# Finitely generated k-Algebra

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?

• What is the codomain of $\phi$? Commented Jul 13 at 12:16
• oh, sorry, it is A, edited my post Commented Jul 13 at 12:21
• So the confusion is $\phi_A$, how it is defined, because if $k$ is not in $A$, then what happens with a constant polynomial? For example if A is the Polynomials generated by monoms, with degree >1 over $\mathbb{Q}$, then 5 is in $\mathbb{Q}[X_1,...X_n]$, but 5 is not in $A$ Commented Jul 13 at 13:47
• I deleted my comment because I realized I made a silly mistake in what I was trying to say. Christopher Nicol's answer is what I wanted to get at. The only complaint is the case of the $k$-algebra $0$ where $\phi_A$ is zero and does not induce an inclusion on $k$. Commented Jul 13 at 14:17

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 $$1$$.