Polynomial Rings - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T05:12:47Z http://mathoverflow.net/feeds/question/117487 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/117487/polynomial-rings Polynomial Rings chatish 2012-12-29T08:01:06Z 2012-12-29T17:15:05Z <p>Let $R$ and $S$ be non-zero rings with identity. Is it possible to have $R[x] \cong S[[x]]$ ?</p> http://mathoverflow.net/questions/117487/polynomial-rings/117530#117530 Answer by Todd Trimble for Polynomial Rings Todd Trimble 2012-12-29T16:02:32Z 2012-12-29T16:46:09Z <p>Here's a proof that no such <em>commutative</em> rings $R$, $S$ exist. (See the edit for an extension to noncommutative rings.) </p> <p>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. </p> <p>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. </p> <p><b>Edit:</b> 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 <a href="http://ysharifi.wordpress.com/2010/09/02/units-in-polynomial-rings/" rel="nofollow">here</a>). Thus $a$ is nilpotent. </p> http://mathoverflow.net/questions/117487/polynomial-rings/117531#117531 Answer by Steven Landsburg for Polynomial Rings Steven Landsburg 2012-12-29T16:12:30Z 2012-12-29T16:12:30Z <p>Let me write $R[x]=S[[y]]$ to avoid confusion.</p> <p>I. Note that 1+y is a unit. Therefore (thinking of y as an element of $R[x]$), we have $y\in R$.</p> <p>II. Now mod out $y$ on both sides: $\overline{R}[x]=S$.</p> <p>III. This gives $R[x]=\overline{R}[x][[y]]$ </p> <p>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.</p> http://mathoverflow.net/questions/117487/polynomial-rings/117538#117538 Answer by tvector for Polynomial Rings tvector 2012-12-29T17:15:05Z 2012-12-29T17:15:05Z <p>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. </p>