# 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.

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## 4 Answers

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 $\mathbb{Z}_5$.

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Hi there, thank you for such a wonderfully detailed explanation, it was extremely helpful!  I was wondering, should the quadratic in Z5 read (x^2+3x+4) rather than (x^2+2x+4)?  And hence the discriminant be 9-16=-7 congruent to 3 mod 5?  Also, as for the complex factorisation, could you please explain to me why we are going about it by introducing the primitive root of unity (I understand what the primitive root of unity is, etc.)?  Thank you. – user29553 Apr 20 '12 at 15:51
@user29553: Yes, you are right about the error in $\mathbb{Z}_5$. As for the factorization in $\mathbb{C}$, it just so happens that the two roots of complex roots of $x^2+\sqrt[3]{2}+\sqrt[3]{4}$ are given by $$\frac{-\sqrt[3]{2}+\sqrt{\sqrt[3]{4}-4\sqrt[3]{4}}}{2} = \frac{-\sqrt[3]{2}+\sqrt[3]{2}\sqrt{-3}}{2} = \sqrt[3]{2}\left(\frac{-1+\sqrt{-3}}{2}\right)$$ – Arturo Magidin Apr 20 '12 at 16:18
Thanks for the clarification, I understand the use of the primitive cube root of unity now :) – user29553 Apr 20 '12 at 16:59

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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.

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Eisenstein's criterion is all very nice, but I think it should be noted that you don't even need it here. You only need to check for roots in $\mathbb{Z}$ to decide irreducibility over $\mathbb{Z}$ and $\mathbb{Q}$, but a root has to be a divisor of the absolute term 2.  Also, make sure to use Gauss's lemma only on primitive polynomials (obviously $f$ is primitive, since it's monic). – m_l Apr 20 '12 at 14:52
Hi there, thank you for such a prompt reply!  I'm terribly sorry but I have edited my question, what I meant to ask is how do we factorise f(t) into irreducibles over the above given list. – user29553 Apr 20 '12 at 14:52

$t^3-2=(t-\sqrt[3]{2})(t^2+(\sqrt[3]2)t+\sqrt[3]4)$

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2
...over $\mathbb{R}$, yes. – m_l Apr 20 '12 at 14:47
@m_l Thanks for your supplement – Yangzhe Lau Apr 20 '12 at 15:30

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.

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Thank you for explaining this! – user29553 Apr 20 '12 at 15:08
Btw, I now understand how we can factorise this into irreducibles over R, but what about C?  If there is no imaginary part, is it still a valid factorisation in C?  Sorry if that is too basic a question. – user29553 Apr 20 '12 at 15:11
Is an Integer also a Rational Number? Is a Real Number also a Complex Number? Since there are more Complex Numbers than Reals a quadratic which is irreducible over $\mathbb{R}$ might factorise over $\mathbb{C}$, but you will know how to find the roots of a quadratic over $\mathbb{C}$. – Mark Bennet Apr 20 '12 at 15:13
It can be expressed as such, yes.  I think I understand your meaning, thank you :) – user29553 Apr 20 '12 at 15:16
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