$A[x,y] \not \simeq A[x^2, xy, y^2]$

Question

Is it true that for any (commutative, unital) ring $A$ that $A[s,t], A[x^2, xy, y^2]$ cannot be isomorphic as rings?

This is mentioned in passing in Eisenbud-Harris, Geometry of Schemes Exercise $III-12$. Thus far, I've convinced myself that $A[s,t] \not \simeq A[x^2, xy, y^2]$ as $A$ algebras for any $A$.

(To see this, suppose we had such an isomorphism, then by base changing to an algebraic closure of a residue field $A/\mathfrak{m}$, we'd have $k[s,t] \simeq k[x^2, xy, y^2]$, for $k$ algebraically closed. Such an isomorphism in particular implies $k[x^2,xy,y^2]$ has 2 $k$-algebra generators $p,q$. From here, I stuck them inside $k[x,y]$. Writing $\mathfrak{m} = (x,y)$, we have $\mathfrak{m} ^2 = (x^2, xy, y^2) \subseteq (p,q)$, so their ideal is supported either at the origin or is (1). I ruled out each case with some linear algebra in $k[x,y]/\mathfrak{m} ^3$-I can spell out details if there's interest.)

Thanks in advance!

Answer 1

It is not true without some more hypotheses on $A$. Set $$A=\mathbb{Z}[x_1,x_2,x_3,\dots][y_1^2,y_1z_1,z_1^2,y_2^2,y_2z_2,z_2^2,y_3^2,y_3z_3,z_3^2,\dots]$$ Then $A[x,y]\simeq A \simeq A[x^2,xy,y^2]$.

Answer 2

You don't have to work so hard to show that they can't be isomorphic as $A$-algebras. You can use a degree argument. First reduce, as you say, to the case that $k$ is a field. Then: two elements can't generate $k[x^2, xy, y^2]$ as an $k$-algebra because, with the grading inherited from $k[x, y]$, its degree-$2$ part has dimension $3$ but its degree-$1$ part has dimension $0$. A generator of degree $0$ is redundant and can be thrown out, so at least three generators are necessary, each of degree at least $2$.