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Thread: Irreducible polynomium

  1. #1
    Junior Member
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    Irreducible polynomium

    Hi everyone, I got stuck in this exercise, where I am to show that the polynomial $\displaystyle g(x)=x^3+x+1$ is irreducible over $\displaystyle \mathbb{Q}[x]$.

    In this case you can't use Eisenstein, so what was it that the alternative looked like?

    I tried to use polynomial division, assuming it had a root and then tried to reach a contradiction, but that just got me where I started:
    $\displaystyle x^3+x+1=(x-a)(x^2+ax+1+a^2)+1+a+a^3$ and then you still had to argue, that $\displaystyle 1+a+a^3$ couldn't be a nonzero, rational number.
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  2. #2
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    Hi

    A polynomial in $\displaystyle \mathbb{Q}[X]$ whose degree is 2 or 3 is reducible if and only if it has a root. Moreover, roots in $\displaystyle \mathbb{Q}$ of a monic polynomial $\displaystyle P\in\mathbb{Q}[X]$ are integers which divide $\displaystyle P(0).$

    Therefore in your case, if $\displaystyle X^3+X+1$ has a root in $\displaystyle \mathbb{Q}, $ it's 1 or -1. Since they're not roots, $\displaystyle X^3+X+1$ has no root and is irreducible over $\displaystyle \mathbb{Q}[X].$
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  3. #3
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    Ahh yes, I can see that.
    Thanks a lot.
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