In my lecture notes for algebraic geometry, an algebra over a field $k$ is defined as a (unital and commutative) ring, together with a ring homomorphism $\lambda:k\to R$ (such a homomorphism preserves $1$). Using $\lambda$, we can define a scalar multiplication $k\times R\to R$ by $c\cdot f=\lambda(c)f$.
On Wikipedia, a finitely generated $k$-algebra $R$ is defined as an algebra over $k$ for which there exist $a_1,\dots,a_n \in R$ such that the evaluation homomorphism $\phi_a:k[x_1,\dots,x_n]\to R$ is surjective. However, in my lecture notes, it defines a finitely generated $k$-algebra as an algebra over $k$ for which there exists a surjective homomorphism $\psi:k[x_1,\dots,x_n]\to R$ (there is no requirement that such a homomorphism is an evaluation homomorphism). Are these two definitions equivalent?
Best Answer
Sure. By the universal property of the polynomial algebra, any $k$-algebra homomorphism $k[x_1,\dotsc,x_n] \to R$ is the evaluation homomorphism for some elements $a_1,\dotsc,a_n \in R$, namely the images of the variables.
Albeit, the conceptually correct definition is even different. If $R$ is any sort of algebraic structure (group, abelian group, monoid, algebra, ring, Lie algebra, lattice, ...), it is called finitely generated if there are finitely many elements that generate $R$, and this means that $R$ is the smallest subalgebra containing these elements. You don't need free algebraic structures to formulate (and prove) that. For example, you want to know what finitely generated groups are without necessarily using free groups. Also, in the case of finitely generated field extensions: there is no free field extension.