Past exam of the algebra 2 course of the University of Oxford / 2015 / 3

For a nonzero polynomial $f={\sum }_i{a}_i{t}^i\in \mathbb{Z}[t]$, its polynomial content is the positive greatest common divisor of its coefficients:For nonzero $g\in \mathbb{Q}[t]$, choose a positive integer $n$ with $ng\in \mathbb{Z}[t]$ and define its rational polynomial content byTo check independence of the denominator, write $g=f/n=h/m$ with $f,h\in \mathbb{Z}[t]$ and $n,m>0$. Then $mf=nh$, while multiplication of all coefficients by a positive integer givesDividing by $mn$ proves $c(f)/n=c(h)/m$, as required.

For multiplication, write $g_i=f_i/n_i$. The integer-coefficient content identity permitted in the source givesThusWe will also use the normalization $g/c(g)\in \mathbb{Z}[t]$, with content one: if $g=f/n$, then $g/c(g)=f/c(f)$. It is a primitive polynomial. For a nonzero rational constant $r$, the same definition gives $c(r)=|r|$.

The set $\tilde{P}$ contains zero. It is closed under negatives because $-f\in P$ whenever $f\in P$. If $f,g\in P$ and $n,m>0$, thensince $P$ is an ideal of $\mathbb{Z}[t]$. Also, any $q\in \mathbb{Q}[t]$ can be written $q=h/d$ with $h\in \mathbb{Z}[t]$ and $d>0$, givingThis proves that $\tilde{P}$ is an ideal of $\mathbb{Q}[t]$.

Clearly $P\subseteq \tilde{P}\cap \mathbb{Z}[t]$, by taking denominator one. Conversely, if $h\in \tilde{P}\cap \mathbb{Z}[t]$, then $nh\in P$ for some positive integer $n$. The condition $P\cap \mathbb{Z}=\{0\}$ implies $n\notin P$. Since $P$ is a prime ideal, $nh\in P$ forces $h\in P$. HenceIn particular, $\tilde{P}$ is nonzero because it contains the nonzero ideal $P$, and it is proper: otherwise its intersection with $\mathbb{Z}[t]$ would contain $1$. This is the concrete extension of an ideal and contraction of an ideal calculation for the localization that inverts nonzero integers.

The nonzero ideal $\tilde{P}$ is principal because $\mathbb{Q}[t]$ is a principal ideal domain. Choose a nonzero generator $h\in \mathbb{Q}[t]$. Part (a) givesThe nonzero rational scalar $c(h)$ is a unit in $\mathbb{Q}[t]$, so this rescaling does not change the generated principal ideal:Part (b) now places $f$ in $P$, since $f\in \tilde{P}\cap \mathbb{Z}[t]$. Thus $(f{)}_{\mathbb{Z}[t]}\subseteq P$.

For the reverse inclusion, let $g\in P$ be nonzero. The displayed equality supplies $q\in \mathbb{Q}[t]$ with $g=fq$. Multiplicativity of rational polynomial content givesSince $g$ has integer coefficients, $c(g)$ is a positive integer. The normalization from part (a) gives $q/c(q)\in \mathbb{Z}[t]$, so multiplying it by the integer $c(q)=c(g)$ proves $q\in \mathbb{Z}[t]$. Consequently $g\in (f{)}_{\mathbb{Z}[t]}$. The zero polynomial belongs as well, provingThis is the divisibility descent in Gauss lemma for polynomials: dividing an integer-coefficient polynomial by a primitive polynomial cannot introduce a denominator into an exact quotient. Here it yields the description of prime ideals of the integer polynomial ring above zero. The generator has positive polynomial degree, since a primitive constant would be a unit and would put $1$ in $P$.

Take $I=(2,t)\subseteq \mathbb{Z}[t]$. The ring homomorphismis surjective. Its kernel of a ring homomorphism is exactly $I$: a polynomial maps to zero precisely when its constant coefficient is even, in which case it has the form $2a+th(t)$ with $a\in \mathbb{Z}$ and $h\in \mathbb{Z}[t]$. By the first isomorphism theorem for rings,This quotient ring is a field, so $I$ is a maximal ideal and in particular a prime ideal.

Suppose $I=(h)$ were a principal ideal. Because $2\in I$, the nonzero polynomial $h$ divides the constant $2$, forcing its polynomial degree to be zero: $h=d$ for a nonzero integer $d$. But $t\in I$ also makes $d$ divide the coefficient $1$ of $t$, so $d=\pm 1$. This would give $I=\mathbb{Z}[t]$, contradicting the properness of the kernel of a ring homomorphism $\ker \theta$. ThereforeThere is no conflict with part (c): $I\cap \mathbb{Z}=2\mathbb{Z}\ne \{0\}$. The zero-intersection hypothesis is exactly what excludes this example.