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Martin Sleziak
Martin Sleziak
3:48 PM
$T$ and $T^2$ generate $\CC[T]$ and they have a root in common :-) — Mariano Suárez-Álvarez Feb 7 '13 at 0:34
The macro is defined in Mariano's answer:
7
A: Showing two polynomial rings over $\mathbb{C}$ aren't isomorphic

Mariano Suárez-ÁlvarezSuppose there is an isomorphism $\phi:\def\CC{\mathbb C}\CC[X,Y]/(X^2-Y^3)\to\CC[T]$, and let $f=\phi(X)$ and $g=\phi(Y)$. Then $f^2=g^3$. It follows from this equality in $\CC[T]$ that $f$ and $g$ have exactly the same zeros. Moreover, if $a$ is one of those zeroes and $m$ and $n$ are the multip...

The other comments found by the first query are fine:
$\newcommand{\CC}{\mathbb{C}} \newcommand{\RR}{\mathbb{R}}$As fields, $\RR^2$ is isomorphic to $\CC$ (because it's not even a field). But they are isomorphic as $\RR$-vector spaces, and therefore, as $\RR$-modules, as abelian groups, and as sets. So one must be specific about what kind of isomorphic-ness one wants. Does that help? — Henry Swanson May 13 '14 at 22:13
$\newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}}$ Well, if $\RR$ is a subset of $\CC$, then $\RR \times \RR$ can be seen as a subset of $\CC \times \CC$, as the pairs of points that only involve real numbers. — Pedro M. Jan 28 '15 at 15:11
@Joald Not by itself. For example $[(12),(1234)]$ fixes this rectangle: $$ABAB\\BABA\\CCCC$$ So if we did not also have the restriction of having 6 of each colour, then there would have been solutions when a pair of rows were swapped. There are also solutions if there were 3 colours, 4 of each. — Jaap Scherphuis Aug 24 '17 at 8:47
 

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