Prove that \[ x^4+x^3+x^2+6x+1 \] is irreducible over \(\mathbb Q\).
Solution
Let \[ f(x)\coloneqq x^4+x^3+x^2+6x+1. \] Its reduction modulo \(3\) is \[ \bar f(x)=x^4+x^3+x^2+1\in\FF_3[x]. \]
1. The polynomial \(\bar f\) has no linear factor in \(\FF_3[x]\).
Direct evaluation gives \[ \bar f(0)=1, \qquad \bar f(1)=1, \qquad \bar f(2)=2 \] in \(\FF_3\). Thus \(\bar f\) has no root in \(\FF_3\), hence no linear factor.
2. The polynomial \(\bar f\) has no factorization into two monicquadratic polynomials over \(\FF_3\).
Suppose \[ \bar f(x) = (x^2+ax+b)(x^2+cx+d), \qquad a,b,c,d\in\FF_3. \] Comparing constant terms gives \[ bd=1. \] Hence either \[ b=d=1 \] or \[ b=d=2. \] In either case \(b=d\neq0\). Comparing the coefficients of \(x^3\) gives \[ a+c=1, \] whereas comparing the coefficients of \(x\) gives \[ ad+bc=b(a+c)=0. \] Since \(b\neq0\), this forces \(a+c=0\), contradicting \(a+c=1\).
3. The reduction \(\bar f\) is irreducible in \(\FF_3[x]\).
A reducible polynomial of degree \(4\) over a field has a factor of degree \(1\) or \(2\). Step 1 excludes degree-\(1\) factors, and step 2 excludes a factorization into two degree-\(2\) factors. Hence \(\bar f\) is irreducible.
4. The polynomial \(f\) is irreducible over \(\QQ\).
The polynomial \(f\) is monic and lies in \(\ZZ[x]\). If it were reducible over \(\QQ\), Gauss’s lemma would give a factorization into two nonconstant monic polynomials in \(\ZZ[x]\). Reducing that factorization modulo \(3\) would give a nontrivial factorization of \(\bar f\) in \(\FF_3[x]\), contradicting step 3.
5. Q.E.D.
Step 4 is the required irreducibility statement.