Intuition behind isomorphism of algebraic varieties

Solution 1:

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.)

Solution 2:

Let $X=\mathbf{V}(z-xy)=\{(x,y,z)\in \Bbb{R}^3\mid z-xy=0\}\subseteq \Bbb{R}^3, W=\mathbf{V}(\{0\})=\Bbb{R}^2\subseteq \Bbb{R}^2$. $\phi:X\to W, \phi(x,y,z)=(x,y), \phi^{-1}(x,y)=(x,y,xy)$.

Let's see another example. Note that $X=\mathbf{V}(y^5-x^2)=\{(x,y)\in \Bbb{R}^2\mid y^5-x^2=0\}\subseteq \Bbb{R}^2$ is not isomorphic to $W=\mathbf{V}(\{0\})=\Bbb{R}\subseteq \Bbb{R}$ because $\phi:X\to W, \phi(x,y)=(x), \phi^{-1}(x)=(x,x^{5/2})$ and $x^{5/2}$ is not a polynomial.

These two example are from Cox's Ideals, Varieties, and Algorithms. (Example 7 and last example in Chapter 5, Section 4.)