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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?

Bill Dubuque
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RFZ
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1 Answers1

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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.

goblin GONE
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  • Thanks for detailed reply! One more question: Am I right that my second question is true even in commutative ring $R$? – RFZ Jun 16 '18 at 19:35
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    @RFZ, yes, this is true in any commutative ring with LCM's. Again, in this case we get a poset whose elements are the principal ideals. The existences of LCM's translates into the existence of joins. Since $a \vee \top = a$ holds in any poset with joins, the desired identity follows. Try having a look at the wikipedia page on meets and joins; it should be pretty enlightening. Joins are a bit like suprema from real analysis; if I have a poset $P$ with elements $a$ and $b$, their join, if it exists, is the smallest element $p \in P$ with $a \leq p$ – goblin GONE Jun 17 '18 at 09:21
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    and $b \leq p$. That is, it's the least upper bound. A simple example is this: if $S$ is a set and $\mathcal{P}(S)$ is the set of all subsets of $S$, then $\mathcal{P}(S)$ has joins and they coincide with unions. In other words, I'm saying that $A \cup B$ is the smallest subset $C$ with $A \subseteq C$ and $B \subseteq C$. This is the most basic example, the poset of principal ideals of $\mathbb{Z}$ is arguably the second most basic example; in this case, joins coincide with lcm's. – goblin GONE Jun 17 '18 at 09:21