# Irreducibility and Splitting Fields

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.

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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Your 4th sentence says «suppose $f$ is neither irreducible nor ...» and your 5th sentence starts with «Then $f$ is irreducible...» :) –  Mariano Suárez-Alvarez Mar 12 '11 at 1:53
@ Mariano: Thanks. It's fixed now. –  Nana Mar 12 '11 at 2:34
@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 $b=e=1$. –  Arturo Magidin Mar 12 '11 at 5:54
@Arturo: Would it still be wrong if I suppose that $F=\mathbb{F}_2$ ? –  Nana Mar 12 '11 at 9:34
@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. –  Arturo Magidin Mar 13 '11 at 20:33

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$). Which means that... - Maybe using the cubic classical formula? – awllower Mar 12 '11 at 3:06 which means the$f$either splits or is irreducible in$F$. Thanks. – Nana Mar 14 '11 at 2:03 @Nana: "... which means it has all roots in$F$, hence$f$splits. So, either$f$is irreducible, or splits." – Arturo Magidin Mar 14 '11 at 2:04 I like your approach, it's not one I would have thought of. Here's an idea that seems to work except when characteristic of F is 3. If$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. - Very nice!${}$– Mariano Suárez-Alvarez Mar 12 '11 at 2:21 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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