Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 11E (Groups, Rings and Modules)

An algebraic integer is a complex number satisfying a monic polynomial with integer coefficients. Let $f\in \mathbb{Q}[x]$ be the monic minimal polynomial of $\alpha$. It is irreducible over $\mathbb{Q}$. If $H\in \mathbb{Z}[x]$ is a monic polynomial vanishing at $\alpha$, then $f$ divides $H$ over $\mathbb{Q}$. Gauss lemma for polynomials implies that monic rational factors of a monic integer polynomial have integer coefficients. Thus $f\in \mathbb{Z}[x]$, and its irreducibility over $\mathbb{Q}$ implies irreducibility in $\mathbb{Z}[x]$.

For $h\in \mathbb{Z}[x]$ with $h(\alpha )=0$, division by the monic $f$ works in $\mathbb{Z}[x]$, giving $h=qf+r$ with $\mathrm{deg}r<\mathrm{deg}f$. Since $r(\alpha )=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. ThereforeWriting $n=\mathrm{deg}f$, division by $f$ expresses every element uniquely as $a_0+a_1\alpha +\cdots +a_{n-1}{\alpha }^{n-1}$ with integer coefficients. A relation among these powers would be a nonzero vanishing polynomial of degree less than $n$. Hence $1,\alpha ,\dots ,{\alpha }^{n-1}$ form a basis of the free module over $\mathbb{Z}$ $\mathbb{Z}[\alpha ]$.

The polynomial $f(x)=x^5+2x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is divisible by two, and the constant coefficient is not divisible by four. Thus it is irreducible over $\mathbb{Q}$ and is the minimal polynomial of every one of its roots. Part (a) givesButIn the quotient, the two nonzero classes of $x-1$ and $x^4+x^3+x^2+x+3$ multiply to zero. They are nonzero because their nonzero representative degrees are less than five. The quotient is not an integral domain, so $(5)$ is not a prime ideal.

For ring homomorphisms preserving the identity, an image $\beta$ of $\alpha$ in the Gaussian integers would satisfy $f(\beta )=0$. The minimal polynomial of $\beta$ over $\mathbb{Q}$ would then be $f$, of degree five. This is impossible because $\beta \in \mathbb{Q}(i)$ has degree at most two. Thus there is no unital homomorphism $\mathbb{Z}[\alpha ]\to \mathbb{Z}[i]$. If homomorphisms are instead allowed not to preserve the identity, the zero map exists; every nonzero homomorphism into this integral domain is unital, because the image of one is a nonzero idempotent and hence one.

The multiplicative field norm on this ring isIf an element is a unit, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore the units are exactly $1$ and $-1$.

The standard failure of unique factorization isThere is no element of norm two or three: $b\ne 0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm \sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are irreducible elements. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.