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 ideal 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 $$a_{m+1} = \sum_{j=1}^m u_j a_j$$ 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:


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

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.