0

A is a DVR,

enter image description here

Could anyone help me to explain these statements? I have no idea how to get these conclusions.

Peter
  • 213

1 Answers1

2

There are lots of statements here (I count five); it's not clear which ones you want explanations for. Nevertheless let me try to answer:

  1. "there is a least integer $k$...": by the Least Integer Principle.

  2. "It follows that $\mathfrak{a}$ contains...": if $v(y) \geq k$, then $v(y/x) \geq 0$, so $y/x \in A$. Therefore $y = (y/x)x$ belongs to $\mathfrak{a}$.

  3. "therefore the only ideals in $A$ are...": the previous step showed that $\mathfrak{m}_k \subseteq \mathfrak{a}$, but by definition of the number $k$, we also have $\mathfrak{a} \subseteq \mathfrak{m}_k$, so in fact $\mathfrak{a} = \mathfrak{m}_k$.

  4. "These form a single chain...": just look at the definition of $\mathfrak{m}_k$.

  5. "Therefore $A$ is Noetherian...": by the Ascending Chain Condition characterisation of Noetherian rings.

  • Thank you very much for your answer. But I still don't get the first statement. How to use the Least Integer Principle? And why we need $\mathfrak a$ in this situation? – Peter Oct 25 '13 at 16:15
  • 1
    I don't understand your second question: the whole point is to study $\mathfrak{a}$, in particular to show it's equal to one of the ideals $\mathfrak{m}_k$. For the first question, another way to phrase it is this: look at the set $S$ of non-negative integers defined by $S=\left{ v(x) \vert x \in \mathfrak{a} \right}$. Then $S$ has a least element, and we call that element $k$. –  Oct 25 '13 at 16:19
  • Well, thanks again – Peter Oct 25 '13 at 16:30
  • @Peter: please let me know if there's still something you didn't understand, and I'll try to explain more. –  Oct 25 '13 at 16:31