Question about discrete valuation rings

Question

Suppose $R$ is a domain, not a field, and there exists an irreducible $t$ such that every non-zero element of $R$ can be written as $ut^n$, with $u$ a unit. Apparently $R$ is then Noetherian, a PID in fact. I'm trying to understand a proof of that. It says that, if $m$ is the maximal ideal generated by $t$, thus containing any given proper ideal $a$, and $m^k$ is the ideal generated by $t^k$, there is a maximum $k$ for which $a \subset m^k$. Why is that?

Answer 1

The first step is to show that every element can be written uniquely as $ut^n$ (i.e. $u,n$ are unique).

Once you know that, you show that every (non-zero) ideal is some power of $\mathfrak{m}$. Specifically, $\mathfrak{a} = \mathfrak{m}^k$ if $k$ is the smallest power such that there is some element $ut^k \in \mathfrak{m}$.

Then what's left is to show that $\mathfrak{m}^{k} \not\subseteq \mathfrak{m}^{k+1}$.

None of the first steps are particularly hard, so I encourage you to try to prove them by yourself.

From the second and third parts, if $\mathfrak{a} = \mathfrak{m}^k$ then $\mathfrak{m}^k$ is the largest power of $\mathfrak{m}$ that contains $\mathfrak{a}$.

Answer 2

For $x\ne 0\in R$, since we know that $x=ut^n$, with $u$ a unit, define $v(x)=n$. Define $v(0)=\infty$.

Then for an ideal $a$, define $v(a)=\min\{v(x): x\in a\}$, which exists since the natural numbers union infinity is well ordered. Then for $k > v(a)$, $a\not\subseteq m^k$, since if $v(a)=n$, then there is $x\in a$ with $x=ut^n$, and $ut^n\not\in m^k$ for $k > n$. Conversely, if $k\le v(a)$, then $a\subseteq m^k$, since every element of $a$ is divisible by $t^{v(a)}$, and thus divisible by $t^k$. Thus $v(a)$ is this maximum $k$ with $a\subseteq m^k$.

Answer 3

Hint:

Note that $\;\bigl(At^n\bigr)_{n\in\mathbf N}$ is a decreasing sequence, and show that $\;\displaystyle\bigcap_{n\in\mathbf N}At^n=\{0\}$.