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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!

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    $\begingroup$ What part of Eisenbud and Harris are you referring to? I only see sections of the form III.x.y where $x \le 3, y \le 8$. $\endgroup$ Nov 2, 2012 at 7:32
  • $\begingroup$ @QiaochuYuan Geometry of Schemes Exercise III-12 in III.2.1. I'll edit that $\endgroup$ Nov 2, 2012 at 7:38

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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 $A[x,y]\simeq A \simeq A[x^2,xy,y^2]$.

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    $\begingroup$ haha oh man that's hilarious-a true swindle. thanks a ton! $\endgroup$ Nov 2, 2012 at 7:41
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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 $2$.

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  • $\begingroup$ Thanks man-perhaps I'm misunderstanding, but how does this argument rule out 2 non-homogeneous generators? $\endgroup$ Nov 2, 2012 at 7:49
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    $\begingroup$ @uncookedfalcon: homogeneity doesn't matter. The point is that if $p, q$ are two generators of degree $2$ then the generators only span a $2$-dimensional subspace of the elements of degree $2$, since any nontrivial product has degree at least $4$. If either $p$ or $q$ has degree greater than $2$ then you can only get a $1$-dimensional subspace, and if either $p$ or $q$ has degree less than $2$ then it has degree $0$ and is a constant, so can be discarded. $\endgroup$ Nov 2, 2012 at 18:15

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