Does $\,0\mid 0?\ $ Is $\,\text{lcm}(a,0)=0?$

Question

1) Let $R$ be a commutative ring. Is it correct to claim that $0\mid 0$?

I guess that it is true since by definition: We claim that $a\mid b$ if there exists $c\in R$ such that $b=ac$. In our case $0=0\times 0$ so $0\mid 0$.

2) Let $R$ be an integral domain. Suppose that $\text{lcm}(a,b)$ exists! Is true that $\text{lcm}(a,0)=0.$

Indeed, $a\mid 0$ and $0\mid 0$. Let $a\mid m$ and $0\mid m$ then $0\mid m$ and we have that $\text{lcm}(a,0)=0$.

Is my reasoning correct?

Answer 1

This is correct.

By the way, this all makes more sense from the viewpoint of lattice theory; maybe try to get your hands on Davey and Priestley's book, which is where I learned the material.

Once you know the basics of lattice theory, simply observe the following:

• The principal ideals in a ring $R$ form a poset $\mathrm{PrinId}(R)$, where we define $aR \leq bR$ to mean $bR \subseteq aR$, which is equivalent to $a \mid b$.
• The principal ideal $1R$ is the smallest element of this poset, and $0R$ is the greatest element.
• This means in particular that $0 \mid 0$ is true, because this translates to the order-theoretic statement $0R \leq 0R$, which is true because we're in a poset.
• If $R$ happens to be a GCD domain, then $\mathrm{PrinId}(R)$ will be a lattice. The $\mathrm{gcd}$ in this lattice coincides with the lattice-theoretic meet and the $\mathrm{lcm}$ coincides with the lattice theoretic join.
• This means in particular that $\mathrm{lcm}(a,0) = a$, because this translates to the lattice theoretic statement $a \vee \top = a$, where $\top$ represents the largest element of the poset, and this fact is true in every lattice, not just $\mathrm{PrinId}(R)$.

Anyway, long story short, you should learn some lattice theory and it'll all fall into place pretty quickly.