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Let $X \subset \mathbb A^n$, $W \subset \Bbb A^m$ be two algebraic sets. A function $\phi:X \rightarrow W$ is a morphism if there exist $m$ polynomial functions $f_1,\ldots,f_m \in K[X]$ such that for every $x \in X$ one has $ \phi(x)=(f_1(x),\ldots,f_m(x)) \in W$.

We say two algebraic sets are isomorphic if there exists a bijective morphism between them whose inverse is also a morphism. Now, now my question is what does ismorphism of two algebraic sets mean intuitively? The general meaning of isomorphism is that they are "equivalent".So, in what geometrical sense two isomorphic algebraic sets can be considered "equivalent"? In other words.what geometrical properties are same for two isomorphic algebraic sets? Also, can a Zariski open set in one Zariski topology be isomorphic to a Zariski closed in another Zariski topology?

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Say $V$ and $W$ are isomorphic algebraic sets. Intuitively, it should mean that the set of equations defining $V$ can be transformed into the set of equations defining $W$. For instance, consider two curves $f(x,y) =0$ and $g(x,y)=0$ in $\mathbf{A}^2$. These being isomorphic means that there is an isomorphism $k[x,y]/(f)\to k[x,y]/(g)$. –  Harry Dec 22 '12 at 16:02
    
To answer your last question, observe that $\mathbb{A}^1$ is open in itself, but is isomorphic to quite a few Zariski-closed subsets of $\mathbb{A}^2$ -- for example, the $x$-axis. –  Paul VanKoughnett Dec 22 '12 at 16:49
    
are there any other examples where one set is just open but not closed? –  Mohan Dec 22 '12 at 16:52
    
Yes, I've edited it now. –  Mohan Dec 22 '12 at 19:14
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up vote 4 down vote accepted

The final part of the question, as amplified in Mohan's comment, has an affirmative answer. $\mathbb A^1-\{0\}$ is open and not closed in $\mathbb A^1$ but its image, under the isomorphic embedding $x\mapsto(x,0)$ is closed and not open in $\mathbb A^2-\{(0,0)\}$.

As for the earlier parts of the question, isomorphism means that the two algebraic sets are not merely in one-to-one correspondence any old way but rather in a way that respects the algebraic structure. By "the algebraic structure" here, I mean the information telling which functions on the algebraic set are polynomials. The definition of isomorphism ensures that, if you have a polynomial function on one of the two isomorphic algebraic sets, then transporting that function to the other set, by composing with the bijection $\phi$ or its inverse, produces a polynomial function there. (Both my explanation and the definition in the question seem to need some caution in finite characteristic, where a polynomial isn't determined by its values.)

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so aren't there any geometrical properties which are invariant under isomorphism? –  Mohan Dec 22 '12 at 17:33
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Yes: all of them! –  Georges Elencwajg Dec 22 '12 at 19:00
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To amplify Georges's comment (which I upvoted): Properties like "open" and "closed" are not properties of an algebraic set but rather properties of a pair, consisting of an ambient algebraic set and an algebraic subset. They would be preserved by isomorphisms of such pairs. Isomorphisms of (single, not paired) algebraic sets preserve the properties of those single sets. –  Andreas Blass Dec 22 '12 at 22:47
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