Show that $\mathbb{F}[x^2,y^2,xy]$ is not polynomial

Question

$\mathbb{F}[x^2,y^2,xy]$ is the polynomials in two variables whose terms all have even degrees. Of course, this generating set $x^2,y^2,xy$ is not algebraically independent, but I need to show that no generating set can be.

There is a hint to work with systems of parameters. I've shown that the extension $\mathbb{F}[x^2,y^2,xy] \subset \mathbb{F}[x,y]$ is finite (with secondary generators x and y), so if there existed an independent generating set it would be a system of parameters for $\mathbb{F}[x,y]$. I'm having trouble figuring out what to do with this, though, since there is nothing wrong with $\mathbb{F}[x,y]$ having a system of parameters.

EDIT: All systems of parameters have the same cardinality. For $\mathbb{F}[x,y]$, this is cardinality 2 (for example, {x,y} is a system of parameters. So the problem reduces to showing that a set of two elements cannot generate $\mathbb{F}[x^2,y^2,xy]$. This seams reasonable enough, but I'm still not sure how to show this. Is it true in general that all minimal generating sets of an algebra have the same cardinality?

Answer 1

If $\mathbb{F}[x^2,y^2,xy]$ were a polynomial ring, it would be a UFD. Then since $x^2$, $y^2$ and $xy$ all have the minimum possible positive degree (2), they would all be irreducible. This is a contradiction since then $x^2y^2$ can be factorized both as $x^2\cdot y^2$ and as $(xy)^2$.

Addendum: It's not true in general that all minimal generating sets of an algebra have the same cardinality. For example, $\{x\}$ and $\{x^2, x^2+x\}$ are both minimal sets of generators for the $\mathbb{Q}$-algebra $\mathbb{Q}[x]$ over $\mathbb{Q}$.

Answer 2

In a polynomial ring, if you pick a maximal ideal $I$ and compute $\dim I/I^2$ you always get the same number.

In your ring the ideal $I=(x^2,xy,y^2)$ has $I/I^2$ of one dimension and all other such quotients of another.

Answer 3

Another nice answer might encorporate the celebrated Chevalley-Shephard-Todd theorem, that states that the invariant algebra of any finite group action is polynomial if and only if the group acts as a complex reflection group.

In our case, we have the invariants for the action of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{F}^2$ that sends $(x,y)$ to $(-x,-y)$. Now this is not a complex reflection group action (since it only fixes the origin), hence the ring of invariants $\mathbb{F}[x^2,xy,y^2]$ is not polynomial.

Answer 4

Your ring has infinite homological dimension. It is isomorphic to $k[x,y,z]/(xy-z^2)$. This is a double cone with a singularity at the origin. Yet a polynomial ring has finite homological dimension.