Proving that an isomorphism between varieties implies an isomorphism between their coordinate rings, and vice-versa

Let X and Y be two algebraic sets in \Bbb{A}^n. Today I will try and prove that X\cong Y\iff \Bbb{C}[x_1,x_2,\dots,x_n]/I(X)\cong \Bbb{C}[x_1,x_2,\dots,x_n]/I(Y). Not writing a full proof of this before has eaten away at my existence for way too long.

First let us assume that X\cong Y. This means that there exist polynomial maps f_1(x_1,x_2,\dots,x_n),f_2(x_1,x_2,\dots,x_n),\dots,f_n(x_1,x_2,\dots,x_n) from X to Y. Hence, for any polynomial that is zero on the image of X in Y (we shall call it f(X)), we get a polynomial that is zero on the whole of X. Hence, we get a map from I(f(X))\to I(X). In this case, f(X)=Y. Hence, we have a map \phi from I(Y)\to I(X). Similarly, as there also exists a map from Y to X, let the map induced from I(X) to I(Y) be called \psi.

Let us explore the isomorphism between X and Y. Let us take (a_1,a_2,\dots,a_n)\in X. Let the maps between X and Y be f:X\to Y. Similarly, the map between Y and X is f^{-1}:Y\to X. By definition, f\circ f^{-1}=id_Y and f^{-1}\circ f=id_X. We can see that f^{-1}\circ f ((a_1,a_2,\dots,a_n))=(a_1,a_2,\dots,a_n). This is obviously not true for all points in \Bbb{A}^n. If we take an arbitrary point (y_1,y_2,\dots,y_n), then we get a point of the form (y_1+c_1,y_2+c_2,\dots,y_n+c_n) back such that c_1,c_2,\dots,c_n=0 iff (y_1,y_2,\dots,y_n)\in X. When is this possible? This is possible if and only if all the c_i‘s are evaluations of polynomials in I(X) at the point (y_1,y_2,\dots,y_n). The same argument works for the action of f\circ f^{-1} on an arbitrary point (z_1,z_2,\dots,z_n).

Now let us explore the map from I(Y) to I(X). As we can guage from the explanation above, any polynomial f(y_1,y_2,\dots,y_n)\in I(Y), when acted on by \psi\circ\phi, gives us f(y_1+d_1,y_2+d_2,\dots,y_n+d_n), where the d_i‘s belong to I(Y). Similarly, any polynomial g(x_1,x_2,\dots,x_n)\in I(X), when acted on by \phi\circ\psi, gives us g(x_1+c_1,x_2+c_2,\dots,x_n+c_n), where the c_i‘s belong to I(X). Thus, we don’t really have an isomorphism from I(X) to I(Y). However, we clearly have an isomorphism from \Bbb{C}[x_1,x_2,\dots,x_n]/I(X)\cong \Bbb{C}[x_1,x_2,\dots,x_n]/I(Y)! Hence, we have proved one side of the assertion.

We shall now try and prove the converse. Let us assume that \Bbb{C}[x_1,x_2,\dots,x_n]/I(X)\cong \Bbb{C}[x_1,x_2,\dots,x_n]/I(Y). Let f:\Bbb{C}[x_1,x_2,\dots,x_n]/I(X)\to \Bbb{C}[x_1,x_2,\dots,x_n]/I(Y) and g:\Bbb{C}[x_1,x_2,\dots,x_n]/I(Y)\to \Bbb{C}[x_1,x_2,\dots,x_n]/I(X). It is then clear that (g\circ f)(x_i)=x_i+p_i(x_1,\dots,x_n), where p_i(x_1,\dots,x_n) is a polynomial in I(X). Similarly, (f\circ g)(x_i)=x_i+p'_i(x_1,\dots,x_n), where p'_i(x_1,\dots,x_n) is a polynomial in I(Y). Hence, using the maps f and g, we can construct an isomorphism between X and Y.

Now both sides have been proven,

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Graduate student

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