endgroup$ Dec 29, 2012 at 16:18
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$\begingroup$ Thanks very much for that suggestion, Martin. I have edited my answer to take it into account. $\endgroup$– Todd Trimble ♦Dec 29, 2012 at 16:48
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1$\begingroup$ @Todd: Nice solution. Can you generalize this to skew polynomials or Ore extensions !? $\endgroup$– user30230Dec 29, 2012 at 17:32
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$\begingroup$ @shatich: perhaps someone else can see such a generalization, but to be honest I had to google Ore extension, so I'd have to sit down and think it over. It might be worth opening another question if you're interested. $\endgroup$– Todd Trimble ♦Dec 29, 2012 at 18:07
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1$\begingroup$ Another suggestion: In the end of the argument, just use the commutative case with $Z(R)[x]$. $\endgroup$ Dec 30, 2012 at 12:11
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Thanks to Martin Brandenburg suggestion, if $R[x] \cong S[[x]]$ then their centers are isomorphic too. So without loose of generality we can assume that $R$ and $S$ are commutative. In commutative case we know $J(R[x]) = Nil(R[x])$. This means that elements in the Jacobson radical of $R[x]$ are all nilpotent. On the other hand $x \in J(S[[x]])$ and $x$ is not nilpotent. This shows that $R[x]$ and $S[[x]]$ can not be isomorphic.
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$\begingroup$ In your argument you seem to be assuming that $R[x]$ is Jacobson. There is no reason for this to be true if $R$ is not Jacobson. $\endgroup$ Dec 29, 2012 at 21:39
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2$\begingroup$ @Qiaochu Yuan: I cann't see that, could you please tell exactly where he/she used that assumption ?? $\endgroup$– user30230Dec 29, 2012 at 22:51
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$\begingroup$ I think Qiaochu's objection arises from $J(R[x])=\mathrm{Nil}(R[x])$. But this, in fact, holds for every commutative ring $R$, and uses the description of the units $R[x]^* = R^* + \mathrm{Nil}(R) (x)$. @tvector: Your proof uses $Z(R[x])=Z(R)[x]$ and $Z(S[[x]])=Z(S)[[x]]$, right? $\endgroup$ Dec 30, 2012 at 12:17
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$\begingroup$ @Martin: Exactly, yes. $\endgroup$– user30276Dec 30, 2012 at 16:45
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Let me write $R[x]=S[[y]]$ to avoid confusion.
I. Note that 1+y is a unit. Therefore (thinking of y as an element of $R[x]$), we have $y\in R$.
II. Now mod out $y$ on both sides: $\overline{R}[x]=S$.
III. This gives $R[x]=\overline{R}[x][[y]]$
IV. The ring on the right contains the element $\sum(xy)^n$. Thus so does the ring on the left (thought of as a subring of the appropriate completion). It follows that $y$ is nilpotent in the ring on the left. But it's clearly not nilpotent in the ring on the right. Contradiction.
answered Dec 29, 2012 at 16:12
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$\begingroup$ I don't think that step I is correct. See also Todd's post for the units of $R[x]$. $\endgroup$ Dec 29, 2012 at 16:22
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