Image of an Algebraic Group Homomorphism Is Closed
Let \(f : G \to H\) be a homomorphism of linear algebraic groups over an algebraically closed field. Then \(\ker(f)\) is a closed normal subgroup of \(G\), and \(f(G)\) is a closed algebraic subgroup of \(H\).
If \(G\) and \(H\) are defined over a subfield \(F\) and \(f\) is defined over \(F\), then \(f(G)\) is a closed \(F\)-subgroup of \(H\).
Proof
The kernel is the inverse image of the closed point \(e_H\), hence is closed; normality is immediate from the homomorphism identity.
For the image, let \(\overline{f(G)}\) be its Zariski closure in \(H\). Because multiplication and inversion on \(H\) are regular, the closure of a subgroup is again a closed subgroup. The image \(f(G)\) is constructible by Chevalley's theorem and is stable under left and right multiplication by its own elements. A constructible subgroup of an algebraic group contains a nonempty open subset of its closure; translating this open subset by elements of the subgroup covers the closure. Hence \(f(G)=\overline{f(G)}\), so the image is closed.
The statement over \(F\) follows because the defining ideal of the closure is stable under the \(F\)-structure and the homomorphism is defined over \(F\).