Solution. In \(\mathbb {Z}_6\), \(2\cdot 3 = 6 \equiv 0\) with neither factor zero, so \(2\) and \(3\) are zero divisors.

In \(\mathbb {Z}_7\), suppose \(ab\equiv 0\) with \(a\not \equiv 0\). Then \(7\mid ab\), and since \(7\) is prime Euclid’s lemma gives \(7\mid a\) or \(7\mid b\); the first is excluded, so \(b\equiv 0\). Hence there are no zero divisors.

The criterion is that \(\mathbb {Z}_n\) is a field if and only if \(n\) is prime. If \(n = ab\) with \(1<a,b<n\) then \(a\) and \(b\) are zero divisors and no field has those; if \(n\) is prime then every non-zero \(a\) has \(\gcd (a,n)=1\), so \(ua+vn=1\) for some integers and \(u\) is the inverse of \(a\).

This is the same statement as \(\mathbb {F}[x]/\langle p(x)\rangle \) being a field exactly when \(p\) is irreducible, with primality and irreducibility playing the same role.

Solution. Let \(u\in A\) be a unit, so \(u^{-1}\) exists in \(R\). Since \(A\) absorbs multiplication by elements of \(R\), \[1 = u^{-1}u \in A .\] Then for any \(r\in R\), again by absorption, \(r = r\cdot 1 \in A\). Hence \(A = R\).

The consequence worth drawing is that a field has no ideals but \(\{0\}\) and itself, since every non-zero element is a unit. That is why quotient constructions on fields are uninteresting, and why the interesting quotients in this course are of polynomial rings rather than of the coefficient field.