Irreducibility of polynomials This is a very basic question, but one that has frustrated me somewhat.
I'm dealing with polynomials and trying to see if they are irreducible or not.  Now, I can apply Eisenstein's Criterion and deduce for some prime p if a polynomial over Z is irreducible over Q or not and I can sort of deal with basic polynomials that we can factorise easily.  
However I am looking at the polynomial $t^3 - 2$.
I cannot seem to factor this down, but a review book is asking for us to factorise into irreducibles over a) $\mathbb{Z}$, b) $\mathbb{Q}$, c) $\mathbb{R}$, d) $\mathbb{C}$, e) $\mathbb{Z}_3$, f) $\mathbb{Z}_5$, so obviously it must be reducible in one of these.
Am I wrong in thinking that this is irreducible over all? (I tried many times to factorise it into any sort of irreducibles but the coefficients never match up so I don't know what I am doing wrong).
I would really appreciate if someone could explain this to me, in a very simple way.
Thank you.
 A: Note that since your polynomial $f(t)=t^3-2$ is a cubic, it's either irreducible or has a linear factor, and hence a root. This should simplify things somewhat.
Over $\mathbb{Z}$, $f$ is irreducible by Eisenstein's criterion with $p=2$. Thus $f$ is irreducible over $\mathbb{Q}$ by Gauss's lemma.
Over $\mathbb{R}$, $f$ has a root (namely $\sqrt[3]{2}$) and so is reducible. Similarly over $\mathbb{C}$.
Over $\mathbb{Z}_3$ and $\mathbb{Z}_5$ (which I assume mean $\mathbb{Z}$ modulo $3$ and $5$, not the $3$ and $5$-adics), there's only a few possible roots to check. In $\mathbb{Z}_3$, $0^3=0$, $1^3=1$, $2^3=8=2$, so $t=2$ is a root. In $\mathbb{Z}_5$, $0^3=0$, $1^3=1$, $2^3=8=3$, $3^3=27=2$, so $3$ is a root.
A: Over $\mathbb{Z}$, since the polynomial is primitive (no common factors of the coefficients) and of degree $3$, it either has a root or is irreducible. Since the polynomial has no rational roots, it is irreducible over $\mathbb{Z}$.
Over $\mathbb{Q}$, we just need to check for roots. There aren't any, so the polynomial is irreducible over $\mathbb{Q}$ as well.
Over $\mathbb{R}$, the polynomial has at least one real root, $\sqrt[3]{2}$. This gives
$$f(x) = x^3-2 = (x-\sqrt[3]{2})(x + \sqrt[3]{2} + \sqrt[3]{4}).$$
Now we need to check if the quadratic is reducible or irreducible over $\mathbb{R}$. The discriminant is
$$\left(\sqrt[3]{2}\right)^2 - 4\sqrt[3]{4} = \sqrt[3]{4}-4\sqrt[3]{4}=-3\sqrt[3]{4}\lt 0.$$
Since the discriminant is negative, the quadratic is irreducible over $\mathbb{R}$. So this gives you the factorization into irreducibles in $\mathbb{R}$.
To get the factorization in $\mathbb{C}$, just factor the quadratic:
$$f(x) = (x-\sqrt[3]{2})(x-\omega\sqrt[3]{2})(x-\omega^2\sqrt[3]{2})$$
where $\omega=\frac{-1+i\sqrt{3}}{2}$ is a primitive cubic root of unity. You can get this either by using the quadratic formula on $x^2+\sqrt[3]{2}x+\sqrt[3]{4}$, or by noting that the three roots of $x^3-2$ are the three complex cubic roots of $2$. If $\alpha$ and $\beta$ are two cubic roots of $2$, then $\alpha/\beta$ is a cubic root of $1$ (just cube it to see it equals $1$; if $\alpha\neq \beta$, then $\beta=\alpha\omega$ or $\beta=\alpha\omega^2$. 
Over $\mathbb{Z}_3$, we have the "freshman's dream": $(a+b)^3 = a^3+b^3$, because the characteristic is $3$. Since $2^3\equiv 2\pmod{3}$, we get
$$x^3-2 = x^3-2^3 = (x-2)^3$$
so the factorization into irreducibles is $x^3-2 = (x-2)(x-2)(x-2)$.
Over $\mathbb{Z}_5$, we have $3^3\equiv 2\pmod{5}$, so $x-3$ divides $x^3-2$. We have 
$$x^3-2 = (x-3)(x^2+3x+4).$$
Now we check the quadratic. The discriminant is $9-16 = -7 \equiv 3\pmod{5}$. Since $3$ is not a square modulo $5$, the discriminant is not a square in $\mathbb{Z}_5$, so the quadratic is irreducible. This gives you the factorization in $\mathbb{Z}_5$. 
A: $t^3-2=(t-\sqrt[3]{2})(t^2+(\sqrt[3]2)t+\sqrt[3]4)$
A: If you find a root $t=a$ then $t-a$ is a factor of the original polynomial. This means (equating coefficients of $t^3$):
$$t^3-2 = (t-a)(t^2+bt+c)$$
Equating coefficients we get that $ac=2$ (constant term) and $-a+b=0$ (quadratic term), so you can compute $b$ and $c$ and complete the factorisation. You know the linear term will work out because you have checked that $a$ is a root, but this can be used to check your arithmetic.
