$f:X\rightarrow Y$ is a polynomial map between two algebraic varieties. $X$ and $Y$. $f_{*}:k[Y]\rightarrow k[X]$ is the corresponding algebraic homomorphism between the coordinate ring. Assume $f_{*}$ is injective and $k[X]$ is integral over $k[Y]$. I want to show that if $Z\subset X$ is Zariski closed, then $f(Z)$ is Zariski Closed.
My approach: Since $X$,$Y$,$Z$ are closed, there are ideals $I_{X}$,$I_{Y}$ and $I_{Z}$ correspond to these sets, with $I_{X}\subset I_{Z}$. Since $f_{*}$ is injective, $ker(f_{*})=0$ implies $f_{*}^{-1}(I_{X})=I_{Y}$. Then $I_{Y}\subset f_{*}^{-1}(I_{Z}):=I$. Now it is routine to check that $I$ is a ideal, so $f(Z)$ is closed.
Is this proof valid? I feel like $f_{*}^{-1}$ is not well defined.If the proof is not valid, can anyone provide a proof?
Your proof makes no reference to the fact that $ f : X \to Y $ is a finite map, i.e. that $ k[X] $ is integral over $ k[Y] $. This condition is essential; for instance, the projection of $ V((XY - 1)) \subset \mathbb A^2 $ down to its first coordinate is the classic example of a map under which the image of Zariski closed subsets is not Zariski closed. This corresponds to the ring extension of $ k[X, X^{-1}] $ over $ k[X] $, which of course is not integral.
The problem is that you can't guarantee the preimage of $ I(Z) $ under the map $ f^* : k[Y] \to k[X] $, in other words $ I(f(Z)) $, has the property that $ V(I(f(Z)) = f(Z) $. In general, this is only equal to the closure of $ f(Z) $, and so this property is equivalent to saying that $ f(Z) $ is closed in the first place.
The property that's vital to prove this statement is the going up property of integral extensions. Suppose $ V(I(f(Z)) $ strictly contains $ f(Z) $, i.e. there is another point at which all elements of $ V(I(f(Z)) $ vanish; and suppose without loss of generality that $ Z $ is irreducible, and thus so is $ f(Z) $. In this case, this point corresponds to a maximal ideal $ \mathfrak m $ containing $ I(f(Z)) $, and by the going-up property the chain $ I(f(Z)) \subset \mathfrak m $ lifts to a chain $ I(Z) \subset \mathfrak m' $ in $ k[X] $ for some maximal ideal $ \mathfrak m' $. Then, $ \mathfrak m' $ corresponds to a point in $ Z $, and this is a contradiction since it lifts a point not in $ f(Z) $ under the extension $ f^* : k[Y] \to k[X] $.