Hilbert's Basis Theorem

Theorem
Let $A$ be a Noetherian ring.

Let $A \sqbrk x$ be the ring of polynomial forms over $A$ in the single indeterminate $x$.

Then $A \sqbrk x$ is also a Noetherian ring.

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

Let $f = a_n x^n + \cdots + a_1 x + a_0 \in A \sqbrk x$ be a polynomial over $x$.

For a polynomial $f = a_n x^n + \cdots + a_1 x + a_0 \in A \sqbrk x$, we call $a_n$ the leading coefficient of $f$.

Let $I \subseteq A \sqbrk x$ be an ideal of $A \sqbrk x$.

We will show that $I$ is finitely generated.

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

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

If $\ideal {f_1, \ldots, f_i} = I$ then stop choosing elements.

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

Since $A$ is Noetherian, the ideal $\ideal {a_1, a_2, \ldots} \subseteq A$ 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 $I$.

not.

Then our process chose an element $f_{m + 1}$, and $\displaystyle a_{m + 1} = \sum_{j \mathop = 1}^m u_j a_j$ 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:


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

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

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

Thus $I = \ideal {f_1, \ldots, f_m}$ is finitely generated, and we are done.