# Polynomial Existence and Irreducibility over Rational #'s

• RJLiberator
In summary: Yes, I know that there are polynomials with no roots in Q which are NOT irreducible. In summary, the division algorithm is not necessary to find irreducible polynomials.
RJLiberator
Gold Member

## 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?

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

Last edited:
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.

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!)

Samy_A and 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.

RJLiberator said:
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
RJLiberator said:
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].

RJLiberator said:
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.

fresh_42 and 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.

RJLiberator said:
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?

## 1. What is a polynomial?

A polynomial is an algebraic expression that consists of variables and constants, as well as the operations of addition, subtraction, and multiplication. It can also include exponents, but not division or square roots.

## 2. What does it mean for a polynomial to exist over rational numbers?

A polynomial exists over rational numbers if all of its coefficients and exponents are rational numbers. In other words, all of the terms in the polynomial can be expressed as a ratio of two integers.

## 3. How do you determine if a polynomial is irreducible over rational numbers?

A polynomial is irreducible over rational numbers if it cannot be factored into two or more polynomials with rational coefficients. To determine this, you can use the Rational Root Theorem and the Factor Theorem.

## 4. Can a polynomial be both reducible and irreducible over rational numbers?

No, a polynomial cannot be both reducible and irreducible over rational numbers. It can only be one or the other. If a polynomial is reducible, it means it can be factored into two or more polynomials with rational coefficients. If it is irreducible, it cannot be factored any further.

## 5. Why is the study of polynomial existence and irreducibility over rational numbers important?

Polynomials are used in many areas of mathematics and science, so understanding their properties and behavior is essential. Knowing if a polynomial is irreducible over rational numbers can also help in solving equations and finding roots. Additionally, the study of polynomial existence and irreducibility is important in fields such as number theory, cryptography, and computer science.

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