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

Solution 1:

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 $\CC[X]$.

Solution 2:

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.

Solution 3:

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