this is false. fix the mistake(s) you made and then i'll help you.
since your polynomial is monic and of degree 3, in order to show that it's irreducible over we only need to prove that it has no root in i.e. the equation has no integer
Use this to show that the polynomial is irreducible in
Eisenstein's creiterio ??
solution. suppose it has integer solutions. then and hence we must have both and which cannot happen and you're done.
it is correct now. one side of the statement holds for any polynomial of degree at least 2: if a polynomial of degree at least 2 has a root then it has to be divisible by and thus it
is reducible. conversely, suppose is a reducible polynomial of degree 2 or 3. then it can be factored into non-constant polynomials of smaller degrees. but how can you factor a
polynomial of degree 2 into non-constant polynomials of smaller degrees? there's one way only: two polynomials of degree 1. so suppose then obviously
(and ) would be a root of now suppose that the degree of is 3. how can be factored into non-constant polynomials of smaller degrees? there are two ways: three
polynomials of degree 1 or one polynomial of degree 1 and one polynomial of degree 2. in either case has a factor of degree 1, say and thus would be a root of