# Showing the polynomial $x^4 + x^3 + 4x + 1$ is irreducible in $\mathbb{Q}[x]$.

Showing the polynomial $$f(x) = x^4 + x^3 + 4x + 1$$ is irreducible in $$\mathbb{Q}[x]$$.

I have two question relating to this which I've bolded below.

(*) $$f$$ has degree $$4 \ge 2$$ so if $$f$$ has a root then $$f$$ is reducible.

Well the rational root test says that if $$\exists$$ a rational root $$\frac{p}{q}$$ with $$p, q$$ coprime then $$p|a_0$$ and $$q|a_n$$.

Hence the potential rational roots are $$\pm 1$$. Neither of these are roots so $$f$$ does not have a root. But (*) only says that if $$f$$ has a root then $$f$$ is reducible, it does not imply that if $$f$$ does not have a root then $$f$$ is irreducible. Is this correct?

Now we can't apply Eisenstein's irreducibility criterion theorem as there is no prime $$p$$ such that

$$p\mid a_0, a_1,..., a_{n-1}$$

$$p \nmid a_n$$

$$p^2 \nmid a_0$$

So where do we go from here? One other fact I am aware of is that if $$f$$ is primitive, which it is, $$f$$ irreducible in $$Q[x] \iff f$$ is irreducible in $$Z[x]$$. But I don't see how that can help me here. So how do I proceed from here?

it does not imply that if f does not have a root then f is irreducible. Is this correct?

Yes, that is correct. For polynomials of degree $> 3$, the absence of roots in $\mathbb{Q}$ is only a necessary, but not a sufficient condition for irreducibility.

Aside from trying a substitution $x \mapsto x - a$ to achieve a form where Eisenstein's criterion is applicable, you can consider the corresponding polynomial over $\mathbb{Z}/(p)$ for a prime $p$. If $f$ is reducible, so is its image $\overline{f} \in \left(\mathbb{Z}/(p)\right)[X]$, if it has the same degree as $f$ (since $f$ here is monic, that is the case). So if $\overline{f}$ is irreducible in any $\left(\mathbb{Z}/(p)\right)[X]$, then $f$ itself is irreducible.

Note: if $\overline{f}\in \left(\mathbb{Z}/(p)\right)[X]$ is reducible, that does not imply that $f$ is reducible.

If we look at the example over $\mathbb{Z}/(2)$, we have $\overline{f}(x) = x^4+x^3+1$. It is readily seen that that has no zero in $\mathbb{Z}/(2)$, so if it were reducible, both factors would have to be quadratic. Since the constant term is $1$, a factorisation would have the form $$x^4+x^3+1 = (x^2+ax+1)(x^2+bx+1).$$ But then the coefficients of $x^3$ and $x^1$ would be equal ($a+b$). So $\overline{f}$ is irreducible.

• Nice explanation, I will have to go through that in detail to take it all in. One follow-on question regarding the substitution that lets us use Eisenstein - I had already tried that with $x \to x + 1$ and it didn't work...And I thought it was not feasible to try keep trying different $a$'s in $x \pm a$. So how do you pick a $\pm a$ to try in the substitution? Nov 15 '13 at 13:50
• Finding a substitution to get the polynomial into Eisenstein form is generally not easy - I'm not even sure it is always possible. I don't know a better method than trial-and-error in general, if one doesn't hit a lucky one in a few trials, I would advise trying the reduction modulo a prime. Nov 15 '13 at 14:16

Another way, instead of trying to apply Eisenstein's, is to assume that you can indeed factor it $(x^2+bx+c)(x^2+dx+e)=x^4+x^3+4x+1$ and arrive at a contradiction. Gauss' lemma makes your work a lot easier too.

Apply Eisenstein's criterion for $$p=3$$ to $$f(x-1)=x^4-3\,x^3+3\,x^2+3\,x-3$$

If $$f$$ can be factored, both factors must have degree $$2$$ becuase there is no integer root. Because of $$f(-11)=13267,f(-3)=43, f(-2)=1, f(-1)=-3, f(0)=1, f(1)=7,f(4)=337$$ one of these degree $$2$$-factors is $$1$$ at three x-values (or -1). So it is identical to $$1$$ (or $$-1$$). So $$f$$ is irreducible. See this link ($$13267,43,3,7,337$$ are primes)

But it is sufficient that $$f(-11)=13267$$ is a prime because $$11 \gt 2+\max\{\frac{|a_i|}{a_4} \mid i =0,1,2,3\}$$ See this and this

Hint: try a substitution $x \mapsto x - a$ for some $a$; if the resulting polynomial is irreducible (using Eisenstein, for examples), your original one is as well.

• How do you know what $a$ to try though? Obviously I would start with $1$, but it seems a bit of a hit and miss approach? And furthermore, why even pick $-a$ instead of $+a$...what are we 'looking for' when we are picking an $a$? Nov 15 '13 at 13:44
• You could just work out $(x-a)^4 + (x-a)^3 + 4(x-a) + 1$. And there is no reason for prefering $-a$ over $+a$; $+a$ probably is a bit easier to work out. Nov 15 '13 at 14:06
• Yes, but how do you pick the $a$ to use when working it out? Nov 15 '13 at 14:48
• Just with $a$ a variable, I mean. Then look at the result and figure out candidate $a$'s by looking at easy-looking coefficients - try to make them prime for instance. It's still trial-and-error, though. Nov 15 '13 at 15:07

As you noted the polynomial $$f$$ is irreducible in $$\mathbb Q[x] \iff$$ is irreducible in $$\mathbb Z[x]$$. Note also that $$f$$ has only one root inside the unit circle and all other roots outside the unit circle (use Rouche's theorem).
Now if we suppose that $$f$$ is reducible with $$f(x)=g(x)h(x)$$ (where $$g(x),h(x)\in\mathbb Z[x], \ \deg f,\deg g>0$$) then one of the polynomials $$g,h$$ has all of its roots outside the unit circle and constant term $$\pm1$$ (which is the product of its roots). This is a contradiction.