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 $R$.

Also, the first statement after the definition is false. In fact, it is precisely what answers your question.

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.

  • $\begingroup$ 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? $\endgroup$ – Lele99_DD Aug 26 '19 at 15:23
  • $\begingroup$ @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 $1 \cdot 0 = 0$. $\endgroup$ – G. Chiusole Aug 26 '19 at 15:25
  • $\begingroup$ 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? $\endgroup$ – Lele99_DD Aug 26 '19 at 15:30
  • $\begingroup$ @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 $\{0\}$. $\endgroup$ – G. Chiusole Aug 26 '19 at 15:32
  • $\begingroup$ 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. $\endgroup$ – Lele99_DD Aug 26 '19 at 15:41

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