Solving $x^{3} - 2 = 0$ and the field $\mathbb{Q}(\sqrt[3]{2})$

Question

Let's start with a simple polynomial $f(x) = x^{3} - 2 \in \mathbb{Q}[x]$. It is known that $f$ is irreducible in $\mathbb{Q}[x]$ and hence without any further information (i.e. working in field $\mathbb{Q}$ and ring $\mathbb{Q}[x]$) we can not distinguish the $3$ roots of $f$. The distinction between roots is only possible once we have the splitting field of $f$ available.

Now consider the quotient $\mathbb{Q}[x]/(f(x))$ which is a field (because $f$ is irreducible) and this field contains a member $x + (f(x))$ which is a root of $f(x)$. Moreover this field is isomorphic to $\mathbb{Q}(\sqrt[3]{2})\subset \mathbb{R}$.

Why is this extension field obtained using quotient by $(f(x))$ give us the real root $\sqrt[3]{2}$ and not one of the other two roots? I mean while taking the quotient with $f$ we are just getting an extension field which depends on $\mathbb{Q}$ and $f(x)$ and we don't know for sure which of the three roots it will contain.

The same question however does not apply to all irreducible polynomials. Thus if we take $f(x) = x^{4} + x^{3} + x^{2} + x + 1$ (or more simply $x^{2} + 1$) then $\mathbb{Q}[x]/(f(x))$ contains all the roots of this polynomial.

Update: For the simple polynomial $x^{2} + 1$ I can show that both of its roots are in the extension field $\mathbb{Q}[x] / (x^{2} + 1)$. With some reasonable effort we can show the same for $f(x) = x^{4} + x^{3} + x^{2} + x + 1$. I think we can also show that for $f(x) = x^{3} - 2$ the field $\mathbb{Q}[x]/(f(x))$ has only one root (I have not yet proved this for myself but I am sure I can).

My real question is that even if we assume that this field has only one of the roots how do I know that this particular root is special and other two roots are more of a similar nature (ie this one is real and others are complex conjugates)?

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It appears I have been misunderstood so I add more formalism and details. Let $K$ be the splitting field of $x^{3} - 2 \in \mathbb{Q}[x]$. So $K$ has three elements $a, b, c$ each of which is a cube root of $2$ in $K$. Also we have $$K \supset \mathbb{Q}[x]/(x^{3} - 2) = L \supset \mathbb{Q}$$ My point is that the roots $a, b, c \in K$ are not indistinguishable from each other. Precisely we have one of these (say $a$) in $L$ and rest two ($b, c$) in $K$. Also there is automorphism $\sigma$ of $K$ which fixes $\mathbb{Q}$ and $\sigma(b) = c$. But there is no automorphism of $K$ which fixes $\mathbb{Q}$ and sends $a$ to $b$ (or $c$). I know this only by working in field $\mathbb{C}$. How do I know this without working in $\mathbb{C}$?

Thanks to DonAntonio, I understood the flaw in my reasoning. Also sorry to other people who answered for bearing with my nonsense comments. What I gather from the overall discussion is this:

Summary: The cubic $x^{3} - 2$ has three roots and they all lie in splitting field $K$. The intermediate field $L = \mathbb{Q}[x]/(x^{3} - 2)$ contains only one of the roots (see last part of egreg's answer as to why $L$ can't have all the roots) and it is not possible to determine via algebra as to which of $a, b, c$ lie in $L$. Thus $a, b, c$ are indeed indistinguishable and any one (say $a$) can be supposed to lie in $L$ and the rest lie in $K$. And we can then prove that $K = L(b) = L(c)$ and $L = \mathbb{Q}(a)$ and $L$ is of degree $3$ over $\mathbb{Q}$ and $K$ is of degree $2$ over $L$. I think it is now easy to get the Galois group and show that it is indeed $S_{3}$.

Answer 1

Why is this extension field obtained using quotient by $(f(x))$ give us the real root $\sqrt[3]{2}$?

It doesn't. The field $K=\mathbb{Q}[x]/(x^3-2)$ contains a cube root of $2$, that is, an element $b$ such that $b^3=2$. There is no canonical embedding of $K$ in $\mathbb{C}$ such that $b\mapsto\sqrt[3]{2}$.

Actually there are exactly three embeddings of $K$ in $\mathbb{C}$, one for each cube root of $2$ in $\mathbb{C}$.

And, yes, $K$ has a single element whose cube is $2$, because it is isomorphic to $\mathbb{Q}(\sqrt[3]{2})$.

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The Galois group of $x^3-2$ (over $\mathbb{Q}$) has six elements (it is $S_3$). This is different from the cases of $x^2+1$ and $x^4+x^3+x^2+x+1$, because their Galois groups are cyclic, so when one root is added, also all the other roots are. Since the group $S_3$ is generated by at least two elements, you need two roots for generating the splitting field.

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Let's work in the field $K$, where $b$ is a root of the polynomial $x^3-2$. Then we have, over $K$, $$ x^3-2=(x-b)(x^2+bx+b^2) $$ for some $u$ and $v$ in $K$. Can a root of $x^2+bx+b^2$ belong to $K$? No, because the discriminant is $-3b^2$ (work it out).

Answer 2

Well, $\Bbb Q[X]/(X^4+X^3+X^2+X+1)$ is a Galois extension of $\Bbb Q$ but $\Bbb Q[X]/(X^3-2)$ isn't. A Galois extension will contain all zeros of its defining polynomial; a non-Galois extension won't.

Answer 3

So you want to know why the Galois group of $K=\Bbb Q(a,b,c)$ permutes $a$, $b$, $c$ arbitrarily without working in $\Bbb C$?

By Eisenstein's criterion $X^3-2$ and $X^2+3$ are irreducible over $\Bbb Q$. Let $a$ and $s$ be zeros of them in some extension field. Let $K=\Bbb Q(a,b)$. It's easy to prove that $|K:\Bbb Q|=6$. Moreover $K\cong \Bbb Q[X,Y]/\langle X^2-2,Y^2+3\rangle$ Let $b=(-1+s)a/2$ and $c=(-1-s)a/2$. Then in $K$ we have $$X^3-2=(X-a)(X-b)(X-c).$$

An automorphism of $K$ must send $a$ to one of $a$, $b$ and $c$ and $s$ to one of $s$ and $-s$. Since $K\cong \Bbb Q[X,Y]/\langle X^2-2,Y^2+3\rangle$ any such pair of choices induces an homomorphism from $K$ to $K$ and on dimension grounds this must be an isomorphism.

The automorphism mapping $a$ to $b$ and fixing $s$ maps $b$ to $c$ and $c$ to $a$.

The automorphism fixing $a$ and mapping $s$ to $-s$ swaps $b$ and $c$.

Combining these we get automorphisms of $K$ permuting $a$, $b$ and $c$ arbitrarily.

Note that throughout I have made no use of $\Bbb C$ or $\Bbb R$.