Let $\mathbb{R}$ be a commutative ring with $1$ and $M$ = $(a)$ be a maximal ideal of $R$. I need to show that $a$ is irreducible.

Question

Let $R$ be a commutative ring with $1$ and $M$ = $(a)$ be a maximal ideal of $R$. I need to show that $a$ is irreducible.

I am thinking along the lines that an ideal $M$ is maximal in $R$ iff $R/M$ is a field but I did not get anywhere. Any help would be appreciated.

Also, is the converse true?

Answer 1

The proof by contrapositive is rather straightforward, if I am not mistaken

Suppose that $a$ is not irreducible. Then it means that $a = xy$ for for some non-unit elements $x, y \in R$.

But then, $(a) = (xy) \subsetneq (x)$ (you might want to formally prove the fact that the last inclusion is strict), which is a contradiction.

More details here

Answer 2

Hint: A reducible element $a$ is the product of two non-units $a=bc$ and so $(a)\subset (b)$.