Finitely generated k-Algebra

Question

If A is a communative, associative k-Algebra, it is finitely generated as an Algebra, if there exists, $a_1,...,a_n$, so that the morphism: $$ \phi_A: K[X_1,...,X_n] \mapsto A\\ f \to f(a_1,...,a_n) $$ is surjectiv.
A k-Algebra is a vector space, with a map $$ A \times A \mapsto A $$ so that for, $ x,y,z \in A$ and $a,b \in k$ $$ (x+y)*z=xz+yz\\ x*(y+z)=xy+zy\\ (ax)*(by)=(ax)(by) $$ So what i don't unterstand is that by definition k does not have to be in A(?). But that would mean the morphismen $\phi$ would not work. So what am I missing?

Answer 1

Usually an algebra is a ring with unit. In this situation you have an element $1 \in A$. And your canonical morphism from $k$ to $A$ is

$ \alpha \longrightarrow \alpha \cdot 1. $

So the axioms you used to define a $k$-algebra imply we haceva ring morphism. As $k$ is a field and our morphism is non $0$, it is injective. So we have a canonic inclusion of $k$ in $A$, and now it is straightforward that $\phi_A$ maps constant polynoms to the image of $k$ by my morphism.

However if you don't have a unit in your ring, you may not be able to find a canonical copy of $k$ in $A$. For instance assume $A=k^n$ with the multiplication being $0$. In this context you won't be able to find a canonical image of $k$ in $A$. But this pathological cas is not important in your context. Indeed we can take the existence of $\phi_A$ as a definition of finitely generated algebra. So $A$ is a quotient of a ring with unit, thus $A$ has a $1$.