A ring homomorphism from $k[G]$ to $k$ is an evaluation map?

Question

Given an affine algebraic group $G$ over an algebraically closed field $k$, I was wondering is a ring homomorphism from $k[G]$ to $k$ always an evaluation map (evaluated at some unique $g \in G$)? Any comments would be appreciated. Thank you.

Answer 1

An affine algebraic group $G$ is in particular an affine variety, therefore we know that $k[G]\cong k[X_1,\ldots,X_n]/\langle f_1,\ldots,f_m\rangle$ for some $n,m\in\mathbb{N}$ and some $f_1,\ldots,f_m\in k[X_1,\ldots,X_n]$. Set $I_G:=\langle f_1,\ldots,f_m\rangle$ and $x_i:=X_i+I_G$. Now, in light of the universal property of the ring of polynomials and in light of the universal property of the quotient, any $k$-algebra map $\phi:k[G]\to k$ is the factorization through the quotient of a $k$-algebra map $\Phi:k[X_1,\ldots,X_n]\to k$ which is uniquely determined by the $n$-tuple $(\phi(x_1),\ldots,\phi(x_n))=(\Phi(X_1),\ldots,\Phi(X_n))$.

On the one hand, for every $f_i$ we have $$f_i(\phi(x_1),\ldots,\phi(x_n))=f_i(\Phi(X_1),\ldots,\Phi(X_n))=\Phi(f_i(X_1,\ldots,X_n)) =0$$ and so $(\phi(x_1),\ldots,\phi(x_n))$ is a point of $G$. On the other hand, it is clear now that (up to the first isomorphism we wrote) $$\phi(\gamma) = \Phi(p(X_1,\ldots,X_n))=p(\Phi(X_1),\ldots,\Phi(X_n))=p(\phi(x_1),\ldots,\phi(x_n))=\gamma(\phi(x_1),\ldots,\phi(x_n))$$ is the evaluation at the point $(\phi(x_1),\ldots,\phi(x_n))\in G$, where $p(X_1,\ldots,X_n)$ is the polynomial function representing $\gamma\in k[G]$.

Notice moreover that I never used that $G$ is a group. This is true for every affine algebraic variety.