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How I can check if the polynomial $x^4 + x^3 + x^2 + x +1$ is Irreducible or Not over $Z$ ?

Here, Eisenstein's criteria isn't applicable neither Mod P irreducible Test.

How should I proceed?

user458361
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3 Answers3

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Actually Eisenstein's Criterion is often used in this case after substituting, say, $x=y+1$. Then the polynomial becomes $y^4+5y^3+10y^2+10y+5$ which is obviously irreducible by Eisenstein for $p=5$, and this implies the original polynomial is irreducible as well.

sharding4
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  • Really? Can we do like this? – user458361 Jun 28 '17 at 22:16
  • Sure. If you take a look at the linked post here -> https://math.stackexchange.com/q/87609/254075 called Eisenstein with a twist, it shows that same trick can be used to show the $p^{th}$ cyclotomic polynomial is irreducible for any $p$. – sharding4 Jun 28 '17 at 22:27
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In this particular case you know how it factors over $\mathbb{C}$, and from that the only way in which it factors over $\mathbb{R}$: $$\left(x-e^{2\pi i/5}\right)\left(x-e^{-2\pi i/5}\right)\left(x-e^{4\pi i/5}\right)\left(x-e^{-4\pi i/5}\right)$$

$$\left(x^2-2\cos(2\pi/5)+1\right)\left(x^2-2\cos(4\pi/5)+1\right)$$

Since $2\cos(2\pi/5)=\frac{1}{2}(\sqrt{5}-1)\not\in\mathbb{Q}$, you can conclude it is not reducible over $\mathbb{Z}$.

Suppose you did not know the value of $2\cos(2\pi/5)$ or $2\cos(4\pi/5)$ with radicals. You can still say that the only way to factor this polynomial over $\mathbb{R}$ (up to unit scalars) is

$$\left(x^2-2\cos(2\pi/5)+1\right)\left(x^2-2\cos(4\pi/5)+1\right)$$

Now, if the polynomial factors over $\mathbb{Z}$, then this must be its factorization (well, or negate both factors), since the only thing you could play with would be scalars. But scaling to give one of these factors the leading term $nx^2$, the other would have to have leading term $\frac{1}{n}x^2$, with $\frac{1}{n}$ not an integer.

So the only way the polynomial factors over $\mathbb{Z}$ is if $2\cos(2\pi/5)$ is an integer. It's clearly positive and less than $2$. So it would have to be $1$. But $\cos(2\pi/5)$ is not equal to $1/2$, because $\cos(2\pi/6)$ is equal to $1/2$.

2'5 9'2
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    how do you know what $2\cos(2π/5)$ is, just curious – Saketh Malyala Jun 28 '17 at 00:34
  • One answer is just knowing that's what it is. Like knowing that $\cos(\pi/6)=\sqrt{3}/2$. But if you want to derive why, angle addition gives: $\cos(5x)=\cos(x+4x)=\cdots=16\cos^5(x)-20\cos^3(x)+5\cos(x)$. So with $x=2\pi/5$, you have that $\cos(2\pi/5)$ is a solution to $1=16X^5-20X^3+5X$. Moving terms around and factoring out the obvious factor, $0=(X-1)(16X^4+16X^3-4X^2-4X+1)$. $X=\cos(2\pi/5)$ is clearly not $1$. So it's a solution to $0=16X^4+16X^3-4X^2-4X+1$. This is a perfect square: $0=(4X^2+2X-1)^2$. There are two solutions from this, and only one is positive: $(-1+\sqrt{5})/4$. – 2'5 9'2 Jun 28 '17 at 04:19
  • oh well. i wouldve never gotten anywhere close to that. but thank you :) – Saketh Malyala Jun 28 '17 at 04:26
  • I edited to offer a solution that does not require knowing the radical value of $\cos(2\pi/5)$. – 2'5 9'2 Jun 28 '17 at 04:36
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For any prime p, the pth cyclotomic polynomial

$$p(x) = \frac{x^p-1}{x-1}= x^{p-1}+ x^{p-2}+...x+1$$

Is Irreducible over Q. And If its Primitive polynomial than its also Irreducible over Z.

user458361
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