# irreducible polynomials over the field of rationals

• Jan 12th 2012, 10:28 AM
oblixps
irreducible polynomials over the field of rationals
Let $\displaystyle p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$ be a polynomial with rational coefficients such that a_n and a_0 are nonzero. Show that p(x) is irreducible over the field of rational numbers if and only if $\displaystyle q(x) = a_0 x^n + a_1 x^{n-1} + ... + a_{n-1} x + a_n$ is irreducible over the field of rational numbers.

i am having trouble starting this problem. i know that irreducible means that the polynomial cannot be factored into a product of smaller polynomials. So p(x) = q(x)g(x) + r(x), it will always have a nonzero remainder since it is irreducible. So I am showing the first direction, if p(x) is irreducible then q(x) is irreducible. I tried a contradiction argument assuming q(x) is reducible so there exists g(x) such that q(x) = k(x)g(x). but i can't think of any way to connect that to p(x) and (hopefully) show that p(x) is irreducible as well. Is this an efficient approach? Help is greatly appreciated.
• Jan 12th 2012, 12:18 PM
alexmahone
Re: irreducible polynomials over the field of rationals
Quote:

Originally Posted by oblixps
Let $\displaystyle p(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$ be a polynomial with rational coefficients such that a_n and a_0 are nonzero. Show that p(x) is irreducible over the field of rational numbers if and only if $\displaystyle q(x) = a_0 x^n + a_1 x^{n-1} + ... + a_{n-1} x + a_n$ is irreducible over the field of rational numbers.

i am having trouble starting this problem. i know that irreducible means that the polynomial cannot be factored into a product of smaller polynomials. So p(x) = q(x)g(x) + r(x), it will always have a nonzero remainder since it is irreducible. So I am showing the first direction, if p(x) is irreducible then q(x) is irreducible. I tried a contradiction argument assuming q(x) is reducible so there exists g(x) such that q(x) = k(x)g(x). but i can't think of any way to connect that to p(x) and (hopefully) show that p(x) is irreducible as well. Is this an efficient approach? Help is greatly appreciated.

Note that $\displaystyle q(x)=x^np\left(\frac{1}{x}\right)$. Don't know if that will help in any way.
• Jan 12th 2012, 02:54 PM
oblixps
Re: irreducible polynomials over the field of rationals
i've noticed that for a polynomial over Q, if we switch the first and last coefficients, the 2nd and 2nd to last, etc. and if the polynomial is reducible then we seemingly do the same thing for each of its factors.

so for one example $\displaystyle x^2 + 5x + 6$ factors as $\displaystyle (x+3)(x+2)$ and $\displaystyle 6x^2 + 5x + 1$ factors as $\displaystyle (3x + 1)(2x + 1)$.

I assume for contradiction that q(x) is reducible or that q(x) = k(x)g(x). Then i know that q(1/x) = k(1/x)g(1/x) and thus $\displaystyle x^n q(\frac{1}{x}) = x^n k(\frac{1}{x})g(\frac{1}{x})$ and i know this is a polynomial since the original polynomial was of degree n and the lowest "degree" after substituting 1/x will be "-n" but multiplying by x^n will make sure all the exponents are positive. I also know that $\displaystyle x^n q(1/x) = p(x)$ so $\displaystyle x^n k(\frac{1}{x})g(\frac{1}{x}) = p(x)$ which means that p(x) is the product of 2 polynomials of smaller degree and thus p(x) is reducible giving a contradiction.

Is this proof correct, or is it maybe missing details, has wrong arguments, etc.?
• Jan 12th 2012, 03:18 PM
alexmahone
Re: irreducible polynomials over the field of rationals
Quote:

Originally Posted by oblixps
I assume for contradiction that q(x) is reducible or that q(x) = k(x)g(x). Then i know that q(1/x) = k(1/x)g(1/x) and thus $\displaystyle x^n q(\frac{1}{x}) = x^n k(\frac{1}{x})g(\frac{1}{x})$ and i know this is a polynomial since the original polynomial was of degree n and the lowest "degree" after substituting 1/x will be "-n" but multiplying by x^n will make sure all the exponents are positive. I also know that $\displaystyle x^n q(1/x) = p(x)$ so $\displaystyle x^n k(\frac{1}{x})g(\frac{1}{x}) = p(x)$ which means that p(x) is the product of 2 polynomials of smaller degree and thus p(x) is reducible giving a contradiction.

Is this proof correct, or is it maybe missing details, has wrong arguments, etc.?

Seems okay, except that you don't need to use proof by contradiction. Just prove that p(x) is reducible over the field of rational numbers if and only if q(x) is reducible over the field of rational numbers, and the required result follows immediately.