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I'm reading the basics of birational geometry in Shafarevich's "Basic Algebraic Geometry, 1" third edition. In theorem 1.8 he proves that

Every irreducible affine variety $X\subseteq \mathbb{A}^n$ is birationally equivalent to some hypersurface $Y\subseteq \mathbb{A}^m$ for some $m$. (In fact, $m=n+1$)

I'm wondering whether the same statement still holds true or not if we replace "birationally equivalent" with "isomorphic". So, my question is:

Is every irreducible affine variety $X\subseteq \mathbb{A}^n$ isomorphic to some affine hypersurface $Y\subseteq \mathbb{A}^m$ for some $m$?

I don't think this is true (it seems "too good to be true"), but I don't know how can I disprove such statement. Can you provide me with an example of an affine variety which is never isomorphic to an affine hypersurface? (I can work out the details myself, I just can't think of an elementary way or example to do this). I'd be surprised if the statement is true, if so, how could I prove it? (it should be an important result in such case)

Related question: Any quasi-affine variety is isomorphic to a closed subvariety in $\mathbb{A}^n\setminus \mathbb{A}^m$ for some $m,n$.

Regards

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It is not true, because for a hypersurface $Y \subset \mathbb{A}^m$ one has $$ \dim T_y(Y) \le m = \dim(Y) + 1, $$ which is not always true for an affine variety.

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    $\begingroup$ Thanks, I am a differential geometer by training, I'm just learning the basics of classical algebraic geometry by myself. I haven't read much about dimension or algebraic tangent spaces yet but I can follow your argument. These results all remind me of Whitney's embedding theorem or Chow's lemma in complex geometry $\endgroup$ Apr 21 at 19:51

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