I just found out about finitely generated $k$-algebras (where $k$ is a field). So it is an algebra $A$ for which we have a finite set of elements $(a_1,...,a_n)$ such that every element in $A$ can be expressed as $p(a_1,...,a_n)$ where $p$ is a polynomial $p \in k[x_1,...,x_n]$. This is pretty abstract for the moment so I am trying to illustrate with some examples. So I understand $k[x_1,...,x_n]$ itself is an example where the generators are $x_1,...,x_n$. It seems that in this particular case, the generators are even algebraically independent. What would be a nice example where the set of generators are not algebraically independent?
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1$\begingroup$ Any quotient of your example; a simple example would be $k[x,y]/(xy)$. $\endgroup$– ServaesCommented Jan 27, 2019 at 0:16
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1$\begingroup$ Maybe this is to concrete for you, but how about $\mathbb{Q}[\sqrt{2}, \pi]$? The generators $\sqrt{2}, \pi$ are not algebraically independent, because $$ (\sqrt{2})^2 + 0 \cdot \pi - 2 =0.$$ $\endgroup$– Severin SchravenCommented Jan 27, 2019 at 0:20
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$\begingroup$ @SeverinSchraven The $\pi$ in that example is redundant, why include it? $\endgroup$– jgonCommented Apr 19, 2019 at 0:17
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$\begingroup$ @jgon Because I like it. Also because I think it is better from a pedagocial point of view to to include two elements when talking about algebraic dependence. Feel free to post an example which is more to your liking. $\endgroup$– Severin SchravenCommented Apr 19, 2019 at 1:34
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Consider $A = k[x,y]/(y-x^2)$. This is a finitely generated $k$-algebra where the generators, i.e. the images of $(x,y)$ in the quotient, are not algebraically independent. Can you see why not?
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3$\begingroup$ But the quotient is basically just $k[x] $. Better perhaps to use $y^3-x^2$ so it isn't trivial. $\endgroup$ Commented Jan 27, 2019 at 1:19
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$\begingroup$ @Mr.chip Yes, thank you. The generators are $\bar{x}$ and $\bar{y}$ but there is an algebraic relation $\bar{y}=\bar{x}^2$ so they are not algebraically independent. So basically every finitely generated $k$-Algebra is isomorphic to a quotient of $k[x_1,...,x_n]$ if $A$ has generator $a_1,...,a_n$? $\endgroup$ Commented Jan 27, 2019 at 12:23
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$\begingroup$ Also I think that the definition of finitely generated algebras looks suspiciously different to the definition of finitely generated modules... $\endgroup$ Commented Jan 27, 2019 at 12:29