= Commutative ring
{title2=$ab=ba$}
{wiki}
A <commutative ring> is a <ring> in which <multiplication> satisfies $ab=ba$ for every pair. The <integers> and the <polynomial ring> $\mathbb Z[X]$ are examples. So is $\mathbb Z/n\mathbb Z$, the <quotient ring> whose elements are <residue classes> in <modular arithmetic>. We represent these by $0,1,\ldots,n-1$ and reduce each sum or product to its remainder. All following calculations modulo $6$ use this convention.
In $\mathbb Z/6\mathbb Z$, $2\cdot3=0$, although neither factor is zero. Thus $2$ and $3$ are <zero divisors>: nonzero elements whose product with another nonzero element can vanish. Also $2\cdot1=2\cdot4=2$. Removing the common factor $2$ would falsely assert $1=4$. The entire row for <multiplication> by $2$ is
$$
2\cdot(0,1,2,3,4,5)=(0,2,4,0,2,4).
$$
The repetitions explain the <linear congruences> directly. The <equation> $2x=2$ has exactly the two solutions $x=1,4$, whereas $2x=1$ has none. A common factor can erase information, and a nonzero coefficient does not guarantee that an <equation> has a solution.
\Image[/residue-ring-multiplication.png]
{title=Multiplication modulo 6 repeats outputs, while every nonzero row modulo 5 is a permutation}
A <unit in a ring> is an element $u$ with a <multiplicative inverse>, meaning $uv=vu=1$ for some $v$. In $\mathbb Z/6\mathbb Z$, the <units> are exactly $1$ and $5$, since $5^2=1$. The other nonzero elements, $2,3,4$, are <zero divisors>; for $4$, use $4\cdot3=0$. In a nonzero <commutative ring>, a <unit> cannot be a <zero divisor>: multiplying $ub=0$ by its <multiplicative inverse> forces $b=0$. But a nonzero element need not be a <unit> or a <zero divisor>: the <integer> $2$ is neither in $\mathbb Z$.
By contrast, in $\mathbb Z/5\mathbb Z$, every row for a nonzero multiplier is a <permutation> of $0,1,2,3,4$. Each such row contains $1$ exactly once, giving a <multiplicative inverse>: $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, and $4^{-1}=4$. Hence this <commutative ring> is a <field>. For example, $2x=1$ now has the unique solution $x=3$. Reading the tables reveals which algebraic steps the arithmetic permits.
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