Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 3 / 10G (Groups, Rings and Modules)

Assume that $p$ is a prime element. If $p=ab$, then $p$ divides $ab$, so $p$ divides $a$ or $b$. If $p\mid a$, write $a=pc$; cancellation in the integral domain gives $1=bc$, so $b$ is a unit. The other case is symmetric. Hence every prime element is irreducible.

Suppose $p$ is irreducible andThen $p=ab$ for some $b\in R$. Irreducibility says that either $a$ is a unit, in which case $(a)=R$, or $b$ is a unit, in which case $(a)=(p)$. Thus no proper ideal lies strictly between $(p)$ and $R$, so

For any commutative ring with identity, the maximal ideal quotient criterion states that an ideal $I$ is maximal exactly when $R/I$ is a field. Applying it to $I=(p)$ proves

Every field is an integral domain, so if $R/(p)$ is a field then it is an integral domain.

By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.