# Is there a quick way to compute the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$?

Is there a way to find the degree of the splitting field of $x^3+x+1$ over $\mathbb{Q}$?

Just analyzing the roots shows that the polynomial is separable, so I suppose the splitting field would be a Galois extension. However, the roots are not easy to get a handle on, so it's not obvious to me what roots would need to be adjoined to $\mathbb{Q}$ to get the degree.

What is the right way to do this? Thanks.

• It's certainly irreducible over $\mathbf Q$, and since we're in characteristic zero this implies separable. – Dylan Moreland Jan 22 '12 at 0:59
• A good thing to look up is the discriminant of a cubic. – Dylan Moreland Jan 22 '12 at 1:09
• @Dylan thanks, I'll read into that. – Waldott Jan 22 '12 at 1:10

The degree is at least 3 and at most $6=3!$. So you only have to decide whether it's 3 or 6. Since $x^3+x+1$ has only one real root, the other two roots are complex conjugates and so conjugation is an automorphism of the splitting field. Since conjugation has order 2, the degree is 6.

You can avoid Galois theory. The complex roots are roots of a quadratic. Since they cannot be in the real field generated by the real root, the splitting field must be a quadratic extension of the real field and so has degree $2\cdot3=6$.

• Wonderful, thank you lhf. – Waldott Jan 22 '12 at 1:08
• Why can't it be 4 or 5? – Matt Jan 22 '12 at 1:13
• @Matt, it must divide 6. – lhf Jan 22 '12 at 1:14
• Galois theory not needed. Let $\alpha$ be a root, let $K$ be a splitting field, then degree of $\alpha$ over the rationals is 3, so degree of $K$ over the rationals is a multiple of 3 by the tower law. – Gerry Myerson Jan 22 '12 at 1:20
• @GerryMyerson Of course - the splitting field contains $\mathbb Q(\alpha)$ as a subextension, which is of degree 3 (since the polynomial is irreducible). Thanks – Matt Jan 22 '12 at 1:24

If $f(x) = x^3 + ax^2 + bx + c = (x - r_1)(x - r_2)(x - r_3)$ is an irreducible (it is important to check this first) cubic and $K$ is its splitting field, then $K$ has degree either $3$ or $6$. Moreover, the following are equivalent:

The equivalence of the first two conditions follows from the fact that $K$ has Galois group either $C_3$ or $S_3$ and the fact that the Galois group of a Galois extension has the same size as the degree of the extension. The equivalence to the third extension follows from the fact that $$(r_1 - r_2)(r_2 - r_3)(r_3 - r_1)$$

is a square root of the discriminant in $K$ and invariant under $C_3$ and not under $S_3$, and lies in $\mathbb{Q}$ if and only if it is invariant under the Galois group by the fundamental theorem of Galois theory.

A sufficient, but not necessary, condition for $K$ to have Galois group $S_3$ is that it has exactly one real root. There are at least two ways to see this:

• The subfield of $K$ generated by the real root embeds into $\mathbb{R}$, so cannot contain the other two roots. Thus it must be a proper subfield of $K$.
• Complex conjugation acts as a transposition, so gives an element of order $2$ in the Galois group.

It also suffices to find a prime $p$ such that $f(x)$ factors into the product of a linear and quadratic factor $\bmod p$ by Dedekind's theorem.

• Thanks for this nice explanation! – Waldott Jan 24 '12 at 21:44

Just some culture, in case of curiosity. If you take any prime $q$ that has an integral expression as $$q = 2 x^2 + x y + 4 y^2,$$ such as $2, \; 5, \; 7\; 19,$ then $$z^3 + z + 1 \equiv 0 \pmod q$$ has no roots in integral $z.$

For the special case of $31,$ we have $$z^3 + z + 1 \equiv (z -3)(z - 14)^2 \pmod {31}.$$ For any other prime $$p = x^2 + x y + 8 y^2,$$ such as $47, \; 67, \; 131, \; 149,$ then $$z^3 + z + 1 \equiv 0 \pmod p$$ has three distinct roots in integral $z$ and factors as three distinct linear factors.

For any prime $r$ with $(-31|r) = -1,$ such as $3, \; 11, \; 13, \; 17,$ then $$z^3 + z + 1 \equiv 0 \pmod r$$ has a single non-repeated root, so the cubic factors as a linear times a quadratic.

Well, why not. It turns out that every integer $n,$ positive or negative or $0,$ has an expression in integers as $$n = x^2 + x y + 8 y^2 + z^3 + z,$$ where we deliberately strip off the 1. The difficult question is, what integers $n$ have an expression in integers as $$n = 2 x^2 + x y + 4 y^2 + z^3 + z \; ?$$ Certainly not all, $n = \pm 1$ do not work. The first few, in absolute value, that do not work are $$\pm 1, \; \pm 869, \; \pm 25171, \; \pm 21118439, \; \pm 611705641,$$ these being the odd integers $u$ with $$27 u^2 - 31 v^2 = -4.$$ The first few even values of $u$ are $$30 = 3^3 + 3,$$ $$729090 = 90^3 + 90,$$ $$17718345150 = 2607^3 + 2607,$$ so these are easily expressed as $2 x^2 + x y + 4 y^2 + z^3 + z$ with both $x,y = 0.$

Well, somebody did mention the discriminant of a cubic, we have $$\mbox{disc}_z \left( z^3 + z + 1 \right) = -31$$ and $$\mbox{disc}_z \left( z^3 + z + u \right) = -4 - 27 u^2 = -31 v^2.$$ So there.

• Thanks, this is an interesting post to add. – Waldott Jan 24 '12 at 21:45