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Please help me to prove that these polynomials are irreducible over $\mathbb{Q}$:

1) $x^4+5x^3+7x^2-6$

2) $x^4-4x^3+8x^2-8x+9$

Unfortunately , it takes infinite amount of time to prove that given polynomials are irreducible according to Eisenstein's criterion. So could you recommend and show other ways, that requires less time , because such exercises will come in final exam where I should solve a problems in a short time.

Bill Dubuque
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Nicko
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3 Answers3

8

Actually Eisenstein is quick and easy here since the required shift is obvious as I explain below. The remark shows how this is a special case of general method to discover such Eisenstein shifts.

Hint $ $ for $(2)\!:\, \bmod\color{#c00}2\!:\ f(x) \equiv x^{ 4}\!+\!1\equiv (x\!+\!1)^{4} \ $ is a $\rm\color{#0a0}{prime\ power}$.

So Eisenstein works on $\,g(x) = f(x\!-\!1) \equiv x^4\ $ by $\,\color{#c00}2^2\!\nmid\! g(0)\! =\! f(-1)\equiv 1\!+\!9\equiv 2\pmod{\!\color{#c00}2^2}$

Remark $\ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are $\rm\color{#0a0}{prime\ powers\ (\equiv x^n)} \bmod p,\,$ and products of primes always factor uniquely. The same works for its shift $\,(x-c)^n,\,$ so we seek primes $\,p\,$ such that, mod $\,p,\,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). The only primes $\,p\,$ that can yield such powers are those dividing the discriminant (here by Alpha = $\,\color{#c00}2^9\cdot 3\cdot 5^2)$. Indeed, if $\,f\equiv a (x-c)^n,\,\ n> 1\,$ then $\,f\,$ and $\,f'\,$ have a common root $\,x\equiv c,\,$ hence their resultant $\, R(f,f')\equiv 0.\,$ But this is, up to sign, the discriminant of $\,f\,$ (presumed monic).

Below is a less trivial example, from this deleted question.

$ f(x)= x^4\!+\!123\:\!x^3\!-\!653\:\!x^2\!-\!386\:\!x+684\,$ $\Rightarrow\, f(x\!-\!c)= x^4\!+(123\!-\!4c)\:\!x^3+\,\cdots$

so $\,p\mid 123\!-\!4c\,\Rightarrow\:\!\bmod p\!:\ 4c\equiv 123,\,$ so $\,\color{#0af}{4(x\!-\!c)\equiv X\!-\!123},\,\ X = 4x,\,$ so

$g(X) := 4^4f(\color{#0af}{x\!-\!c}) = 4^4 f(\color{#0af}{\frac{X-123}4}) \equiv X^4 -\color{#c00}i\:\! X^2 + \color{#c00}j\:\! X - \color{#c00}k,\,$ $\ \scriptsize\begin{align}\color{#c00}i &= 101222\\ \color{#c00}j&=17432440\\ \color{#c00}k&=841514019\end{align}$

thus we infer $\,g(X)\equiv\color{#0a0}{X^4}\!\!\iff\! \gcd(\color{#c00}{i,j,k})\!=\!\color{#c00}{107}\equiv 0,\,$ so $\,4c\!\equiv\! 123\!\equiv \!16\!\iff\! \color{darkorange}{c\!\equiv\! 4}$

and indeed $f(x\!-\!\color{darkorange}4) = \color{#0a0}{x^4} + \color{#c00}{107}(x^3\!-\!19\:\!x^2\!+\!98\:\!x\!-\!148)\,$ is Eisenstein at $\,\color{#c00}{p=107}$.

Bill Dubuque
  • 282,220
4

The first polynomial is not irreducible since we have $$ x^4+5x^3+7x^2-6= (x^2 + 3x + 3)(x^2 + 2x - 2). $$ For the second, we see by the rational root test that there is no root. Hence the only way to decompose it is into two quadratic factors. By writing these as equations in the integer coefficients we obtain a contradiction (by Gauss Lemma we may assume that the coefficients are integers). Hence the second polynomial is irreducible.

Dietrich Burde
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1

Try with contradiction, say $$x^4-4x^3+8x^2-8x+9 = (x^2+ax+b)(x^2+cx+d)$$ or $$x^4-4x^3+8x^2-8x+9 = (x^3+ax^2+bx+c)(x+d)$$

Second option is valid only if polynomal has (positive) integer roots. The canidates are only $1,3,9$ and non works, so you have to try with first option.

nonuser
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