I know that to prove something is a a Unique factorization Domain i need to show that the factorization is unique. So i start like that.// Proof: Let $P$ be a principal ideal domain, and let $$r \in P$$ where $r \not=0$ and $r$ is not a unit. So that means r is reducible so we can write r in to a product of two other elements in $P$. $$r=ab$$. if $a$ and $b$ are irreducible then we done suppose that $a$ or $b$ or both of them are reducible then we can write $$a=a_1a_2$$ again with this ideal of if these elements are irreducible then we done, but if they are not then they are reducible thus $$a_1=a_3a_4$$ and so on.... Now my question is how do i show that this will eventually stop. I was think that maybe i need to use that the union of of a chain of ideal is an ideal because i just prove that statement but i know that does not show how this will become stationary. Any idea of how to proceed because i feel like i am close to finish this problem thanks
Asked
Active
Viewed 4,324 times
3
-
Not a unit does not mean reducible. – lhf May 19 '15 at 16:15
-
yeah but i did not want to say prime directly. Basically what i am trying to say that any element can not be one or zero – user146269 May 19 '15 at 16:16
1 Answers
7
The main point here is that to divide is to contain, which means that if $a \mid b$ then $(a) \supseteq (b)$.
A descending chain of divisors $$\cdots \mid d_n \mid \cdots \mid d_1 $$ corresponds to an ascending chain of ideals $$\cdots \supseteq (d_n) \supseteq \cdots \supseteq (d_1) $$
Now, consider $D=\bigcup_i (d_i)$. Then $D$ is an ideal and so $D=(d)$ for some $d$.
This means that the chain stops at the ideal $(d_n)$ that contains $d$.
In turn, this implies that $d$ is irreducible and you can continue your argument.
lhf
- 216,483