symmetric monoidal (∞,1)-category of spectra
monoid theory in algebra:
A zero-divisor is something that, like zero itself, when multiplied by something possibly nonzero still produces zero as a product.
Let $M$ be a absorption monoid (such as a commutative ring or any ring).
An element $x$ of $M$ is a non-zero-divisor if, whenever $x \cdot y = 0$ or $y \cdot x = 0$, then $y = 0$. An element $x$ is a zero-divisor if there exists $y \ne 0$ such that $x \cdot y = 0$ or $y x = 0$.
In constructive mathematics, we want $\ne$ to be a tight apartness relation on $M$ in the definition of zero-divisor. We also say that $x$ is a strong non-zero-divisor if, whenever $y \ne 0$, then $x y \ne 0$ and $y x \ne 0$. (The notion of (weak) non-zero-divisor makes sense even without any apartness relation.)
If $M$ is (or may be) non-commutative, then we may distinguish left and right (non)-zero-divisors in the usual way.
By this definition, zero itself is a zero-divisor if and only if $M$ is non-trivial. (Some authorities will differ on this point, but if you think about it, this is clearly the correct definition, by the same principle that the trivial ring is not a field, $1$ is not a prime number, etc. See too simple to be simple.)
An integral domain is precisely a commutative ring (whose multiplicative monoid is an absorption monoid by definition) in which zero is the unique zero-divisor of the multiplicative monoid of the commutative ring (or constructively, in which the strong non-zero-divisors are precisely the strong non-zero elements in the multiplicative monoid, that is those elements $x$ such that $x \ne 0$).
The non-zero-divisors of any absorption monoid $M$ form a monoid under multiplication, which may be denoted $M^{\times}$. Note that if $M$ happens to be a field, then this $M^{\times}$ agrees with the usual notation $M^{\times}$ for the group of invertible elements of the multiplicative monoid $M$, but $M^{\times}$ is not a group in general. (We may use $M^{\div}$ or $M^*$ for the group of invertible elements.)
If $I$ is any ideal of $M$, then we can generalise from a zero-divisor to an $I$-divisor. In a way, this is nothing new; $x$ is an $I$-divisor in $M$ if and only if $[x]$ is a zero-divisor in $M/I$. Ultimately, this is related to the notion of divisor in algebraic geometry.
Last revised on December 8, 2022 at 03:41:16. See the history of this page for a list of all contributions to it.