I do not have an idea on how to approach this problem. Similar problems that I found online try to show that there does not exist a linear factor of the given equation in $\mathbb{Q}$. But my question is, how is that sufficient to prove that the polynomial is irreducible? It could also be a product of two quadratic polynomials which do not have a solution in $\mathbb{Q}$. Any help is appreciated!
Prove that the polynomial $x^4 -5x^2+x+1$ is irreducible over the ring $\mathbb{Z}[x]$.
abstract-algebrairreducible-polynomialsring-theory
Related Solutions
Answer to question 1: Consider de polynomial $(x^2+1)^2\in\Bbb Q[x]$. It has no rational roots and it is reducible over $\Bbb Q$. Your reasoning fails.
The absence of roots only guarantees that there are no factors of degree $1$.
Your polynomial has degree $4$, since it has no rational roots, it has no linear factors (that is factors of degree $1$), but it could have two factors of degree $2$.
Assume it does and try to reach a contradiction. If you do, then it is irreducible over $\Bbb Q$.
Answer to question 2: The factors don't need to be monic polynomials, but they can be. If you find a factorization in which the factors aren't monic, you just have to multiply by a certain constant to make the factors monic.
Answer to question 3: If you find a factorization $(x-\alpha )q(x)$, where $q(x)$ is a polynomial with rational coefficients of degree $3$, then the polynomial will have a rational root, namely $\alpha$ and you have estabalished it doesn't.
Answer to question 4: The coefficients need not be integers, but (the second) Gauss's lemma allows us to assume the coefficients are integers, making the calculations much simpler.
Looking at questions like these modulo a prime is not guaranteed to work but it is definitely something that should be in your toolbox.
- The theoretical justification comes from a corollary to Gauss' lemma and states that a monic polynomial $f(x)\in\Bbb{Z}[x]$ is irreducible in $\Bbb{Q}[x]$ if and only if it is irreducible in $\Bbb{Z}[x]$.
- Should it happen that $f(x)=g(x)h(x)$ non-trivially in $\Bbb{Z}[x]$, then both $g$ and $h$ can be assumed to be monic. Reducing that equation modulo a prime $p$ tells that $\overline{f}(x)=\overline{g}(x)\overline{h}(x)$ in $\Bbb{Z}_p[x]$. As $g$ and $h$ were monic, we have $\deg g=\deg \overline{g}$ and $\deg h=\deg \overline{h}$.
- Checking for irreducibility in $\Bbb{Z}_p[x]$ is often easier because, at least in principle, we can try all the alternatives as the field of coefficients is finite. Should it happen that $\overline{f}$ is irreducible then we can immediately conclude that $f$ is also irreducible.
- Factoring in $\Bbb{Z}_p[x]$ is also easier (for small $p$ it is feasible to build a list of low degree irreducible polynomials and use those much the same way you use a list of small primes to factor smallish integers). Even if $\overline{f}$ is not irreducible we may still get information about the degrees of $g$ and $h$. Behold!
Let's look at your example polynomial $f(x)=x^5+2x+1$ modulo primes $p=2$ and $p=3$. Either choice of $p$ leads to a proof of irreducibility.
- When $p=2$ we get $\overline{f}=x^5+1=(x+1)(x^4+x^3+x^2+x+1)$. Here it is not immediately obvious that the degree four polynomial is irreducible, but it does follow from the facts that A) it has no zeros modulo two (only two choices to test!), B) $x^2+x+1$ is the only irreducible quadratic in $\Bbb{Z}_2[x]$, and e.g. long division shows that the quartic is not divisible by that either. How does this help us? What we have at this point is that if $f$ is not irreducible it must be a product of a linear factor and a quartic factor. But the rational root test tells you that $f(x)$ does not have a linear factor. We are done - $f$ must be irreducible! Here the key was that a modulo $2$ consideration showed that $f$ cannot have a quadratic factor!
- As Lulu pointed out, $f(x)$ is irreducible modulo $p=3$. It is, of course, easy to see that none of $f(0),f(1),f(2)$ are congruent to $0$ modulo $3$, so $f(x)$ has no linear factors. To exclude the possibility of a quadratic factor takes a bit more work here than it did with $p=2$. Anyway, with a bit of work we can show that $p_1(x)=x^2+1$, $p_2(x)=(x-1)^2+1$ and $p_3(x)=(x+1)^2+1$ are all the irreducible monic quadratics in $\Bbb{Z}_3[x]$. It takes a bit of work to show that none of them are factors of $f(x)$. We can do that in one swoop if we use the fact that $$(x-1)(x+1)p_1(x)p_2(x)p_3(x)=x^8-1$$ (this comes from a general related result, ask if you want to know more), and then compute with Euclid's algorithm that $\gcd(x^8-1,f(x))=1$ in $\Bbb{Z}_3[x]$. Anyway, we then know that $f$ is irreducible modulo three, and we are done.
For comparison, the methods in other answers are all very fine. They work well here as well as in many other cases. They work in many cases where modular techniques may fail. But, modular tricks are good to know. For example, instead of your $f(x)$ consider the polynomial $$ f(x)=x^5+2x+7^n, $$ where $n$ is either a parameter or some ridiculously large natural number. AFAICT all the other answers fail or lead to uncomfortably many cases. But, the above proof using reduction modulo $3$ goes through verbatim. The modulo $2$ argument survives if you can do the part with the rational root test and show that $f(\pm 7^\ell)\neq0$ for all $\ell, 0\le\ell\le n$. One term will usually dominate the others.
Best Answer
Write $f(x):=x^4-5x^2+x+1$. Since $f$ is monic, whenever $f=gh$ in $\mathbb{Z}[x]$, we may assume without loss of generality that both $g$ and $h$ are monic. You already know that $\operatorname{deg}(g)=1$ is impossible. Assuming $\operatorname{deg}(g)=\operatorname{deg}(h)=2$, we have $$x^4-5x^2+x+1=(x^2+ax+b)(x^2+px+q)\\=x^4+(a+p)x^3+(q+ap+b)x^2+(aq+bp)x+bq$$ for some integers $a,b,p,q$. Comparing coefficients, either $b=q=1$ or $b=q=-1$, but it is easy to check that both cases lead to contradiction. Hence $f$ is irreducible in $\mathbb{Z}[x]$.
Now use Gauss's lemma to conclude that $f$ is irreducible in $\mathbb{Q}[x]$.