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Let $A$ be a ring. I'm a bit confused with definition of $A$ algebras and I would appreciate any clarification. Let $A \subset B$ rings and suppose $B$ is a finitely generated $A$ algebra. Does this mean there exist $x_1, .., x_n \in B$ for some $n$ such that $B = A[x_1, .., x_n]$ ?

This is not clear to me looking at the definition in Wikipedia https://en.wikipedia.org/wiki/Finitely_generated_algebra for example.

Johnny T.
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No. It only means that there exists $a_1, ... a_n \in B$ such that for every $x \in B$, there is a polynomial $f \in A[x_1, ... x_n]$ such that $f(a_1, ... a_n) = x$.

In other words, the "evaluation at $a_1, ... a_n$" homomorphism $A[x_1, ... x_n] \rightarrow B$ is surjective. By the first isomorphism theorem, this implies $B \cong A[x_1, ... x_n] / I$ for some ideal $I$ of $A[x_1, ... x_n]$. This ideal is the kernel of the evaluation homomorphism.

David Lui
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    I didn't know there was commutative in the definition of finitely generated $A$-algebra. Thus the quaternions are a finitely generated $\Bbb{R}$-module but not a finitely generated $\Bbb{R}$-algebra, which is weird. – reuns Nov 16 '20 at 15:39