Let \(F\) be a field (of arbitrary characteristic). Suppose \(g\) is a nonnegative integer, and polynomials \(a(x),b(x)\in F[x]\) satisfy \(\deg a(x)\leq g\) and \(\deg b(x)=2g+1\). Prove that the polynomial \(y^2+a(x)y+b(x)\) is irreducible over \(F(x)\).
Solution
If instead it factors in \(F(x)[y]\) into polynomials of \(y\)-degree \(\geq1\), then by Gauss’s lemma, it factors in \(F[x][y]=F[x,y]\) into polynomials of \(y\)-degree \(\geq1\). Thus we would have
\[ y^2+a(x)y+b(x)=(y+p(x))(y+q(x)) \]
for some \(p(x),q(x)\in F[x]\). Since \(p(x)q(x)=b(x)\) has odd degree, \(p(x)\) and \(q(x)\) have distinct degrees, so
\[ \deg(p(x)+q(x))=\max(\deg p(x),\deg q(x))\geq(\deg p(x)+\deg q(x))/2=(2g+1)/2>g. \]
This contradicts \(\deg a(x)=\deg(p(x)+q(x))\leq g\).