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,