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I want to show that the following polynomial is irreducible in $\mathbb{Z}[X]$:

$$f(X) = X^4 -4X^3 -4X^2 + 16X - 8.$$

I thought it was irreducible mod 3 but someone pointed out that was wrong.

Any suggestions? Thanks.

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But $\rm\,mod\ 3\,$ it is $\rm\,(x^2+x-1)^2.$ –  Math Gems Feb 17 '13 at 17:23
    
Oops, yeah you're right it is reducible mod 3. But it is X^4 mod 2, right? So that's reducible as well. –  MrReese Feb 17 '13 at 17:24
    
@MrReese Do you know how to prove that this polynomial doesn't have any rational roots without resorting to reduction $\pmod p$ ? –  Git Gud Feb 17 '13 at 17:25
    
@GitGud No rational roots does not imply irreducible. –  Math Gems Feb 17 '13 at 17:38
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@GitGud But the OP seeks to prove irreducibility, not nonexistence of rational roots, so I don't see how the latter pertains. Many students mistakenly conflate the two - which is why I thought it worthy to remark about their difference. –  Math Gems Feb 17 '13 at 18:37
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2 Answers 2

up vote 8 down vote accepted

Let $X=2y$. Substitute and divide through by $8$. We get an expression of shape $P(y)=2y^4+a_3y^3+a_2y^2+a_1y -1$, where the $a_i$ are divisible by $2$.

Let $y=\dfrac{1}{z}$, and consider the polynomial $Q(z)=z^4P(1/z)=-z^4+a_1z^3+a_2z^2+a_3z+2$. This is irreducible over the rationals if and only if $P(y)$ is. Now use the Eisenstein Criterion.

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I don't really understand the substituion $y=\frac{1}{z}$. You don't even end up with a polynomial anymore. I don't get it. –  Git Gud Feb 17 '13 at 17:31
    
@git $\rm\,\ x^4/8\cdot f(2/x)\, =\, -x ^4 + 4x^3 - 2x^2 - 4x + 2\ $ so Eisenstein applies. Alternatively one can use Newton polygons. –  Math Gems Feb 17 '13 at 17:33
    
The polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ is irreducible over the rationals if and only if the polynomial $a_0z^n +a_1z^{n-1}+\cdots +a_n$ is irreducible. –  André Nicolas Feb 17 '13 at 17:35
    
@AndréNicolas I really don't follow... Consider $p(x)=x^2+1$. If you let $\displaystyle x=\frac{2}{z}$ you end up with $\displaystyle p(\frac{2}{z}$)=$(\displaystyle \frac{2}{z})^2+1$ which isn't even a polynomial. –  Git Gud Feb 17 '13 at 17:42
    
Certainly it is not a polynomial. But multiply through by $z^2$. From the irreducible $x^2+1$ you get the irreducible $4+z^2$. –  André Nicolas Feb 17 '13 at 17:45
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Hint $\rm\ \ f(X)\:\ =\:\ a X^n + b p X^{n-1} + c p^2 X^{n-2} +\cdots + d p^{n-1} X + e p^{n-1} ,\ \ p\nmid ae$

$\rm\,\ \Rightarrow\,\ f(pX) = (a p X^n + b p X^{n-1} + c p X^{n-2} +\cdots + d p X + e)\, p^{n-1}$

Thus $\rm\:f(pX)/p^{n-1}\,$ is a reversed Eisenstein polynomial, so irreducible, hence so too is $\rm\,f.$

More generally, see Newton polygons - the master method behind many irreducibility tests.

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That is a nice technique –  Jonathan Feb 17 '13 at 18:20
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