Let and be two algebraic sets in . Today I will try and prove that . Not writing a full proof of this before has eaten away at my existence for way too long.

First let us assume that . This means that there exist polynomial maps from to . Hence, for any polynomial that is zero on the image of in (we shall call it ), we get a polynomial that is zero on the whole of . Hence, we get a map from . In this case, . Hence, we have a map from . Similarly, as there also exists a map from to , let the map induced from to be called .

Let us explore the isomorphism between and . Let us take . Let the maps between and be . Similarly, the map between and is . By definition, and . We can see that . This is obviously not true for all points in . If we take an arbitrary point , then we get a point of the form back such that iff . When is this possible? This is possible if and only if all the ‘s are evaluations of polynomials in at the point . The same argument works for the action of on an arbitrary point .

Now let us explore the map from to . As we can guage from the explanation above, any polynomial , when acted on by , gives us , where the ‘s belong to . Similarly, any polynomial , when acted on by , gives us , where the ‘s belong to . Thus, we don’t really have an isomorphism from to . However, we clearly have an isomorphism from ! Hence, we have proved one side of the assertion.

We shall now try and prove the converse. Let us assume that . Let and . It is then clear that , where is a polynomial in . Similarly, , where is a polynomial in . Hence, using the maps and , we can construct an isomorphism between and .

Now both sides have been proven,

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