Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 12F (Groups, Rings and Modules)

In the polynomial ring $\mathbb{Z}[x]$, the ideal $(x)$ is nonzero and prime: the quotient ring $\mathbb{Z}[x]/(x)$ is isomorphic to the integral domain $\mathbb{Z}$, by evaluation at zero. It is not maximal, since that quotient is not a field. Equivalently $(x)⊊(2,x)⊊\mathbb{Z}[x]$. Thus $(x)$ is a nonzero prime ideal which is not maximal.

Consider the ideal $I=(2,x)$ in $\mathbb{Z}[x]$. It is proper: evaluation at zero followed by reduction modulo $2$ maps the ring onto ${\mathbb{F}}_2$ and has kernel $I$. Suppose $I=(f)$. Then $f$ divides the constant polynomial $2$, so $f$ is a constant integer dividing $2$. Up to sign, the only possibilities are $1$ and $2$. The first would make the ideal the whole ring; the second cannot divide $x$ in $\mathbb{Z}[x]$. Both are impossible. This proves the integer polynomial ring is not a principal ideal domain result: $\mathbb{Z}[x]$ is not a principal ideal domain.

The coefficient identityholds over every field. Both factors have positive degree, including in characteristics $2$ and $3$, because their leading coefficients remain one. Therefore there is no field over which this polynomial is irreducible.

In the quotient ring, the classes of $x-1$ and $x^2+x+1$ are nonzero: neither nonzero polynomial of degree less than three belongs to the ideal generated by $x^3-1$. But their product is the zero class, sinceThe quotient thus contains zero divisors, so it is not an integral domain.

Use the usual convention that a ring homomorphism preserves the identity. Such a map from the quotient ring to $\mathbb{C}$ fixes $\mathbb{Q}$: it fixes integers, and preservation of inverses fixes rational numbers. It is therefore determined by the image $\zeta$ of the class of $x$. The defining relation forces ${\zeta }^3=1$. Conversely, evaluation at any such $\zeta$ annihilates $(x^3-1)$ and hence factors through the quotient. Thus there are exactly three unital homomorphisms, namelyIf the convention permits nonunital ring homomorphisms, add the zero map. These are the only extra possibilities: the image of $1$ is an idempotent in $\mathbb{C}$, hence either zero or one, and image zero forces the whole map to vanish.