Can I prove that this polynomial is irreducible? 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.
thanks, 
 A: 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.
A: 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.
A: If you substitute $y$ by $z-1$ you get the polynomial $z^3-3z^2+3$. Now apply Eisenstein's Criterion at the prime $p=3$.
A: 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.
A: 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 $\mathbb{Q}$. 
