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In fact, it is a simple problem. I just want to know whether there are some interesting proof.

$Z[x_1, x_2, ......, x_{n^2-1}]$ and $Z[y_{11}, ......, y_{1n}, y_{21}, ......, y_{nn}]/(det(y_{ij})-1))$, where $Z$ is integer.

One way to prove is select a prime number,say $p=2$,then localize these two rings, one can count the number of elements in both rings and they are NOT equal.

Question: Is there any other geometric way to "see" they are obviously not isomorphic to each other?

Any related comments are welcome. Thanks

The reason I want to ask is some one argued that the proof I gave above is not natural. He thought this is not a "Grothendieck style proof"

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Why not ask whether the "same" rings with $\mathbb{Z}$ replaced by $\mathbb{Q}$ (or your favourite characteristic zero field) are isomorphic? – Robin Chapman Jul 8 '10 at 6:44
The proof you gave is perefctly natural: it observed that the fibres of the two affine schemes at the closed point of characteristic $p$ are not isomorphic. :-) – Robin Chapman Jul 8 '10 at 6:54
up vote 5 down vote accepted

Does your critic dislike that the argument seems not applicable over general rings? But it is: if there's an isomorphism over some ring $R$ then we can descend to a finitely generated subring and pass to the quotient by a maximal ideal to get such an isomorphism over a finite field, and then count points.

Or maybe your critic would prefer to invoke the fact that any group variety structure on an affine space over a field $k$ is unipotent, which ${\rm{SL}}_n$ is not? Here is a Grothedieck-style proof of this fact about affine space, exactly in the same spirit as the preceding argument: if an affine space over $k$ has a non-unipotent $k$-group structure then by increasing $k$ to an algebraic closure it would (by virtue of being smooth, connected, and affine) contain a nontrivial $k$-split $k$-torus as closed $k$-subgroup. We can then once again descend this property to a subring of $k$ finitely generated over $\mathbf{Z}$ and specialize to a finite field and conclude by counting points: the size of affine space over $k$ of size $q$ is a power of $q$, whereas the nontrivial $k$-split $k$-torus subgroup forces the total number of $k$-points to be divisible by $q-1$, so we get a contradiction as long as $q \ne 2$, and we can certainly always arrange that by increasing the finite field before "counting" anyway.

(There is a more "direct" proof of this general fact about group structures on affine spaces in Springer's book on algebraic groups, but it is kind of complicated. The specialization trick sure makes it easier, at the cost of better algebro-geometric technique to work over a base that is not a field during the middle of that process.)

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Dear Shizhuo, here is a another proof that the two rings are not isomorphic.

By extending the scalars to $\mathbb C$ it suffices to prove that $SL(n, \mathbb C)$ is not isomorphic to affine space $\mathbb A^{n^2-1} _\mathbb C$. But these spaces are not homeomorphic when endowed with their classical topology. Indeed affine space has no cohomology (it is homotopic to a point), whereas the cohomology algebra of $SL(n, \mathbb C)$ is the exterior algebra on $n-1$ variables $\Lambda (e_3, e_5,...,e_{2n-1})$ if $n\geq 2$ [For $n=1$ your two rings are obviously isomorphic!]. References are given on this very site, as an answer to Evgeny Shinder's question on the cohomology of $GL_n$ and $SL_n$.

PS Personally, I much prefer your proof ! But maybe your harsh critic, Mr "some one", will accept the above as satisfying his strange criteria...

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Your argument seems to also prove that $\mathbb{R}$ and $\mathbb{R}^2$ are not isomorphic as groups (they're not homeomorphic!). And there are varieties which are conjugate but whose $\mathbb{C}$-points are not homeomorphic (I know Serre constructed a non-singular projective surface which is an example, not sure about affine examples). Can you explain how to get the reduction to the classical topology (without going through étale cohomology or something like that)? – Tom Church Jul 8 '10 at 15:39
This is all quite straightforward: if we have two affine subvarieties $X$ and $Y$ of complex affine space, and a variety isomorphism $f:X\to Y$, then $f$ is given by a bunch of polynomials so it is continuous as a map on the complex points. The same holds for $f^{-1}$, so the sets of complex points of $X$ and $Y$ are homoemorphic as topological spaces. – Robin Chapman Jul 8 '10 at 17:17
@Tom Church: I am baffled at your claim that my argument proves that $\mathbb R$ and $\mathbb R^2$ are isomorphic : it certainly does nothing of the sort . Also, Serre's example is irrelevant in this context: he starts with a scheme defined over a quadratic number field, which has two embeddings into $\mathbb C$ . Here we have schemes defined over $\mathbb Z$ and I don't even know what you mean by "conjugate" . For the rest, Robin has given a very clear explanation: thanks Robin! – Georges Elencwajg Jul 8 '10 at 19:31
The point I was missing is that an isomorphism between the original rings would give an isomorphism over $\mathbb{C}$ between $\text{SL}_n\mathbb{C}$ and $\mathbb{A}^{n^2-1}$, which would certainly imply that their $\mathbb{C}$--points are homeomorphic. My confusion was that I thought you were arguing that $\mathbb{C}[x_1,\ldots,x_{n^2-1}]$ and $\mathbb{C}[y_1,\ldots,y_{n^2}]/(\text{det}=1)$ are non-isomorphic as rings, which as far as I can tell doesn't follow from your argument, rather than as $\mathbb{C}$-algebras. (Of course this isn't what was asked.) – Tom Church Jul 8 '10 at 19:40
Tom, I'm happy that our little misunderstanding has been cleared. Cheers. – Georges Elencwajg Jul 8 '10 at 20:07

Here is my, admittedly, ad hoc way of proving they are distinct. It comes from trying to make it concrete that the graded pieces have different sizes.

First, you better assume that $n\geq 2$ as these rings are isomorphic if $n=1$.

Assume, by way of contradiction, there is an isomorphism $\varphi:S\rightarrow R$ where $S$ is the second ring (with the determinant relation) and $R$ is the first ring. Let $f_{ij}=\varphi(y_{ij})$. Thus $\det((f_{ij}))=1$.

Let $I_{0}$ be the ideal of $R$ generated by $2$ and $x_{k}^{2}$ for each $k$. Note that as the $y_{ij}$ generate $S$, we must have that $x_{1}$ occurs in one of the $f_{ij}$ (even modulo $I_{0}$) with non-zero support. Let $I_{1}$ be the ideal generated by the same relations as $I_{0}$, except we add the relation $x_{1}=f_{ij}-x_{1}$. Note that $R/I_{1}$ is isomorphic (naturally) to $\mathbb{Z}[x_2,x_3,\ldots, x_{n^{2}-1}]$.

Now, $x_{2}$ occurs with non-zero support in (a different) $f_{ij}'$ (modulo $I_{1}$) as these polynomials still generate $R/I_{1}$. Create a new ideal $I_{2}$ containing $f_{ij}'$, but for which $R/I_{2}$ looks like $\mathbb{Z}[x_3,\ldots, x_{n^{2}-1}]$.

Repeating this process enough times, we can make make the matrix $(f_{ij})$ have both determinant 1 and 0, modulo an ideal $I$, even though $R/I$ is not the zero ring. This gives you the needed contradiction.

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