Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

8 Answers

up vote 6 down vote accepted

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.

share|improve this answer
    
Well...if only I had a few different ways to think about the solution ;) –  AsinglePANCAKE Feb 7 '13 at 13:49
add comment

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 $\CC[X]$.

share|improve this answer
add comment

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, $A\not\cong\CC[T]$.

share|improve this answer
1  
+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
add comment

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 $A\not\cong\CC[T]$.

share|improve this answer
4  
Only four answers? Why stop now? –  JSchlather Feb 7 '13 at 0:43
add comment

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 $B\not\cong\CC[T]$.

share|improve this answer
add comment

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

share|improve this answer
    
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
add comment

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...

share|improve this answer
    
$T$ 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
add comment

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 $A\not\cong B$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.