Hilbert's Basis Theorem

Theorem
Let $A$ be a Noetherian ring.

Let $A \left[{x}\right]$ be the ring of polynomial forms over $A$ in the single indeterminate $x$.

Then $A \left[{x}\right]$ is also a Noetherian ring.

Proof
From the definition, a Noetherian ring is also a commutative ring with unity.

Let $f = a_n x^n + \cdots + a_1 x + a_0 \in A \left[{x}\right]$ be a polynomial over $x$.

For a polynomial $f = a_n x^n + \cdots + a_1 x + a_0 \in A \left[{x}\right]$, we call $a_n$ the leading coefficient of $f$.

Let $I \subseteq A \left[{x}\right]$ be an ideal of $A \left[{x}\right]$.

We will show that $I$ is finitely generated.

Let $f_1$ be an element of least degree in $I$, and let $\left({g_1, \ldots, g_r}\right)$ denote the generated by the polynomials $g_1, \ldots, g_r$.

For $i \ge 1$, if $\left({f_1, \ldots, f_i}\right) \ne I$, then choose $f_{i+1}$ to be an element of minimal degree in $I \backslash \left({f_1, \ldots, f_i}\right)$.

If $\left({f_1, \ldots, f_i}\right) = I$ then stop choosing elements.

Let $a_j$ be the leading coefficient of $f_j$.

Since $A$ is Noetherian, the ideal $\left({a_1, a_2, \ldots}\right) \subseteq A$ is generated by $a_1, a_2, \ldots, a_m$ for some $m \in \N$.

We claim that $f_1, f_2, \ldots, f_m$ generate $I$.

Suppose not. Then our process chose an element $f_{m+1}$, and $\displaystyle a_{m+1} = \sum_{j=1}^m u_j a_j$ for some $u_j \in A$.

Since the degree of $f_{m+1}$ is greater than or equal to the degree of $f_j$ for $j = 1, \ldots, m$, the polynomial:


 * $\displaystyle g = \sum_{j=1}^m u_j f_j x^{\deg f_{m+1} - \deg f_j} \in \left({f_1, \ldots, f_m}\right)$

has the same leading coefficient and degree as $f_{m+1}$.

The difference $f_{m+1} - g$ is not in $\left({f_1, \ldots, f_m}\right)$ and has degree strictly less than $f_{m+1}$, a contradiction of our choice of $f_{m+1}$.

Thus $I = \left({f_1, \ldots, f_m}\right)$ is finitely generated, and we are done.