关于$\mathbb{Q}$上$x^3-2$伽罗瓦群计算的技术问询（基于Dummit和Foote教材）

Great question—this is a core detail in figuring out how Galois groups act on splitting fields, so let's break it down step by step to make it click.

First, let's ground this in key properties of field automorphisms (which are exactly the elements of our Galois group):

• Any automorphism $\sigma$ that fixes $\mathbb{Q}$ must map an element $\alpha$ to another root of its minimal polynomial over $\mathbb{Q}$. Why? If $f(x) \in \mathbb{Q}[x]$ is the minimal polynomial of $\alpha$, then $f(\alpha) = 0$. Applying $\sigma$ (which preserves $\mathbb{Q}$ elements and field operations) gives $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$. So $\sigma(\alpha)$ has to be a root of $f(x)$.
• The primitive 3rd root of unity $\rho$ satisfies $\rho^3 = 1$ and $\rho \neq 1$, so it's a root of $x^2 + x + 1$ (since $x^3 - 1 = (x-1)(x^2 + x + 1)$). This quadratic polynomial is irreducible over $\mathbb{Q}$ (its discriminant is $1 - 4 = -3$, which isn't a square in $\mathbb{Q}$), so it's the minimal polynomial of $\rho$ over $\mathbb{Q}$.

Now let's connect these dots to your question:

• The minimal polynomial of $\rho$ has exactly two roots: $\rho$ and $\rho^2$. You can verify this directly: $(\rho²⁾2 + \rho^2 + 1 = \rho^4 + \rho^2 + 1 = \rho + \rho^2 + 1 = 0$ (since $\rho^3 = 1$, so $\rho^4 = \rho$).
• Could $\sigma(\rho)$ be something like $\sqrt[3]{2}$? Nope—$\sqrt[3]{2}$ is a root of $x^3 - 2$, not $x^2 + x + 1$. That would violate the automorphism rule we just laid out.
• Could $\sigma(\rho) = 1$? Also no—field automorphisms are injective (one-to-one). Since $\rho \neq 1$, $\sigma(\rho)$ can't equal $\sigma(1) = 1$ (that would mean two distinct elements map to the same value, breaking injectivity).

So to wrap up: the only valid targets for $\sigma(\rho)$ are the roots of its minimal polynomial over $\mathbb{Q}$—which are exactly $\rho$ and $\rho^2$.

内容的提问来源于stack exchange，提问作者ponchan