I have a "cool" polynomial that I want to prove is irreducible. Let $$P(X) = X^7 - X - 1 \in (\Bbb Z/7\Bbb Z)[X].$$

I wish to show that if one root of $P$ is in $\Bbb Z / 7 \Bbb Z$ then all roots of $P$ are in there.

I've made a few observations that I'd like (maybe) to develop but I'm not quite sure how I might do so! Here they are;

Suppose that $\alpha$ is a root of $P$ so that $P(\alpha) = 0$. Then $X-\alpha \mid P$. If we divide out by $X-\alpha$ we get

$$P(X) = (X-\alpha)(X^6 + \alpha X^5 + \alpha^2 X^4 + \alpha^3 X^3 + \alpha^4X^2 + \alpha^5 X + \alpha^6 - 1) = \ell(X)s(X).$$

Now suppose that $\beta \neq \alpha$ is another root of $P$. Then $\beta - \alpha \neq 0$ so $\beta$ must be such that $s(\beta) = 0$. If we evaluate $s$ at $\beta$ we get

$$s(\beta) = \beta^6 + \alpha\beta^5 + \alpha^2\beta^4 + \alpha^3\beta^3 + \alpha^4 \beta^2 + \alpha^5 \beta + \alpha^6 - 1.$$

If we define $\pi : \alpha \mapsto \beta$ then $\pi(s(\beta)) = s(\beta)$. In fact it seems if we apply $\pi$ to $P$ then we get $P(X) = (X-\beta)(X^6 + \beta X^5 + \beta^2X^4 + \beta^3X^3 + \beta^4X^2 +\beta^5X +\beta^6 - 1)$ and we know that $\alpha$ is a root of this sextic factor. Denote by $S(X)$ the division of $P$ by $X - \beta$. Then $X-\alpha \mid S(X)$ so that $P(X) = (X-\alpha)(X-\beta)F(X)$ where $F$ is some degree $5$ factor.

Can I use this method to systematically "drag" linear factors out of $P$ based on the assumption that $P$ has a root in $\Bbb Z/7 \Bbb Z$, and use this to prove that $P$ is irreducible?

  • $\begingroup$ It's vacuously true because $P$ has no roots in $\Bbb Z/7$ since $\,P(n) = -1$ by $\,n^7 = n$ by little Fermat. Perhaps you meant to ask something different? $\endgroup$ – Bill Dubuque Sep 3 '17 at 14:04
  • $\begingroup$ @BillDubuque Did you read the whole question? I was trying to prove that $P$ is irreducible using the assumption that if $P$ has one root in $\Bbb Z/7\Bbb Z$, then all of the roots are in there. $\endgroup$ – Edward Evans Sep 3 '17 at 16:44
  • $\begingroup$ @BillDubuque I suppose this could be closed as a duplicate of this old thread. Because I'm also involved in that other thread. I should not use my moderator powervote to close. FWIW other threads have been merged with that one in the past. $\endgroup$ – Jyrki Lahtonen Sep 3 '17 at 17:46
  • $\begingroup$ I'm guessing that you'd like advice about your own attempt to prove $P$ is irreducible, based on a proposed property that "if one root of $P$ is in the field, then all the roots of $P$ are in the field". This is not the definition of a polynomial being irreducible, and it isn't equivalent in general. There are many reducible polynomials with the proposed property, and of course all irreducible polynomials have that property, so perhaps your approach to proving irreducibility needs clarification. $\endgroup$ – hardmath Sep 4 '17 at 23:47

Hint: Show that if $\alpha$ is a zero of $P(x)$ then so is $\alpha+1$. Induction shows that $\alpha+k$ is a zero for all $k$.

See this thread for more discussion and links. There are no roots in $\Bbb{Z}/7\Bbb{Z}$. All seven of them are in the field $\Bbb{F}_{7^7}$.

  • 2
    $\begingroup$ It follows from Galois theory that if $f(x)$ is an irreducible polynomial over $K:=\Bbb{Z}/p\Bbb{Z}$, and $\alpha$ is a zero in an extension field, then all the zeros are in $K(\alpha)$. That has also been covered in an earlier question. $\endgroup$ – Jyrki Lahtonen Sep 3 '17 at 13:41
  • $\begingroup$ There is a free software called SAGE which is very powerful and gave the factors in $GF(7^7)$ in milliseconds $(x + 2z_7) (x + 2z_7 + 1) (x + 2z_7 + 2) (x + 2z_7 + 3) (x +2z_7 + 4) (x + 2z_7 + 5) (x + 2z_7 + 6)$ $\endgroup$ – Raffaele Sep 3 '17 at 17:30

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