Irreductible polynomial 1
Prove that if and then the polynomial is irreductible in .
Solution: Using the following Lemma we see that is irreductible if and only is irreductible. Therefore, we can consider . Since we see that has a root .
Let , and inductively we get . It is easy to see that the coefficients of satisfy the hypothesis of This Problem so the roots of have modulus strictly greater than 1, or has a root which is a root of unity of some order . Then has as a root, and since has integer coefficients, . If the remainder of modulo is not zero, then we reach a contradiction. Therefore which implies . Therefore is a root for ( then ). Then . Contradiction. Therefore has only roots of modulus strictly greater than 1.
Suppose now that is not irreductible. Then . Now, at least one of has all its roots with modulus greater than 1. Suppose without loss of generality that has this property. Then . But, on the other hand so . The contradiction obtained, proves that in fact, is irreductible.