# Galois theory: splitting field of cubic as a vector space

I'm trying to express the splitting field of a cubic equation as a vector space over the rationals. Specifically I am looking for a set of six independent vectors that span the space. If the roots are a,b, and c then I know I need to be able to express all expressions in a,b and c up to second degree: namely,

a, a^2

b, b^2

c, c^2

ab, bc, ca

a^2b, b^2c, c^2a, ab^2, bc^2, ca^2

Obviously there are 15 of these and they are not all independent. Given a and b, I easily get c as a linear term in a, b, and (a+b+c). Likewise c^2 is not independent of a^2 and b^2. And I find that I can generate ab etc as linear combinations of c, c^2, and abc.

So what are my six vectors? I have allowed myself 1, a, b, a^2, and b^2 ....this gives me five and I think they are all linearly independent. I'm only allowed one more and I can't get it to work. I'm inclined to try (a^2b + b^2c + c^2a) because I'm pretty sure I need it, but having done so I can't see how I generate terms like a^2b on their own.

Any ideas? I'd be especially interesteds if there is a basis which is more symmetric in some sense than the arbitrary collection of terms which I'm cobbling together.

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A cubic extension has degree $3$, not degree $6$. Or do you mean the splitting field of a cubic? In that case, it depends on the Galois group of the cubic. – Qiaochu Yuan Jul 3 '11 at 5:01
Have you tried a Gröbner basis algorithm (Buchberger?) on the obvious set of generators? Also, is it immediately clear that you don't need to worry about monomials like $a^2b^2$? – Jyrki Lahtonen Jul 3 '11 at 5:25
@qiaochu Although English is my first language, I am very weak on the terminology in this subject. Yes, I think I meant the splitting field of the cubic; and I know that some cubics have a different Galois group, but I meant the most general case of S3. – Marty Green Jul 3 '11 at 5:45
@Jyrki I don't know anything about the methods you refered to; but yes, I forgot about terms like a^2b^2 when I posted my question. In fact, I get them the same way I got terms like ab; since a^2b^2c^2 is an integer, I divide by c^2 which is the same as multiplying by a linear term in c. I think. – Marty Green Jul 3 '11 at 5:48
@Marty: Sorry about bringing up Gröbner here. Too heavy a tool for this job. – Jyrki Lahtonen Jul 3 '11 at 8:17

The simplest basis I can think of consists of monomials $\mathcal{B}=\{1,a,a^2,b,ab,a^2b\}$. One way to see this goes as follows. Let's first write $$P(x)=(x-a)(x-b)(x-c)=x^3-s_1x^2+s_2x-s_3,$$ with the elementary symmetric polynomials $s_1=a+b+c$, $s_2=ab+bc+ac$ and $s_3=abc$. If $K$ is your base field, I claim that all the monomials $a^ib^jc^k$ can be written as linear combinations of monomials from the set $\mathcal{B}$ with coefficients from the field $L=K(s_1,s_2,s_3)$ (in your case surely $s_1,s_2,s_3\in K$, so the combinations are really $K$-linear).
The first and most obvious thing is to replace everywhere $c$ with $s_1-a-b$. After that we only need to take care of monomials $a^ib^j$. The quantities $a$ and $b$ are zeros of $P(x)$, so we know how to replace $a^i, i\ge 3$ and $b^j, j\ge3$ with lower powers. So we are left with the 9 monomials $\mathcal{B}\cup\{b^2,ab^2,a^2b^2\}$. Substituting $c=s_1-a-b$ to the equation $ab+ac+bc-s_2=0$ gives us a relation $$b^2=ab+a(s_1-a-b)+b(s_1-a)-s_2$$ that allows us to write $b^2$ as an $L$-linear combination of $a^2,ab,a,b$ and $1$. When we do this substitution to a monomial like $ab^2$ or $a^2b^2$ we introduce higher powers of $a$. But we can reduce these to lower powers of $a$ as before. It may feel like $s_3$ wasn't really used, but it did make an appearance, when we reduced the higher powers of $a$ and $b$.
In retrospect the basis is obvious given that this is the usual way of getting a basis for a tower of field extensions. $L(a)/L$ has a basis $\{1,a,a^2\}$ and $L(a,b)/L(a)$ a basis $\{1,b\}$. A basis for $L(a,b)/L$ is gotten by including all the products of elements of the two bases. – Jyrki Lahtonen Jul 3 '11 at 18:41