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Let $k$ be an algebraically closed field, and let $\mathbb{A}^n(k)$ denote the linear $n$-dimensional vector space over $k$. Suppose $X \subset \mathbb{A}^n$, $X' \subset \mathbb{A}^m$ are two varieties. We say that $X, X'$ are isomorphic if there exists a polynomial map $\phi : X \rightarrow X'$ which is bijective and whose inverse is a polynomial map. Note that $\phi$ is polynomial if $\phi(x) = (f_1(x), \cdots, f_m(x))$ where each $f_j$ is a polynomial. I want to show that if $p \in X$ is smooth (the tangent space of $X$ at $p$ has the same dimension as $X$) if and only if $\phi(p)$ is a smooth point of $X'$. It is not clear to me why if $(x_1, \cdots, x_n)$ is an eigenvector of the Jacobian of the generators of $I(X)$ at $p$, that $\phi((x_1, \cdots, x_n))$ must be an eigenvector of the Jacobian of the generators of $I(X')$, which I presume to be the natural approach to show that the two tangent spaces have equal dimension.

Any suggestions?

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That's not the natural approach. Show that the Jacobians of $\phi$ and its inverse are inverses. –  Qiaochu Yuan Mar 16 '11 at 21:13
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A more intrinsic way to express smoothness is via the (sheaf of) ring(s) of functions: in your case, $X$ is smooth at $x$ iff the local ring at $x$ is regular. The isomorphism between $X$ and $X'$ gives an isomorphism between (sheaves of) rings of functions.

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It is, incidentally, worth noting that when the field is not algebraically closed, regularity and smoothness become different (in characteristic $p$); smoothness (a property of morphisms of schemes) is the nicer notion. –  Akhil Mathew Mar 17 '11 at 1:37
    
I think you mean not algebraically closed (and not perfect). –  Plop Mar 17 '11 at 1:46
    
Whoops! Thanks for catching that. –  Akhil Mathew Mar 17 '11 at 3:10
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