1
$\begingroup$

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?

  • 3
    That's not the natural approach. Show that the Jacobians of $\phi$ and its inverse are inverses.2011-03-16

1 Answers 1

3

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.

  • 0
    Whoops! Thanks for ca$t$ching that.2011-03-17