Notes - Rings and Modules HT24, Principal ideal domains

Created: May 06, 2024

Flashcards

Quickly prove that if:

  • $R$ is a PID
  • $I _ 1 \subseteq I _ 2 \subseteq \cdots$ is a sequence of ideals

Then:

  • $I := \bigcup _ {n \ge 0} I _ n$ is an ideal
  • $\exists N \in \mathbb N$ such that $\forall n \ge N$, $I _ n = I _ N = I$.

$I$ is an ideal:

Take $p, q \in I$. Then $\exists k, l \in \mathbb N$ such that $p \in I _ k$ and $q \in I _ l$. Then $\forall r \in R$, $rp \in I _ k \subseteq I$ and if $n := \max \{k, l\}$ then $p, q \in I _ n$ and $p + q \in I _ n \subseteq I$.

Stabilises:

Since $R$ is a PID, $\exists c$ s.t. $I = \langle c \rangle$. But there must be some $N$ such that $c \in I _ N$, hence $I = \langle c \rangle \subseteq I _ N \subseteq I$, so $I = I _ N = I _ n$ for all $n \ge N$.

What does it mean for a (commutative) ring to be Noetherian, and what does it mean for a ring to satisfy the ACCP?
  • Noetherian: Every chain of ideals eventually stabilises (i.e. if $I _ 1 \subseteq I _ 2 \subseteq \cdots$, then eventually this just equals $I$).
  • ACCP: Every chain of principal ideals eventually stabilises.

Suppose:

  • $R$ is a PID containing infinitely many elements
  • $R$ has finitely many units

Quickly show that $R$ contains infinitely many maximal ideals.

Proof 1: For a contradiction, suppose there are finitely many units but also only finitely many maximal ideals. We use the correspondence that

\[x \text{ irreducible} \iff \langle x \rangle \text{ maximal}\]

So if there are finitely many maximal ideals in a PID, there are also only finitely many irreducible elements. Call these $q _ 1, \cdots, q _ k$.

For each $a \in R$, consider

\[aq_1 \cdots q_k + 1\]

then this must be a unit (consider otherwise, then there would be an irreducible factor that isn’t $q _ 1, \cdots, q _ k$).

Since $R$ is in particular an integral domain, we see that $a \mapsto a q _ 1 \cdots q _ k + 1$ is injective since

\[\begin{aligned} & a q_1 \cdots q_k + 1 = b q_1 \cdots q_k + 1 \\\\ \implies& (a - b) q_1 \cdots q_k = 0 \\\\ \implies& a = b \end{aligned}\]

Then this would imply that there are infinitely many units, a contradiction.

Proof 2: Suppose that $\mathcal M$ is the set of proper maximal ideals, and consider

\[I = \bigcap_{M \in \mathcal M} M\]

$I$ is an ideal, and since $R$ is not a field, $I \ne R$.

Suppose that $x \in I$ such $1 - x$ is not a unit. Then $\langle 1 - x \rangle$ is a proper ideal, and so

\[\langle 1 - x \rangle \subset M_i\]

for some $i$. Then $x \in I$, so

\[1 = (1 - x) + x \in M_i + M_i = M_i\]

which contradicts the fact $\langle 1 - x \rangle \subset M _ i$. Then $1 - x$ is a unit for all $x \in I$. But then $I$ is finite. However, $R$ is infinite and an integral domain, so $I$ must equal $\{0\}$.

However, if $\mathcal M$ is finite, then we can enumerate $\mathcal M$ as $M _ 1, \cdots, M _ k$. Each of these ideals is nonzero, and so we can take nonzero elements $x _ i \in M _ i$. But then $x _ 1, \cdots, x _ k \in I$, a contradiction.



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