- #1

RJLiberator

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## 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 fraction

*p*/

*q*of two integers, a numerator

*p*and a non-zero denominator

*q*

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

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?