An irreducible element is not a zero divisor 
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.
 A: 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.)
A: The commutative ring $R=\Bbb{Z}/4$ has the irreducible element $u=2$. However, it is a zero divisor.
