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Algebra-II: Companion notes

Abhay Soman

Section 4.7 Irreducibility criterions

We have already seen the following result for determining whether polynomials of degree \(2\) or \(3\) in one variable over a field is reducible or not.
Since \(r/s\) is a root of \(f\text{.}\) We have that
\begin{equation*} 0=a_0+a_1(r/s)+\cdots+a_n(r/s)^n. \end{equation*}
This implies that
\begin{equation*} 0=a_0s^n+a_1rs^{n-1}+\cdots+a_nr^n\quad\text{i.e.,}\quad -a_0s^n=r(a_1s^{n-1}+\cdots+a_nr^{n-1}). \end{equation*}
Since \(\gcd(r,s)=1\) and \(r\mid -a_0s^n\) we get \(r\mid a_0\text{.}\) Similarly, we have
\begin{equation*} -a_nr^n=s(a_0s^{n-1}+\cdots+a_{n-1}) \end{equation*}
and this implies, by the similar reasoning as earlier, that \(s\mid a_n\text{.}\)
Assume that \(f=gh\in A[x]\) with \(\deg(g)\lt\deg(f)\) and \(\deg(h)\lt\deg(f)\text{.}\) Since \(f\) is monic, we may assume that \(g,h\) are also monic. Then \(f\equiv gh\mod I\) is a factorization of \(f\) in \((A/I)[x]\text{,}\) a contradiction.
Assume that \(f\) is reducible, i.e., there are monic proper factors \(g,h\in A[x]\) such that \(f=gh\text{.}\) Thus \(f\equiv gh\mod\mfp\text{.}\) Since \(a_i\in\mfp\) we have that \(f\equiv x^n\mod\mfp\text{,}\) and hence \(x^n\equiv gh\mod\mfp\text{.}\) Assume that \(g=x^r+b_{r-1}X^{r-1}+\cdots+b_0\) and \(h=x^s+c_{s-1}x^{s-1}+\cdots+c_0\text{.}\) Since \(\mfp\) is a prime ideal, \(A/\mfp\) is an integral domain. Thus, \(x^n\equiv gh\mod\mfp\) implies that either \(b_0\in\mfp\) or \(c_0\in\mfp\text{.}\) Without loss of generality assume that \(b_0\in\mfp\text{.}\) We claim that \(c_0\in\mfp\) as well.
Let \(\overline{b_k}\overline{x}^k\) be the nonzero term of \(\overline{g}\) of the least degree. Thus, \(0\lt k\leq r\text{.}\) Since \(\overline{f}=\overline{g}\overline{h}\) we have \(\overline{a_k}=\sum_{i=0}^{k}\overline{b_{i}}\overline{c_{k-i}}\text{.}\) By our assumption \(\overline{b}_k\) is the nonzero least degree term in \(\overline{g}\text{,}\) therefore \(\overline{b}_i=0\in A/\mfp\) for \(i=0,1,\ldots,{k-1}\) and \(\overline{a_k}=\overline{b}_k\overline{c}_0\text{.}\) Since \(\overline{a}_k=0\in A/\mfp\) and \(\overline{b}_k\neq 0\in A/\mfp\) we must have \(\overline{c_0}=0\in A/\mfp\text{,}\) i.e., \(c_0\in\mfp\text{.}\)
However \(b_0,c_0\in\mfp\) implies that \(a_0\in\mfp^2\text{,}\) a contradiction. Hence, \(f\) must be an irreducible polynomial over \(A\text{.}\)
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