(i) Give an example of a polynomial f(X) ∈ R[X] which is irreducible in R[X] butAnswer for (i)
not in C[X].
(ii) Are there polynomials in R[X] which are irreducible in C[X] but not in R[X]?
(iii) Which of the following polynomials are irreducible in R[X]:
f(X) = X^3+ 5X − 2 , g(X) = X^2 + 3 , h(X) = X^4 − 1 ?
(iv) Which of the following polynomials are irreducible in Q[X]:
f(X) = X^3 − 2 , g(X) = X , h(X) = X^2+ 2X + 1
I think the polynomial X^2 + 1 is a good example for this questionAnswer for (ii)
I don't think so, because C[X] contains R[X]; therefore, if the polynomial was irreducible in C[X] then it should be irreducible in R[X]Answer for (iii)
f(x) // I couldn't factor/reduce it, but maybe it can be reduced?
h(x) can be reduced to (x^2 + 1)(x+1)(x-1) does this mean it is reducible or is it irreducible in R[X] because one can't reduce (x^2 + 1)?Answer for (iv)
f(x) is reducible -- difference of cubes? where 2 = (2^1/3)^3
g(x) is irreducible -- it is linear.
h(x) is reducible -- (x+1)^2 = (x+1)(x+1)
For (i), you are fine.
For (ii), you're right but your argument is a bit flaccid. By definition of irreducible, there are no polynomials with complex coefficients such that the polynomial factors as the product of those, but real polynomials are in particular complex polynomials (as you said), so a factorization there would also be a complex factorization, producing a contradiction.
For (iii) Recall that the only irreducibles over
are linear polynoimals, and quadratics with an imaginary root (because
is algebraically closed and any algebraic extension of
is contained in its algebraic closure. But then you can compare degrees and get a contradiction if any of the degree 3 or higher were irreducible.)
For (iv) Remember you're working over
, so the factor
is not a rational polynomial.