In my course notes there is a lemma that states : "If every two elements of R have a greatest common divisor, then every irreducible element in R is prime."
Now the thing that caused confusion in me, is the first part saying "If every two elements of R have a greatest common divisor". I thought every 2 elements always have a greatest common divisor since the multiplicative identity always divides every element in the ring. So I thought maybe in the definition it leaves out the multiplicative identity, but it does not :
"Given $n ∈ \mathbb{N}$, and elements $a_1, . . . , a_n ∈ R$, we call an element $c ∈ R$ a greatest common divisor of $a_1, . . . , a_n (in R)$ if $c$ divides $a_1, . . . , a_n$ and if for any $d ∈ R$ dividing $a_1, . . . , a_n$ one has $d | c$ "
And by the way, the definition of a prime element in ring:""We call an element $a ∈ R\setminus R^×$ prime (in $R$) if $a \neq 0$ and if for all $x, y ∈ R$ with $a | xy$ we have that $a | x$ or $a | y$.
So my question is, is the first part of the lemma unnecessary and is every prime element in a ring irreducible ?
Thanks for reading my question.
Yes the first part is necessary.
From wikipedia: Consider the ring $\mathbb Z[\sqrt{-3}]$ and let $a=4=2\cdot 2=(1+\sqrt{-3})(1-\sqrt{-3})$, and $b=2(1+\sqrt{-3})$. Clearly $2|a, 2|b$ and $2$ is a maximal divisor. Also $(1+\sqrt{-3})|a$ and $(1+\sqrt{-3})|b$ and $1+\sqrt{-3}$ is a maximal divisor. However they are not associates.Hence gcd($a,b$) does not exist.