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Show that over any field $F$, the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors.


Edited:

This is my attempt: Let $f(x)=x^3-3x+1$. Let $a_1,a_2,a_3$ be the roots of $f$. Suppose char $F\neq 2,3$. Suppose also that $f$ is neither irreducible nor splits in $F$. Then $f$ is reducible which implies that $a_1 \in F$. i.e. $f(x)=(x-a_1)g(x)$, where $g(x)\in F[x]$ is irreducible with deg $g=2$. Let $K$ be the splitting field of $g$. The $K$ is Galois over $F$. So if $\sigma \in $ Aut($K/F$), then $\sigma (a_1)=a_1$ since $\sigma $ fixes $F$ and $a_1 \in F$. Since $\sigma$ permutes the roots of $f$, WLOG suppose $\sigma (a_2)=a_3$. Then
$\sigma(\triangle) = \sigma((a_1-a_2)(a_1-a_3)(a_2-a_3))=-\triangle$.

But $\triangle^2=D(f)=81$, so $\triangle = \pm 9 \in F$, so $\triangle \in F$. Therefore, $9=-9$ $\implies 1=-1 \implies$ char $F =2$ which is a contradiction. So $f$ is either irreducible or splits in $F$.

Next suppose char $F=2$. Well, I'm not exactly sure what I can say about $f$.

I would like to know if my approach is correct and also what to do in the second case. Thanks.

ADDED:
If char $F=2$, then $f=x^3+x+1$. Suppose $b$ is a root of $f$. Then $b^2 $ is also a root, since $f(b^2)=(b^2)^3+b^2+1=(b+1)^2+b^2+1=2b^2+2=0$. Is it enough to conclude that $f$ splits?

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  • $\begingroup$ Your 4th sentence says «suppose $f$ is neither irreducible nor ...» and your 5th sentence starts with «Then $f$ is irreducible...» :) $\endgroup$ Mar 12, 2011 at 1:53
  • $\begingroup$ @ Mariano: Thanks. It's fixed now. $\endgroup$
    – Nana
    Mar 12, 2011 at 2:34
  • $\begingroup$ @Nana: The added material is wrong. Characteristic of $F$ equal to $2$ does not imply that $F$ is the field of two elements. So $be=1$ does not imply $b=e=1$. $\endgroup$ Mar 12, 2011 at 5:54
  • $\begingroup$ @Arturo: Would it still be wrong if I suppose that $F=\mathbb{F}_2$ ? $\endgroup$
    – Nana
    Mar 12, 2011 at 9:34
  • $\begingroup$ @Nana: Even if it were correct then, that wouldn't really do you much good; proving it works for a single field of characteristic two does not seem to me to be making much progress. $\endgroup$ Mar 13, 2011 at 20:33

3 Answers 3

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I like your approach, it's not one I would have thought of. Here's an idea that seems to work except when characteristic of F is 3.

If $x$ satisfies $x^3 - 3x + 1 = 0$, then so does $1 - \frac{1}{x}$. Clearly zero is not a root of the given polynomial, so this makes sense. Also if $x = 1 - \frac{1}{x}$, then $x$ would satisfy the equation $x^2 - x + 1 = 0$, and taken in combination with the cubic polynomial above, we would infer $3 = 0$.

So except for characteristic 3, the formula $1 - \frac{1}{x}$ turns out to cyclically permute the three roots of the cubic polynomial. Thus the polynomial either splits in a field F or is irreducible.

Added: Arturo has nailed the characteristic 3 case.

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There's a slight problem with your argument: from $9=-9$ you can only conclude $1=-1$ if $3$ is invertible. What if $\mathrm{char}(F) = 3$? So you may want to start by positing that $\mathrm{char}(F)\neq 2,3$.

(There is a typo in the fourth sentence: "Then $f$ irreducible" should be "Then $f$ reducible...")

So, you need to deal with $\mathrm{char}(F) = 2$ and with $\mathrm{char}(F) = 3$. If $\mathrm{char}(F) = 3$, then $f(x) = x^3+1 = (x+1)^3$, so there is nothing to do in that case.

What about $\mathrm{char}(F)=2$? Your polynomial is just $f(x)=x^3+x+1$. Suppose $\alpha$ is a root. Try to see if you can construct some other root of the polynomial using $\alpha$. For instance, $\alpha+1$ doesn't work, because you have $$(\alpha+1)^3 + (\alpha+1)+1 = \alpha^3+\alpha^2+\alpha+1+\alpha +1 = \alpha^2+\alpha+1 = \alpha^3+\alpha^2 = \alpha^2(\alpha+1)=\alpha^5\neq 0$$ but maybe some other expression involving $\alpha$ will do?

Added. So, now you have that if $\mathrm{char}(F)=2$, and $b$ is a root of $f(x)$, then so is $b^2$.

Can $b=b^2$? That is, can you simply get the same root again? If $b=b^2$, then $b^2-b=0$, so $b(b-1)=0$. Since you are in a field, either $b=0$ or $b=1$. But neither $b=0$ nor $b=1$ are roots of $f(x)$. So that means that if $b$ is a root of $f$, then $b\neq b^2$, and b^2$ is also a root of $f(x)$.

So if $f(x)$ has at least one root in $F$, then it has at least two roots in $F$ (namely, $b$ and $b^2$). Which means that...

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  • $\begingroup$ Maybe using the cubic classical formula? $\endgroup$
    – awllower
    Mar 12, 2011 at 3:06
  • $\begingroup$ which means the $f$ either splits or is irreducible in $F$. Thanks. $\endgroup$
    – Nana
    Mar 14, 2011 at 2:03
  • $\begingroup$ @Nana: "... which means it has all roots in $F$, hence $f$ splits. So, either $f$ is irreducible, or splits." $\endgroup$ Mar 14, 2011 at 2:04
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Here is one way to do it using almost no theory, just playing with algebra.

Suppose the polynomial has a root $a$ in $F$. If you divide $x^3 - 3x + 1$ by $x-a$ the quotient is the polynomial $$ x^2 + ax + a^2 - 3 $$ in $F(a)[x]$. From the quadratic formula (assuming characteristic $\neq 2$ here, for the moment, I guess) you can see that this will have its roots in $F(a)$ if and only if $12-3a^2$ is a square in $F(a)$.

You may know that any element in $F(a)$ can be written in the form $pa^2 + qa + r$ for some $p,q,r$ in $F$. The idea is to square this symbolic expression, rewrite it as a polynomial in $a$ of degree at most $2$, and then see if values $p, q, r$ in $F$ can be found making this equal to $-3a^2 + 0a + 12$.

Using the identity $a^3 = 3a - 1$ (from the fact that $a$ is a root of the given polynomial) and $a^4 = 3a^2 - a$ (from multiplying the previous identity by $a$) one gets $$ (pa^2 + qa + r)^2 = (3p^2 + 2pr + q^2) a^2 + (-p^2 + 2qr + 6pq) a + (r^2 - 2pq). $$ So the goal is to find $p,q,r$ satisfying $3p^2 + 2pr + q^2 = -3$, $-p^2 + 2qr + 6pq = 0$, and $r^2 - 2pq = 12$. Since we don't know anything about $F$, we might optimistically look for these $p, q, r$ in $\mathbb{Z}$. This system of $3$ polynomial equations in $3$ integer unknowns can be fed to software, and one finds that e.g. $p = 2$, $q = 1$, and $r = -4$ give a solution (no matter what $F$ is!). So the roots of $x^3 - 3x + 1$ are all in $F(a)$, and we can actually write formulas for them: $a$, $\frac{-a + (2a^2 + a -4)}{2}$, and $\frac{-a - (2a^2 + a - 4)}{2}$, which simplify to $a$, $a^2 - 2$, $-a^2 - a + 2$.

Although we assumed characteristic $\neq 2$ to use the quadratic formula, we can immediately check that the single identity $a^3 - 3a + 1 = 0$ is indeed enough to ensure that $$ (x-a)(x - (a^2 - 2)) (x - (-a^2 - a + 2)) = x^3 - 3x + 1. $$ [In detail: expanding, the coefficient of $x^3$ is $1$ on the nose, the coefficient of $x^2$ is $0$ on the nose, and the coefficient of $x$ is a polynomial in $a$ that, when divided by $a^3 - 3a + 1$, has a remainder of $-3$; similarly the constant term is a polynomial in $a$ that, when divided by $a^3 - 3a + 1$, has remainder of $1$.]

So if $x^3 -3x + 1$ has one root in $F$, it splits in $F$ for the reason that we can explicitly write the other two roots as polynomials in the one that we already have.

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