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The task is prove that the next polynomial us irreducible in $\mathbb{Q}$ $$y^3-3y +1 = 0.$$ I don't know the easy form for attack other problems similar.


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Over what field? Presumably $\mathbb Q$? – Alex Becker Aug 28 '12 at 4:31
The question in the title is pretty much impossible for anyone to answer who does not know you closely... – Mariano Suárez-Alvarez Aug 28 '12 at 5:38
up vote 13 down vote accepted

Third degree polynomials are irreducible over a field $F$ if and only if the polynomial does not contain a linear term if and only if the polynomial does not have a root in $F$.

If you are considering it over $\mathbb{C}$, then it is not irreducible since all polynomial have roots over the algebraically closed field $\mathbb{C}$.

If you are considering it over $\mathbb{R}$, then it is still not irreducible. The polynomial is degree $3$ (odd degree). Non-real roots always come in conjugates. Hence at least one root over $\mathbb{C}$ is real. Then over $\mathbb{R}$, you can factor out this linear term.

If you are asking if this third degree polynomial polynomial is irreducible over $\mathbb{Q}$, you need to determines if it has a root in the rationals. You can apply the rational root theorem here. By rational root theorem, the only possible roots are $-1$ and $1$. Plugging both value into the polynomial, you see that neither are roots. Hence the cubic polynomial does not have rational roots.

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thanks for your changes in your answer. – jonaprieto Aug 28 '12 at 4:41

Suppose that $y^3-3y+1$ is not irreducible over $\mathbb Q$. Then it can be factored into a linear and a quadratic term over $\mathbb Q$. Thus it has a rational root (the root of the linear term). Yet by the rational root theorem, any rational root of $y^3-3y+1$ is $\pm 1$ and it is straightforward to check that neither of these works.

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I guess you mean irreducible over $\mathbb Q$. If $y^3-3y+1=0$ were reducible then it could be written as a product of either tree linear factors or as a product of one linear factor and a quadratic factor. In either case it would have a rational root. You can check that it does not have rational roots using the rational root test.

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If you substitute $y$ by $z-1$ you get the polynomial $z^3-3z^2+3$. Now apply Eisenstein's Criterion at the prime $p=3$.

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By Gauss's Lemma a primitive polynomial $p \in \mathbb{Z}[x]$ is irreducible over $\mathbb{Q}$ iff it is irreducible over $\mathbb{Z}$. Reducing the polynomial mod $2$ easily shows that $p$ is irreducible over $\mathbb{Z}$ and therefore over $\mathbb{Q}$.

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