Desperate need of help

• Dec 16th 2009, 07:07 PM
Desperate need of help
Let p be a prime integer.

1. Prove that Z_p is an integral domain.

2. Prove that Z_p[x] is an integral domain.

I know that if a ring is a ED it is a PID and if it is a PID it is a UFD and if it is a UFD it is an integral domain. So this can be proven by proving ED, PID or UFD. However, I'm stuck. I need a clear explanation/proof.
• Dec 17th 2009, 06:06 AM
Black
Quote:

I know that if a ring is a ED it is a PID and if it is a PID it is a UFD and if it is a UFD it is an integral domain. So this can be proven by proving ED, PID or UFD. .

You don't really have to do all of that. To show that a ring is an integral domain, you just need to show that it's commutative, have unity, and have no zero divisors.

1) Let $a,b \in \mathbb{Z}_p$ and $ab=0$. Then $ab=pk$ for some $k \in \mathbb{Z}$, which means that $p|a$ or $p|b$. Therefore, in $\mathbb{Z}_p, a=0$ or $b=0$. $\Box$

2) $\mathbb{Z}_p[x]$ is a commutative ring with unity; therefore, we need to show that there are no zero divisors.

Let $f(x), g(x) \in \mathbb{Z}_p[x]$, where

$f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_0$
$g(x)=b_nx^n+b_{n-1}x^{n-1}+\dots+b_0$,

and $a_m \not=0$ and $b_n \not=0$. Then $f(x)g(x)$ has leading coefficient $a_mb_n$, but since $\mathbb{Z}_p$ is an integral domain, $a_mb_n \not=0$. $\Box$
• Dec 17th 2009, 12:32 PM
....
can you help me prove that they are both unique factorization domains?
• Dec 17th 2009, 03:31 PM
Black
For that one, I think it would be easier to prove that they are both PID's.

1) $\mathbb{Z}_p$ is a field, and the only ideals in a field is {0} and the field itself. Since {0} $=(0)$ and $\mathbb{Z}_p=(1)$ , $\mathbb{Z}_p$ is a PID. $\Box$

2) Let's look at the stronger case: if $K$ is a field, then $K[x]$ is a PID.

Let $I$ be a nontrivial ideal in $K[x]$. Let $g(x)$ be an element in $I$ with minimum degree. We claim that $I=(g(x))$.

Clearly, $(g(x))\subseteq I$. Let $f(x) \in I$. We can write $f(x)$ as $f(x)=g(x)q(x)+r(x)$, where $r(x)=0$ or deg $r(x)$< deg $g(x)$.
Solving for $r(x)$ gives us $r(x)=f(x)-g(x)q(x) \in I \Longrightarrow r(x)=0$ (minimality of $g(x)$). Therefore, $f(x)=g(x)q(x) \in (g(x)) \Longrightarrow I \subseteq (g(x))$. $\Box$
• Dec 17th 2009, 04:18 PM
??
how do you know the only ideal are {0} and the field itself? and how does this mean that every ideal can be written as <a> for some a in the field?
• Dec 17th 2009, 04:37 PM
Black
Quote:

how do you know the only ideal are {0} and the field itself?

Well, you use the fact that if an ideal of a ring contains unity, then the ideal is equal to the ring. Every nonzero element in a field is a unit, so it pretty much follows from there.
• Dec 17th 2009, 04:56 PM
??
you are completely losing me here. can you explain this in a simpler way? isn't a field a PID only if every ideal is generated by only one ideal in the field? if so how is {0} and the field itself generated by the same ideal?
• Dec 17th 2009, 05:12 PM
Black
No, {0} and the field isn't generated by the same element.

Ok, let's say we have a field $F$ and a nontrivial ideal $I$ of $F$. Let $a$ be an element in $I$. Since $a$ is also in $F$, there exists an element $b$ in $F$ such that $ab=1$ (every nonzero element in a field is a unit). Then, by the definition of an ideal, $1=ab \in I$. Since $I$ contains unity, $I=F$. Therefore, the only ideals in the field $F$ are {0} and $F$.

In the case of $\mathbb{Z}_p$, 1 is a generator for $\mathbb{Z}_p$, so $\mathbb{Z}_p=(1)$, and clearly {0} $=(0)$.
• Dec 17th 2009, 05:15 PM