I've been scratching my head lately over this problem in my textbook:
Prove that the polynomial $x^4 + x^3 +x^2 +x +1$ is irreducible over $\mathbb{Q}$.
I've done some research and found this link, but they talk about Eisenstein's criterion, which we haven't covered in our math class just yet. Is there a general strategy where we can show whether a polynomial is irreducible over a field?
My textbook really didn't go into depth on the topic of irreducibility of polynomials, but this Wiki link somewhat helps. Perhaps the rational root theorem may be helpful here, but how would I go about starting this proof?
EDIT
Please see the first comment for my initial strategy at showing the required result.
Best Answer
A couple of strategies:
This may be the only strategy you can use at this point (without Eisenstein's Criterion). If you are able to show that your polynomial (let's say $p(x)$) has no rational roots (by the rational root test for instance), then $p(x)$ has no linear factors in $\Bbb Q[x]$. Thus, since $p(x)$ is quartic, if it is to be reducible, then it must factor as two quadratics. This is because a quartic is either factored as: 4 linear, 1 linear and 1 cubic, 2 linear and 1 quadratic, or 2 quadratics, if it is reducible. So you suppose that $p(x)=(x^2+ax+b)(x^2+cx+d)$ for some $a,b,c,d\in\Bbb Z$. Then you compare the coefficients of the terms of $p(x)$ with the expanded form of that polynomial involving $a,b,c,d$ to see if there is a solution. If there cannot be $a,b,c,d$ satisfying these requirements, then $p(x)$ must be irreducible in $\Bbb Q[x]$.
You can use Eisenstein's Criterion here (although if you haven't learned about it this may be a problem). However, it will not work immediately. You can use the result (may require a proof) that $f(x)$ is irreducible if and only if $f(x+a)$ is irreducible for some $a\neq 0$ in the field you are working with. Thus, you can show that $f(x+a)$ is irreducible for some $a\in\Bbb Z$ by Eisenstein's Criterion, which will allow you to conclude that $f(x)$ is irreducible in $\Bbb Q[x]$.
Another way to show this polynomial is irreducible in $\Bbb Q[x]$, which is a more interesting approach (although not expected) is the following. Using the aforementioned theorem of $f(x)$ being irreducible if and only if $f(x+a)$ is irreducible, one can show that $f(x)=x^{p-1}+x^{p-2}...+x+1$ is irreducible in $\Bbb Q[x]$ for any positive prime $p$. One also needs to use Eisenstein's Criterion here. In this case, for $p=5$, we see that the polynomial in question is irreducible in $\Bbb Q[x]$.
Proof for strategy 3:
I will leave it to you to prove $f(x)$ is irreducible if and only if $f(x+a)$ is irreducible. Define $\phi_p(x)\in\Bbb F[x]$ as $\phi_p(x)=\frac{x^p-1}{x-1}$ for a positive prime $p$. Then we know that $\phi(x)=x^{p-1}+x^{p-2}+...+x+1$. Evaluating at $x+1$, we have: $$\begin{align}\phi_p(x+1)&=\frac{(x+1)^p-1}{x+1-1}\\&=\frac{1}{x}\sum_{n=0}^p\binom{p}{n}x^n-1\\&=x^{p-1}+\binom{p}{1}x^{p-2}+...+\binom{p}{p-2}x+\binom{p}{p-1}\end{align}$$ Note that $p\mid\binom{p}{k}$ for every $k\in\{1,2,...,p-1\}$, but $p\nmid 1$ and $p^2\nmid p=\binom{p}{p-1}$. Thus, by Eisenstein's Criterion and the fact that $f(x)$ irreducible iff $f(x+a)$ irreducible, we have that $x^{p-1}+x^{p-2}+...+x+1$ is irreducible in $\Bbb Q[x]$.