Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a commutative ring, and let $a \in R$ be an irreducible element. Prove that $a$ is not a zero divisor.

I need help proving this. I know that $b \in R$ is a zero divisor if there is $a \in R$, $a$ not equal to zero, such that $ab=0$. Also, an element $p$ is irreducible if whenever $p=ab$, either $a$ or $b$ is a unit.

I just don't know how to properly put the two definitions together to formulate a sound proof for this question. I think the best approach would be a proof by contradiction. So assuming $a$ is a zero divisor and then showing that $a$ in R is in fact reducible which leads to a contradiction. But I still don't know how to show that.

share|improve this question
    
Do you know the definition of an irreducible element, as well as what it means for $a$ to be a zero divisor? –  user61527 Nov 20 '13 at 4:35
    
Thanks, Agnes for taking the time to post that second paragraph. It really helped me (us?) get into your shoes and see what you were working with. Be sure to do that on every post :) –  rschwieb Nov 20 '13 at 14:41
    
user's right: adding a hypothesis that irreducible elements from being zero divisors trivializes your question. Since it changes the solution completely, I'm going to revert that change. Please refrain from making changes that would dramatically affect existing solutions. –  rschwieb Nov 21 '13 at 18:20

1 Answer 1

Based on what you wrote, you won't be able to prove this: there is a counterexample.

Consider the field of two elements $F_2$, and look at the ring $F_2[x]/(x^2)$.

It's got four elements $\{0,1,x,x+1\}$. Two of them, $\{1,x+1\}$ are units.

According to the multiplication in this ring:

$\begin{array}{c|ccccc} \cdot & 0 & 1 & x & x+1 \\ \hline 0 &0 &0 &0 &0 \\ 1 &0 &1 &x &x+1 \\ x & 0 &x & 0 & x \\ x+1 &0 &x+1 &x &1 \\ \end{array}$

So $x$ certainly satisfies this definition of irreducible: the only pairs $a,b$ multiplying to $x$ are $x1$ and $x(x+1)$, and in both cases one of the pair is a unit.

But $x$ is a zero divisor since $x^2=0$!


Just a few more comments on the definition you provided for "irreducible element." While I suppose there's nothing wrong with defining an irreducible element of a commutative ring this way, the definition is usually made in the context of an integral domain. Furthermore, irreducible elements are usually prohibited from being $0$ or a unit (which you did not do in your definition.)

share|improve this answer
3  
A detailed explanation + a nice convincing table + pertinent comments = a perfect answer ! –  Georges Elencwajg Nov 21 '13 at 18:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.