# The Hilbert Basis Theorem

Recall from the Noetherian Rings page that a ring $R$ is said to be a Noetherian ring if it satisfies the ascending chain condition, that is, for all ascending chains of ideals $I_1 \subseteq I_2 \subseteq ... \subseteq I_n \subseteq ...$ there exists an $N \in \mathbb{N}$ such that for all $m \geq N$ we have that $I_m = I_N$. Equivalently, we proved that $R$ is Noetherian if and only if every ideal $I$ is finitely generated, that is, there exists $x_1, x_2, ..., x_n \in I$ such that $I = (x_1, x_2, ..., x_n)$.

We about to prove a very important result known as the Hilbert basis theorem which tells us that if $R$ is a Noetherian ring then the corresponding ring of polynomials of a single variable $x$, $R[x]$, is a Noetherian ring. We first need the following lemma.

We first need to get some notation out of the way. If $F \in R[x]$ then $F$ is of the form:

(1)We define the function $\mathrm{cof} (F)$ to be the leading coefficient of $F$. That is, if $F$ has degree $n$ as above then:

(2)For $m \geq 0$ and for an ideal $I$, we define:

(3)Lemma 1: Let $R$ be a Noetherian ring and let $I$ be an ideal. Then for all $m \geq 0$, $J_m$ is an ideal. |

**Proof:**Let $a \in J_m$ and let $b \in R$. Then $a = \mathrm{cof}(F)$ for some polynomial $F \in I$ with $\deg F \leq m$. Consider the function $bF$. Then $bF \in I$ since $I$ is an ideal. Furthermore, $\deg (bF) = \deg (F) \leq m$. Therefore $\mathrm{cof} (bF) \in J_m$. But $\mathrm{cof} (bF) = ab$. So $ab \in J_m$.

- Now let $a, b \in J_m$. Then $a = \mathrm{cof} (F)$ and $b = \mathrm{cof} (G)$ for some polynomials $F, G \in I$ with $\deg F, \deg G \leq m$. Let $\deg F = s$ and let $\deg G = t$. Then $F$ and $G$ have the form:

- Without loss of generality, assume that $s \geq t$. Define a new polynomial $H$ by:

- Since $F, G \in I$ we have that $H \in I$. Furthermore observe that:

- Therefore $\mathrm{cof}(F + x^{s-t}G) = a + b$. Hence $(a + b) \in J_m$. Thus $J_m$ is an ideal.

Theorem 1 (The Hilbert Basis Theorem): Let $R$ be a Noetherian ring. Then $R[x]$ is a Noetherian ring. |

**Proof:**Let $I \subseteq R[x]$ be an ideal and for each $m \geq 0$ let $J_m$ be defined in terms of $I$.- Consider the following ascending chain of ideals:

- Since $R$ is Noetherian, there exists an $N \in \mathbb{N}$ such that for all $m \geq N$ we have that $J_m = J_N$.

- Also, since $R$ is a Noetherian ring every ideal in $R$ is finitely generated. So for each $m \geq 0$ there exists elements $a_{m_1}, a_{m_2}, ..., a_{m_k}$ such that:

- For each $1 \leq j \leq k$, choose a polynomial $F_{m_j} \in I$ such that $a_{m_j} = \mathrm{cof} (F_{m_j})$. Let $I'$ be defined as the ideal generated by the $F_{m_j}$s for all $m \leq N$. We claim that $I = I'$.

- By definition, we have that $I' \subseteq I$. Now suppose that $I' \not \supseteq I$. Then there exists a function $G \in I$ such that $G \not \in I'$. Let $G$ be chosen of minimal degree in $I'$ and let $\deg (G) = d$. Then $\mathrm{cof} (G) \in J_d$. So $\mathbf{cof} (G)$ has the form:

- Where $\alpha_j \in R$ and $F_{d_j}$ are in the list of generators for $I'$ and where $\deg (F_{d_j}) \leq d$. Let:

- Then $Q \in I$ and $\deg (Q) = d$. But $\mathrm{cof} (Q) = \mathbf{cof} (G)$. Then $G - Q \in I$. But $\deg (G - Q) \leq d - 1$. Since $G \not \in I'$ and $Q \in I'$ we hae that $G - Q \not \in I'$ and is such that $\deg (G - Q) < d$. But this contradicts $G$ having minimal degree in $I$.

- Therefore $I = I'$. Since $I'$ is finitely generated so is $I$. So every ideal in $R[x]$ is finite generated, i.e., $R[x]$ is Noetherian. $\blacksquare$