Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 3 / 10E (Groups, Rings and Modules)

A polynomial over a unique factorization domain is called a primitive polynomial when the greatest common divisor of its coefficients is a unit.

Assume that the primitive polynomial $f$ factors in $F[X]$ as $f=gh$, with both factors of positive degree. Clearing denominators and removing contents writeswhere $a,b\in F^{\times }$ and $g_0,h_0\in R[X]$ are primitive. ThenBy Gauss lemma for polynomials, $g_0{h}_0$ is primitive. Since $f$ is also primitive, comparison of contents forces $ab$ to be a unit of $R$. Absorbing that unit into one factor gives a nontrivial factorization of $f$ in $R[X]$, contrary to its assumed irreducibility. Hence $f$ is irreducible in $F[X]$.

Suppose $f$ is prime in $F[X]$ and divides $gh$ in $R[X]$. It divides one factor in $F[X]$, say $g=fq$ for $q\in F[X]$. Write $q=a{q}_0$ with $q_0\in R[X]$ primitive and $a\in F$. The product $f{q}_0$ is primitive. Sincehas all coefficients in $R$, a denominator of $a$ would divide every coefficient of the primitive polynomial $f{q}_0$; it must therefore be a unit. Thus $a\in R$, so $q\in R[X]$ and $f$ divides $g$ in $R[X]$. Hence $f$ is prime in $R[X]$.

To deduce the final assertion, use the characterization that an integral domain is a unique factorization domain when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.

An irreducible constant of $R$ is prime in $R[X]$. Every irreducible polynomial of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal ideal domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, andThis is the polynomial ring over a unique factorization domain theorem.

Choose a coprime presentationFor a polynomial $r$, write $r^{\ast }(X,Y)=r(Y,X)$. Symmetry of $f$ givesSince $\mathbb{C}[X,Y]$ is a unique factorization domain and $\gcd (p,q)=1$, one has $p\mid p^{\ast }$ and $q\mid q^{\ast }$. Exchanging the variables preserves total degree, sofor nonzero constants $\lambda ,\mu$. The displayed identity gives $\lambda =\mu$, and applying the exchange twice gives ${\lambda }^2=1$.

If $\lambda =-1$, then $p(X,X)=q(X,X)=0$, so $X-Y$ divides both polynomials, contradicting coprimality. Therefore $\lambda =1$, and both $p$ and $q$ are symmetric. Taking $g=p$ and $h=q$ proves the claim, which is the symmetric rational function in two variables result.