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By definition, an $R$-algebra is a ring homomorphism $f: R \to S$. For example, if $R=\mathbb Z$ and $S= \mathbb Z / n \mathbb Z$ then the projection $k \mapsto k \mod n$ is a ring homomorphism so that $\mathbb Z / n \mathbb Z$ is a $\mathbb Z$-algebra. I think the point of an algebra is that it's a bit like a module in that we extend its structure by adding a ring that is acting on it. In the case of modules, we start with an abelian group and in the case of algebras we start with a ring.

Now for my question: I've been trying to come up with a non-finitely generated $R$-algebra but couldn't. Can someone help me and give me an example? Thank you.

Rudy the Reindeer
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    Well, you could look at stuff like $k[x_1, x_2, \ldots]$ over a field $k$, or $\mathbf Q$ viewed as a $\mathbf Z$-algebra. – Dylan Moreland Jul 18 '12 at 09:40
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    @DylanMoreland By $k[x_1, x_2 \dots]$ you denote the ring of "polynomials" (?) in infinitely many variables? Does it have a name? (Thanks, why didn't I think of $Q$ over $Z$! : )) – Rudy the Reindeer Jul 18 '12 at 09:45
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    @DylanMoreland the example with $Q$ isn't finite over $Z$ because I have to have $\frac1p$ for each prime $p$ in the generating set, right? (or perhaps a variation over it since I could have $\frac{1}{pq}$ and $\frac1p$ in it to get $\frac1q$. – Rudy the Reindeer Jul 18 '12 at 09:53
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    @ClarkKent Yes, essentially. Another way to do it is that for some finite set of generators, there is a largest prime occurring in the denominators of the generators, say $p_i$. No larger primes $p_j$ can ever have their reciprocals in the submodule generated by this set. – KReiser Jul 18 '12 at 10:00
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    @KReiser Thank you very much. That's exactly the argument I was looking for but couldn't put it into words. – Rudy the Reindeer Jul 18 '12 at 10:49

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You may find the following example illustrative of the questions involved. If $K$ is a field, then the polynomial ring $K[X]$, although it is a (countably) infinite dimensional vector space over $K$, is finitely generated (by $X$ alone) as a $K$-algebra. However the field $K(X)$ of rational functions (quotients of polynomials) is not finitely generated as a $K$-algebra, since any finite set of generators can only produce finitely many irreducible factors in the denominators. The argument is similar to $\mathbf Q$ being non finitely generated as a $\mathbf Z$-algebra (see the comment by Dylan Moreland), because of the inifiniteness of the set of prime factors one needs for denominators.

One can also prove (can you do it?) that the ring $K[[X]]$ of formal power series is not finitely generated as a $K$-algebra. Nor is $\mathbf R$ finitely generated as $\mathbf Q$-algebra; in these examples the sheer (uncountable) dimension as a vector space already excludes finite generation.

Marc van Leeuwen
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    Great, thank you. I think one would prove it like this: assume $p_1(x), \dots, p_n(x)$ was a generating set for $K[[X]]$. Then we cannot express polynomials of degree larger than $\max p_i$ and in particular not infinite formal power series. Hence it cannot be a generating set. – Rudy the Reindeer Jul 18 '12 at 10:51
  • For $R$ over $Q$ I guess one would argue that $Q[\pi]$ is a transcendental extension of $Q$ that properly contains $R$ hence the degree or $R$ must certainly be larger than that of $Q[\pi]$ which is already infinite. – Rudy the Reindeer Jul 18 '12 at 10:52
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    @ClarkKent: No, I don't think you've got it. In generaing an algebra, multiplication is allowed, so polynomials can generate others of unbounded degree. And in addition you cannot assume a generating set for $K[[X]]$ to consist of polynomials in the first place; they could be any power series. For the second comment, the transcendence of $\pi$ makes $\mathbf Q[\pi]$ infinite dimensional, and isomorphic to $\mathbf Q[X]$, but like the latter it is finitely (indeed singly) generated. But the subfield $\mathbf Q(\pi)$ is indeed not finitely generated over $\mathbf Q$, just like $\mathbf Q(X)$. – Marc van Leeuwen Jul 18 '12 at 12:07
  • Dear Marc. Then how about this: $\frac{1}{(1 - X)} = \sum_{n=0}^\infty X^n$ is in $K[[X]]$, and hence all things of the form $\frac{1}{(k - X)}$ are in $K[[X]]$. So to generate $K[[X]]$ we need to at least be able to represent $\frac{1}{(k - X)}$ for each $k$ in $K$. So if we pick a finite generating set $S$ and then a product $\prod_{k_i \in K, i \in I} \frac{1}{(k_i - X)}$, $k_i$ and $k_j$ pairwise different and $|S| < |I|$ then we cannot represent this product using elements in $S$. – Rudy the Reindeer Jul 18 '12 at 13:26
  • Although then if $K$ is finite my idea is broken so it must be wrong still. – Rudy the Reindeer Jul 18 '12 at 13:27
  • Reading this made me realise that I'm confusing finitely generated algebra with finite algebra. – Rudy the Reindeer Jul 18 '12 at 13:43
  • Ok, I think I understand the difference: If $S$ is a finite $R$-algebra it means that every element in $S$ can be written as $s = \sum_{n=0}^N f(r_n) s_n$ that is, $S=f(R)s_1 + \dots + f(R) s_N$. On the other hand, $S$ is finitely generated if $s = \sum_{n=0}^N f(r_n) s_n^n$ for $s_1, \dots s_K$ some generating set ($N$ here may vary). – Rudy the Reindeer Jul 18 '12 at 13:53
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Take $R$ and $S$ from your example above.

Let $S_0 = S \oplus S \oplus \cdots$, then $f:R\longrightarrow S_0$.

$S_1$ is not finitely generated. To prove this you could consider $$S_1=S\oplus 0 \oplus \cdots \subset S_0$$ $$S_2=S\oplus S\oplus0\oplus\cdots\subset S_0$$ $$...$$ and show that $S_n$ requires $n$ generators.

Zhen Lin
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hayd
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