Show that $\mathbb{F}[x^2,y^2,xy]$ is not polynomial $\mathbb{F}[x^2,y^2,xy]$ 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?
 A: 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 $\mathbb{Q}$.
A: 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.
A: 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.
A: 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.
