1.4 Review of polynomial rings

The ring $F[X]$ of polynomials in the symbol (or “indeterminate” or “variable”) $X$ with coefficients in $F$ is an $F$-vector space with basis $1$, $X$, … , $X^{n}$, … , and with the multiplication

The $F$-algebra $F[X]$ has the following universal property: for any $F$-algebra $R$ and element $r$ of $R$, there is a unique homomorphism of $F$-algebras $\alpha\colon F[X]\rightarrow R$ such that $\alpha(X)=r$.

Division algorithm: given $f(X)$, $g(X)\in F[X]$ with $g\neq 0$, there exist $q(X)$, $r(X)\in F[X]$ with $r=0$ or $\deg(r)<\deg(g)$ such that

moreover, $q(X)$ and $r(X)$ are uniquely determined. Thus $F[X]$ is a Euclidean domain with $\deg$ as norm, and so it is a unique factorization domain.

Let $f\in F[X]$ be nonconstant, and let $a\in F$. The division algorithm shows that

with $q\in$ $F[X]$ and $c\in F$. Therefore, if $a$ is a root of $f$ (that is, $f(a)=0$), then $X-a$ divides $f$. From unique factorization, it now follows that $f$ has at most $\deg(f)$ roots (see also Exercise 1-3).

Euclid’s algorithm: Let $f(X)$, $g(X)\in F[X]$. Euclid’s algorithm constructs polynomials $a(X)$, $b(X)$, and $d(X)$ such that

and $d(X)=\gcd(f,g)$.

Recall how it goes. We may assume that $\deg(f)\geq\deg(g)$ since the argument is the same in the opposite case. Using the division algorithm, we construct a sequence of quotients and remainders

with $r_{n}$ the last nonzero remainder. Then, $r_{n}$ divides $r_{n-1}$, hence $r_{n-2}$,…, hence $g$, and hence $f$. Moreover,

and so every common divisor of $f$ and $g$ divides $r_{n}$: we have shown $r_{n}=\gcd(f,g)$.

Let $af+bg=d$. If $\deg(a)\geq\deg(g)$, write $a=gq+r$ with $\deg(r)<\deg(g)$. Then

and $b+qf$ has degree $<\deg(f)$ because $(b+qf)g=d-rf$, which has degree $<\deg(g)+\deg(f)$.

PARI knows how to do Euclidean division: typing divrem(13,5) in PARI returns $[2,3]$, meaning that $13=2\times 5+3$, and gcd(m,n) returns the greatest common divisor of $m$ and $n$.

Let $I$ be a nonzero ideal in $F[X]$, and let $f$ be a nonzero polynomial of least degree in $I$; then $I=(f)$ (because $F[X]$ is a Euclidean domain). When we choose $f$ to be monic, i.e., to have leading coefficient one, it is uniquely determined by $I$. Thus, there is a one-to-one correspondence between the nonzero ideals of $F[X]$ and the monic polynomials in $F[X]$. The prime ideals correspond to the irreducible monic polynomials.

As $F[X]$ is an integral domain, we can form its field of fractions $F(X)$. Its elements are quotients $f/g$, $f$ and $g$ polynomials, $g\neq 0.$