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Show that the following polynomials are irreducible over $\mathbb{Q}$.

a) $3X^4-8X^3+6X^2-4X+6$

b) $X^3-3X+1$

c) $\frac{1}{5}X^4-\frac{1}{3}X^3-\frac{2}{3}X+1$

At first I found the prime number needed for each and went through each of the three steps. However, upon closer inspection, do I need to alter each polynomial first? For a, I used p=2. However, since we're in $\mathbb{Q}$, then we could find some x such that $2x=3$ correct? Because $\frac{3}{2}$ is in $\mathbb{Q}$. For b I originally said $p=3$ and for c, $p=5$ after multiplying by $15$. Any help is appreciated, thanks.

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    32 days being a member + 11 questions asked = learn already how to properly type mahtematics in this site. – DonAntonio Nov 30 '16 at 00:03
  • A cubic with integer co-efficients is reducible over the rationals iff it has a rational root. For (b), the Rational Roots theorem implies that a rational root must equal $\pm 1$. Since neither of $\pm 1$ is a root of this cubic, it is irreducible over Q – DanielWainfleet Nov 30 '16 at 02:11

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Yes, $\,p=2\,$ works for $(a)$ and clearing denominators yields an obvious choice for $(c)$.

For $(b)$ note $\,f(x\!-\!1) = x^3-3x+3\,$ so $\,p=3\,$ works (note $\,f(x\!-\!1)\,$ is reducible iff $\,f(x)\,$ is). See this answer for the method for finding that change of variables. Note that the criterion doesn't apply to $\,x^3-3x+\color{#c00}1\,$ since $\,3\nmid\color{#c00}1.$

The only alteration of the polynomial in $(c)$ is that you need to scale it by a integer $c\neq 0$ so it has integer coefficients that satisfy the standard statement of Eisenstein's criterion. This then implies that $\,cf\,$ is irreducible over $\,\Bbb Q\,$ hence so too is $\,f.\,$

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