Eisenstein's Criterion with an example

Question

Wolfram Alpha says $x^5 -x^2 +1$ is irreducible over $\mathbb{Z}$. Is there any way to prove it by Eisenstein's Criterion?

I tried to translate this function. I translated the function a couple of times by some integers, but none of the translations worked. Any hint would be appreciated. Thanks so much.

@Andrei Polynomials can have no roots and yet be reducible. Consider $(x^2 +1)(x^2 + 2)$ over $\mathbb{Z}$, for example. — Viktor Vaughn, Mar 23 '20 at 22:59
No, it's not possible to apply Eisenstein's criterion in this case. The discriminant of $f$ is $7 \cdot 431$, and modulo these primes $f$ does not factor as a power of a single irreducible linear factor. Bill Dubuque had a nice post on this, but it unfortunately seems to have been deleted... Keith Conrad has some good notes on Eisenstein polynomials and totally ramified primes, too. — Viktor Vaughn, Mar 23 '20 at 23:04

score 2 · Accepted Answer · answered Mar 23 '20 at 23:07

By here an Eisenstein shift cannot work. But it follows easily from irreducibility mod $2\!:\,$ over $\Bbb F_2$ it has no roots so no linear factors, so if it splits it has an irreducible quadratic factor $g$, therefore in $\, \Bbb F_2[x]/g \cong \Bbb F_{\color{#c00}4}\!:\,$ $\,\color{#c00}{x^3 = 1}\,$ so $\ 0 = f = x^2(\color{#c00}{x^3})-x^2+1 = 1,\,$ contradiction.

Remark $ $ Above is a special case of a general polynomial irreducibility test over finite fields - which is an an efficient analog of the impractical Pocklington-Lehmer integer primality test.

score 1 · Answer 2 · answered Mar 23 '20 at 22:58

1

Hint: Expand $ (x+a)^5 - (x+a)^2 + 1$

Compare the coefficients of $x^4 , x^2$.

Show that $ \gcd ( 5a, 10a^3 - 1) = 1$.

$ $

Hence conclude that Einstein's criterion cannot prove it is irreducible.

answered Mar 23 '20 at 22:58

Calvin Lin

68,864

Thank you very much. – Matha Mota Mar 24 '20 at 10:17

Eisenstein's Criterion with an example

2 Answers2

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