Exercise A0.V.1.03

Proposition

Let $k$ be a field, with $f \in k[x]$ a non-constant polynomial. Let $R$ be a subring of $k[x]$ such that $f \in R$ and $k \subseteq R.$ Define the map $\varphi : k[t] \to R$ to be constant on elements of $k$ and send $t$ to $f.$ This maps gives every such $R$ a $k[t]$-algebra structure. Then

1. $k[x]$ is finitely generated as a $k[t]$-module
2. every subring $R$ (as above) is finitely generated as a $k[t]$-module
3. every subring of $k[x]$ containing $k$ is a Noetherian ring.

Proof

1.

Take the set $S = \{ x^n \mid n \in \mathbb{N}, n < \deg(f) \}.$ Any other power of $x$ can be written by taking a sum of elements of $S$ scaled by an appropriate elements of $k[t]$ using the following procedure.

For each $n \in \mathbb{N}$ we can write it down as $m_1\deg(f) + m_2$ with $m_1,m_2 \in \mathbb{N}$ and $m_2 < \operatorname{deg}(f).$ Define $g_n$ as $\varphi(t^{m_1}/c^{m_1})x^{m_2}$ with $c$ being the leading coefficient of $f.$ It is clear that $g_n$ is a monic polynomial of degree $n.$ Moreover, since $x^{m_2} \in S,$ this doesn’t go beyond our generating set.

Let $n$ be our desired power of $x.$ Let $s_j$ denote our current approximation of $x^n,$ with $s_0 = g_n.$ We can find a better approximation $s_{j+1}$ as follows.

Suppose $u_j \in k$ is the leading coefficient of $s_j - x^n$ and let $d_j = \deg(s_j - x^n).$ If $u = 0,$ then $s_j = x^n$ and we’re done. Otherwise, take $s_{j+1} = s_j - u_jg_{d_j}.$

Note that each step of this procedure strictly decreases the degree of the error $s_j - x_n.$ Since a polynomial has a finite degree, this procedure terminates. This gives us

$$x^n = g_n - \sum_{j=1}u_jg_{d_j},$$

which expands to an $k[t]$-linear combination of elements of $S.$ Any element of $k[x]$ can, in turn, be produced by taking $k$-linear combinations of the powers of $x.$

2.

Since $k$ is a field. it is Noetherian. By Hilbert’s basis theorem $k[t]$ is Noetherian. Any subring $R$ of $k[x]$ containing both $f$ and all $k$-scalars is a $k[t]$-submodule of $k[x]$ with $k[t]$ acting via $\varphi.$ Since $k[x]$ is a finitely generated module over a Noetherian ring, it is itself Noetherian (by III.6.8). Hence, $R$ is finitely generated.

3.

Let $R$ be a subring of $k[x]$ containing $k.$ Suppose $R$ contains no non-constant polynomials. Then $R$ is isomorphic to $k$ and hence Noetherian. Otherwise, it must contain at least one non-constant polynomial $f.$ In such case, the arguments above show that $R$ is finitely generated as a $k[t]$-module.

Take an ideal $I$ of $R.$ As an $k[t]$-module, $I$ is also finitely generated. But note that $k[t]$ acts on $R$ by embedding $k[t]$-scalars in $R$ via $\varphi.$ We can just forget $k[t]$ and use $R$-scalars directly. Hence, $I$ is finitely generated as an $R$-module.

Thus, $R$ is Noetherian.

$\blacksquare$