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A field is a commutative ring with $0\ne 1$ in which every nonzero element has a multiplicative inverse. Examples include the rational numbers, real numbers, and complex numbers. The integers are a ring but not a field: dividing $1$ by $2$ leaves them. For a prime number $p$, the prime field ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$ is a finite example, because every nonzero residue is coprime to $p$ and hence has a modular inverse.

In ${\mathbb{F}}_5$, the nonzero inverse pairs are $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, and $4^{-1}=4$. Thus $2x=3$ has solution $x=3\cdot 3=4$ in this field; the real number solution would be $3/2$. Dividing by a nonzero coefficient is valid in both settings, but it means division according to the chosen field's operations.

A field has no zero divisors: if $ab=0$ and $a\ne 0$, multiplying by $a^{-1}$ gives $b=0$. This short argument supports polynomial factorization arguments and pivot division in Gaussian elimination. It does not hold in every commutative ring: modulo $6$, the nonzero residues $2$ and $3$ multiply to zero. The distinction is visible in the figure under algebraic structure.