# Showing two polynomial rings over $\mathbb{C}$ aren't isomorphic

Im trying to show that the ring of polynomials in one variable over the complex numbers is not isomorphic to the ring over $\mathbb C$ with two variables $x$ and $y$ modulo $\langle x^2-y^3\rangle$. I've shown previously that if the relationship $p^2=q^3$ holds for some $p$ and $q$ in one variable, there exists $r$ such that $p=r^3$ and $q=r^2$. I'm assuming this helps in some way but I'm not precisely sure how.

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Suppose 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 multiplicities of $a$ in $f$ and in $g$, respectively, we have $2m=3n$, so that there is a $k\in\mathbb N$ such that $m=3k$ and $n=2k$. This means that there is a polynomial $h\in\CC[T]$ (which has the same zeroes as $f$ and $g$) such that $f=h^3$ and $g=h^2$.

Now $f$ and $g$ generate $\CC[T]$ (because $\phi$ is surjective), so $h$ also generates $\CC[T]$. It is easy to see that this is only possible if $h$ is of degree $1$. There is an isomorphism $\tau:\CC[T]\to\CC[T]$ which maps $h$ to $T$, so if we consider $\tau\circ\phi$ instead of $\phi$ we can assume that $h=T$.

So we have come to the conclusion that, if there is an isomorphism, $T^3$ and $T^2$ generate $\CC[T]$. This is not true.

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Well...if only I had a few different ways to think about the solution ;) – AsinglePANCAKE Feb 7 '13 at 13:49

If $I$ is a maximal ideal in $\def\CC{\mathbb C}\CC[X]$, then there is an $\alpha\in\CC$ such that $I=(X-\alpha)$, and using this it is easy to see that $\dim_\CC I/I^2=1$.

On the other hand, the ideal $J=(X,Y)\subset A=\CC[X,Y]/(X^2-y^3)$ is maximal and $J/J^2$ is a vector space of dimension $2$.

It follows that $A$ is not isomorphic to $\CC[X]$.

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Let $A=\def\CC{\mathbb C}\CC[X,Y]/(X^2-Y^3)$. Since the polynomial $X^2-Y^3$ is prime, the algebra $A$ is a domain. A little computation shows that $X$ and $Y$ are irreducible (and non-units), so that the element $u=X^2$ has two different factorization as products of irreducible elements, so $A$ is not a unique factorization domain.

On the other hand, $\CC[T]$ is a UFD, as we all know.

It follows that $A\not\cong\CC[T]$.

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Only four answers? Why stop now? – JSchlather Feb 7 '13 at 0:43

The algebra $\def\CC{\mathbb C}\CC[T]$ is integrally closed in its fraction field; this follows from Gauss's lemma, for example.

On the other hand, the element $t=x/y$ of the fraction field $F$ of $A=\CC[X,Y]/(X^2-Y^3)$, which is not if $A$, satisfies the polynomial $f(T)=T^2-Y\in F[T]$. This means that $A$ is not integrally closed and, therefore, $A\not\cong\CC[T]$.

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+1 I think I like this the most, though the one with the UFD is very nice, too. I shall keep this answer in my favourites section: so many approaches. – DonAntonio Feb 7 '13 at 1:16

Recall that if $A$ is a $\def\CC{\mathbb C}\CC$-algebra and $\delta:A\to A$ is a $\CC$-linear map, we way that $\delta$ is a derivation of $A$ if for all $a$, $b\in A$ we have $$\delta(ab)=\delta(a)b+a\delta(b).$$

Let $d:\CC[t]\to\CC[T]$ be the usual derivative map, so that $d(f)=f'$ for all $f\in\CC[T]$. Of course, $d$ is a derivation of $\CC[T]$. Moreover, it is easy to see that if $\delta:\CC[T]\to\CC[T]$ is a derivation, then there exists an $u\in\CC[T]$ such that for all $f\in \CC[T]$ we have $\delta(g)=uf'$.

Using this information, it is easy to see that the unique non-zero derivations of $\CC[T]$ which are diagonalizable are $\epsilon:f\in\CC[T]\mapsto xf'\in\CC[T]$ and its scalar multiples. It is almost immediate that the eigenvalues of this derivation $\epsilon$ are the non-negative integers. It follows that the set of eigenvalues of every non-zero diagonalizable derivation of $\CC[T]$ is an additive subsemigroup of $\CC$ generated by one element.

On the other hand, let $B=\CC[X,Y]/(X^2-Y^3)$. There is exactly one derivation $\eta:B\to B$ such that $\eta(X)=3X$ and $\eta(Y)=2Y$. This derivation is diagonalizable and the set of its eigenvalues is the set $\{n\in\mathbb N:n=0 \vee n\geq2\}$, and this is a subsemigroup of the additive group $\CC$ which is not generated by one of its elements.

It follows that $B\not\cong\CC[T]$.

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So basically you want to show that

$$\Bbb C[t]\ncong \Bbb C[x,y]/\langle x^2-y^3\rangle$$

I think your approach is good and you're pretty close: suppose there's an isomorphism $\,\phi\,$ , and let $\,p,q\in\Bbb C[t]\,$ be s.t. $\,\phi(p)=x+I\;\;,\;\phi(q)=y+I\,$ , with $\,I:=\langle x^2-y^3\rangle\,$ , but then

$$\phi(p^2)=\phi(p)^2=x^2+I\\\phi(y^3)=\phi(y)^3=y^3+I$$

and since both elements above in RHS are the same in the quotient ring and $\,\phi\ ,$ is injective, it must be that $\,p^2=q^3\,$.

But $\,x+I\,\,,\,\,y+I\,$ generate (as algebra) $\,\Bbb C[x,y]/I\,$ , so applying $\,\phi^{-1}\,$ it must be that $\,p\,\,,\,q\,$ generate $\,\Bbb C[t]\,$ , which is impossible since if the degree of one of them is more than $\,1\,$ then they have one root in common, and then every element in $\,\Bbb C[t]\,$ would also vanish at this common root...

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$T$ and $T^2$ generate $\CC[T]$ and they have a root in common :-) – Mariano Suárez-Alvarez Feb 7 '13 at 0:34
That's right, thanks. It's mean to be algebraically independent or, perhaps more accurately, s.t. one of them isn't irredundant as generator of the algebra, as in the case of $\,T\,,\,T^2\,$ . – DonAntonio Feb 7 '13 at 0:37

Yet another argument, the most laborious:

One can show with some calculation that the automorphism group of the $\mathbb C$-algebra $\mathbb C[X,Y]/(X^2-Y^3)$ is isomorphic to $\mathbb C^\times$, which is a nice abelian group.

On the other hand, the automorphism group of $\mathbb C[T]$ is the group $\mathbb C\rtimes\mathbb C^\times$, known as the group of affine linear maps of the line, which is not abelian.

It follows that $\mathbb C[X,Y]/(X^2-Y^3)\not\cong\mathbb C[T]$.

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The magic is in the computation of the automorphism group of the cusp, of course. I'll try to sketch that later. – Mariano Suárez-Alvarez Feb 7 '13 at 8:51

A simple one:

Let $\def\CC{\mathbb C}A=\CC[X,Y]/(X^2-Y^3)$ and let consider the ideal $J=(X,Y)$ of $A$, which is maximal. One can easily see that if $d:A\to A$ is a derivation, then $d(J)\subseteq J$.

On the other hand, let $B=\CC[T]$ and consider the derivation $\delta:f\in B\mapsto f'\in B$ . Then for no maximal ideal $I\subseteq B$ we have $\delta(I)\subseteq I$. indeed, given such an ideal there is an $\alpha\in C$ such that $I=(T-\alpha)$, and then $I\not\ni1=\delta(T-\alpha)\in \delta(I)$.

It follows that $A\not\cong B$.

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