# GCD of {0} in integral domains

Quoting Hungerford's Algebra:

A nonzero element $$a$$ of a commutative ring $$R$$ is said to divide an element $$b \in R$$ (notation: $$a \mid b$$) if there exists $$x \in R$$ such that $$ax = b$$.

Let $$X$$ be a nonempty subset of a commutative ring $$R$$. An element $$d \in R$$ is a greatest common divisor of $$X$$ provided:

1. $$d \mid a$$ for all $$a \in X$$;
2. $$c \mid a$$ for all $$a \in X$$ $$\implies$$ $$c \mid d$$.

In particular, the greatest common divisor (from now on I use "gcd") of any nonempty subset of a commutative ring $$R$$ must be a nonzero element of $$R$$ (else, writing $$d \mid a$$ would make no sense). Now what's the gcd of $$\{0\}$$ in an integral domain? Condition 1 is satisfied by every element $$d \in R$$. In fact, given $$d \in R$$, we have that $$d0 = 0$$ for all $$d \in R$$, hence $$d \mid 0$$ by definition. What can we deduce from condition 2?

Since every element $$c \in R$$ satisfies the condition $$\forall a \in X:c\vert a$$, condition 2 tells us that $$d \in R$$ needs to be an element such that $$\forall c \in R: c \vert d$$. There exists only one such element in $$R$$.
According to the definition of Hungerford, a GCD of a subset is indeed defined to be non-zero since a divisor is defined to be non-zero. That given, it is sensible to simply define the GCD of $$\{0\}$$ as $$0$$. Note, however, that the definition varies among authors as is also mentioned here.
• The first statement after the definition is false. Why? By definition the gcd divides every element of $X$, thus by the previous definition the gcd must be a nonzero element. Where am I wrong? – Lele99_DD Aug 26 '19 at 15:23
• @Lele99_DD The nonempty subset $\{0\}$ is precisely the only exception for that statement. In that case $d \vert a$ i.e. $0 \vert 0$ makes sense since $1 \cdot 0 = 0$. – G. Chiusole Aug 26 '19 at 15:25
• So the solution is that the definition provided by Hungerford works for every nonempty subset $X \neq \{0\}$, whereas we define the greatest common divisor of $\{0\}$ to be $0$? Just a matter of definitions? – Lele99_DD Aug 26 '19 at 15:30
• @Lele99_DD No,no, it's not a matter of definition. The GCD of $\{0\}$ is not defined to be $0$ - there is a proof for it: 1. $\forall a \in X: 0 \vert a$ since $0 \vert 0$ since there exists a $c \in R$ such that $c\cdot 0 = 0$ for examples $c = 1$. 2. for any $c \in R$ s.t. $c \vert 0$ (which is true for any $c \in R$) we have $c \vert 0$ since $0 \cdot c = 0$. Hence $0$ is the GCD of $\{0\}$. – G. Chiusole Aug 26 '19 at 15:32
• Following Hungerford's definition of divisibility, writing $0 \mid a$ makes no sense because "a nonzero element $a$ of a commutative ring $R$ is said to divide". I guess what you're trying to say is true only if we use different definitions and I'd prefer to stick with Hungerford's. – Lele99_DD Aug 26 '19 at 15:41