### 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,