The Hilbert Basis Theorem

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:

\begin{align} \quad F(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \end{align}

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

\begin{align} \quad \mathrm{cof}(F) = a_n \end{align}

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

\begin{align} \quad J_m = \{ \mathrm{cof} (F) : F \in I, \deg F \leq m \} \end{align}
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:
\begin{align} \quad F(x) &= ax^s + a_{s-1}x^{s-1} + ... + a_1x + a_0 \\ \quad G(x) &= bx^t + b_{t-1}x^{t-1} + ... + b_1x + b_0 \end{align}
  • Without loss of generality, assume that $s \geq t$. Define a new polynomial $H$ by:
\begin{align} \quad H(x) &= F + x^{s-t}G) \end{align}
  • Since $F, G \in I$ we have that $H \in I$. Furthermore observe that:
\begin{align} H(x) &= [ax^s + a_{s-1}x^{s-1} + ... + a_1x + a_0] + x^{s-t}[bx^t + b_{t-1}x^{t-1} + ... + b_1x + b_0] \\ &= ax^s +a_{s-1}x^{s-1} + ... + a_1x + a_0 + bx^s + b_{t-1}x^{s-1} + ... + b_1x^{s-t+1} + b_0x^{s-t} \\ \end{align}
  • 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:
\begin{align} \quad J_1 \subseteq J_2 \subseteq ... \subseteq J_n \subseteq ... \end{align}
  • 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:
\begin{align} \quad J_m = (a_{m_1}, a_{m_2}, ..., a_{m_k}) \end{align}
  • 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:
\begin{align} \quad \mathrm{cof} (G) = \sum \alpha_j \mathrm{cof} (F_{d_j}) \end{align}
  • 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:
\begin{align} \quad Q(x) = \sum \alpha_j x^{d - \deg(F_{d_j})}F_{d_j} \end{align}
  • 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. $\blacksquare$
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