Polynomial Existence and Irreducibility over Rational #'s

[RJLiberator]

Homework Statement

Prove that for all n ≥ 1, there exists a polynomial f(x) ∈ ℚ[x] with deg(f(x)) = n such that f(x) is irreducible in ℚ[x].

Homework Equations

In mathematics, a rational number is any number that can be expressed as the quotient or fractionp/q of two integers, a numeratorp and a non-zero denominatorq


f(x) is called irreducible if it is not reduicble.
f(x) is reducible if there exists g(x)*h(x) such that f(x) = g(x)*h(x) and deg(g(x)) < deg(f(x)), deg(h(x)) < deg(f(x)).

http://mathworld.wolfram.com/IrreduciblePolynomial.html

The Attempt at a Solution

Is it possible to use the quadratic discriminant here to show when no roots exist?

b^2 - 4ac ≥ 0
So we say whenever b^2-4ac < 0 then f(x) is irreducible.

And it would seem clear that you could always make a=c and a^2 > b^2...
Ouch. Nvm, this doesn't work for higher degree terms actually.

In that case, hm.

Could we work with cases, possibly induction?

Case n = 1. Then by definition f(x) is irreducible.
Induction step: n+1 can be irreducible.
Then no, this is not working.Is the division algorithm necessary?

Theorem 7 Let f be a polynomial with rational coefficients in lowest terms so that the numerators of the coefficients are relatively prime (i.e., have no common prime factor). Let d be the least common multiple (lcm) of the denominators of the coefficients of f. Let g = df. Then g has integer coef- ficients. Furthermore, f is irreducible over the rationals if and only if g is irreducible over the integers

Am I going in the right direction by observing the division algorithm?

 

[fresh_42]

Have you ever thought about unity roots or just powers of 2, e.g.?

 

[RJLiberator]

Well, with powers of two we use the quadratic discriminant and see that b^2 - 4ac >= 0.

So we have that when b^2-4ac < 0 we have an irreducible polynomial.

 

[fresh_42]

This special discriminate is only for 2nd degree polynomials.
You only have to point out one irreducible polynomial for each degree.
So taking algebraic roots that are not in ℚ will do the job, e.g. roots of $2^n$, e.g.
Even the roots of $1^n$ will be probably all but at most two outside of ℚ, which has to be proved.
(Att.: the all but 1 or 2 fact reduces the degree of the searched polynomials!)

 

[RJLiberator]

Hm. I understand your point that the discriminant is only for second degree polynomials.

So I only have to point out one irreducible polynomial for each degree, I agree.

Algebraic roots that are not in Q will do the job, roots of 2^n. I'm not really following what you mean here.

(Att.: the all but 1 or 2 fact reduces the degree of the searched polynomials!)

I'm struggling to understand this hint, I can understand it is important.

 

[fresh_42]

Hm. I understand your point that the discriminant is only for second degree polynomials.

So I only have to point out one irreducible polynomial for each degree, I agree.

Algebraic roots that are not in Q will do the job, roots of 2^n. I'm not really following what you mean here.
I'm struggling to understand this hint, I can understand it is important.

Sorry, I have been too sloppy on that. I meant $x^n -2 = 0$ or $x^{n-1}-1=0$.
This delivers $n$ roots, and with Euler's Formula it is easy to show that these roots are not in ℚ. However, $±1$ might be roots of the latter polynomial which are in ℚ! So you have to divide the polynomial by $x±1$ to get the irreducible part. But this irreducible part will then of course be only of degree $n-1$ or $n-2$.

 

[RJLiberator]

Okay, so what we are saying is:

Take an example polynomial x^n - 2 = 0
and show that it delivers n roots, say with Euler's Formula.
Then show that these roots are NOT in Q.

 

[fresh_42]

Okay, so what we are saying is:

Take an example polynomial x^n - 2 = 0
and show that it delivers n roots, say with Euler's Formula.
Then show that these roots are NOT in Q.

That was the plan, yes.

 

[RJLiberator]

So let's try this:

Observe: x^n-2 = 0.
x^n = 2
x = nthroot(2)

Therefore, for every n this is not possibly in Q.

We observe now x^1 is irreducible by definition.
So now, I've shown that there exists a reducible polynomial for every degree n in Q[x].

 

[Dick]

So let's try this:

Observe: x^n-2 = 0.
x^n = 2
x = nthroot(2)

Therefore, for every n this is not possibly in Q.

We observe now x^1 is irreducible by definition.
So now, I've shown that there exists a reducible polynomial for every degree n in Q[x].

You do know that there are polynomials with no roots in Q which are NOT irreducible, yes? Give me an example.

 

[RJLiberator]

x^2+1 = 0

I think I solved this problem in class today. So help is no longer needed, it was the more exhausting proof of my latest homework assignment.

 

[Dick]

x^2+1 = 0

I think I solved this problem in class today. So help is no longer needed, it was the more exhausting proof of my latest homework assignment.

Ok, but how did you prove it? x^2+1 IS irreducible over Q and has no roots. (x^2+1)(x^2+1) also has no roots. Is it irreducible?