How do you distinguish between a finitely generated $k$-algebra and a finitely generated module over $k$? Suppose $k$ is a field. Consider the polynomial ring $k[x]$. It was mentioned that this is finitely generated as a $k$-algebra "because it is just finitely generated by $x$." However, $k[x]$ is not finitely generated as a $k$-module "because you need an infinite number of powers of $x$ to span it as a module."
I'm having problems understanding what's written above, and would appreciate some help understanding the distinction.
 A: Let $R$ be a commutative ring.
In an $R$-module, you generate new elements from old ones by $R$-linear combinations.  In an $R$-algebra, you generate new elements from old ones by $R$-linear combinations of products*, or equivalently by polynomials with coefficients in $R$.
For example, the $R$-submodule generated by elements $x_1, \ldots, x_k$ of an $R$-module is $\sum_{i=1}^k Rx_k$. That is the smallest $R$-submodule containing all $x_i$. The $R$-subalgebra generated by elements $x_1, \ldots, x_k$ of an $R$-algebra is $R[x_1,\ldots,x_k]$. That is the smallest $R$-subalgebra containing all $x_i$.  As a special case, for a single element $x$ of an $R$-module, the $R$-submodule it generates is $Rx$, while for a single element $x$ of an $R$-algebra, the $R$-subalgebra it generates is $R[x]$.  Do you see the difference?
A: Suppose that $M$ is a module over a ring $R$. We say that $M$ is finitely-generated as an $R$-module if there exist finitely many elements $m_1, m_2, ..., m_r\in M$ such that every $m\in M$ can be expressed as
$$
m = c_1 m_1 + c_2 m_2 + \cdots + c_r m_r
$$
with $c_i\in R$ (these coefficients, of course, depend on $m$).
In your example, $M=k[x]$ and $R=k$. If $k[x]$ were finitely generated as $k$-module, this would mean that there exist finitely many polynomials $f_1, ..., f_r\in k[x]$ such that every polynomial $f\in k[x]$ can be written as a linear combination of $f_1, ..., f_r$ with coefficients in k. But this is impossible since we can choose any $f$ with $\deg(f) > \max\{\deg(f_1), \deg(f_2), ..., \deg(f_r)\}$, for which such a combination is impossible.
On the other hand, suppose that $M$ is also an $R$-algebra (which just means that $M$ is also a ring containing $R$ as a subring). We say that $M$ is finitely-generated as an $R$-algebra if there exist finitely many $m_1, ..., m_{r}\in M$ such that every element $m\in M$ can be expressed as a polynomial in $m_1, ..., m_r$ with coefficients in $R$, i.e.
$$
m = \sum_{I} c_{I} m_{1}^{i_1} m_{2}^{i_2} \cdots m_{r}^{i_r}
$$
where the sum runs over all multi-indices $I=(i_1, ..., i_r)$ and $c_I\in R$. It is clear that being generated as an $R$-algebra is easier to hold than being generated as an $R$-module, because we allow taking arbitrary polynomials in $m_i$ (with coefficients in $R$) instead of just linear forms as we saw in the previous definition.
Now, when $M=k[x]$ and $R=k$, then by definition, every element of $M$ is a polynomial in the generator $x$, so $M=k[x]$ is finitely generated as a $k$-algebra with a single generator.
