Example of non-finitely generated $R$-algebra 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.
 A: 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.
A: 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.
