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This is Wikipedia's definition of a finitely generated algebra:

A finitely generated algebra (also called an algebra of finite type) is an associative algebra $A$ over a field $K$ where there exists a finite set of elements $a_1,…,a_n$ of $A$ such that every element of $A$ can be expressed as a polynomial in $a_1,…,a_n$, with coefficients in $K$.

My question is, shouldn't the last part "with coefficients in $K$" read "with coefficients in $\alpha(K)$" where $\alpha : K \rightarrow A$ is the homomorphism that makes $A$ a $K$-algebra?

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    It is customary to assume the product is directly by the scalars of $;K;$ , just as with linear spaces (and an algebra over $;K;$ is a linear space over $;K;$) . If you want to be very formal and the definition mentions that homomorphism, then I guess you could do what you mention at the end....But observe that many define "algebra over a field" as a linear space over that field that is also a ring. This level of formality is usually more than enough. – DonAntonio Jun 23 '19 at 11:54

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It depends on what is meant. Both are actually valid, in the following sense : if you're looking completely internally to $A$, then it makes more sense to say $\alpha (K)$.

However, if you have a more global view on all $K$-algebras, then you see that for any $K$-algebra $A$ and any polynomial $P\in K[X_1,...,X_n]$, there is an associated map $A^n\to A$, and you can say that $(a_1,...,a_n)$ generate $A$ if for any $a\in A$ there exists $P\in K[X_1,...,X_n]$ (so a polynomial with coefficients in $K$) such that the image of $(a_1,...,a_n)$ under the associated map is $a$.

The thing is that this associated map is in some sense "independent" of the algebra $A$, and it completely determines $P$ if you allow yourself to look at all $K$-algebras, so it makes sense to say that it is a polynomial with coefficients in $K$.

(PS : I'll add my personal opinion here, but I think that's not the best definition of finitely generated algebra; to me the subalgebra of $A$ generated by a subset $S$ should be the intersection of all subalgebras of $A$ containing $S$, so literally "the smallest subalgebra containing $S$", and then the description with polynomials should be a theorem/characterization, not the definition)

Maxime Ramzi
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