This is taken from Lang's Algebra book chapter 5 exercise 5
If $\alpha$ is a complex root of $X^6 + X^3 + 1$, find all homomorphisms $\sigma: \mathbb{Q}(\alpha) \rightarrow \mathbb{C}$.
My (probably incorrect) solution is as follows:
Each homomorphism corresponds to an extension generated by $r, r^2, …,r^5$ where $r$ is a root of $X^6 + X^3 + 1$. There are a total of $6$ complex roots of this equation and thus there are a total of $6$ different homomorphisms. Letting $\alpha_1, \alpha_2,…,\alpha_6$ be the roots the homomorphisms are defined as $\alpha \mapsto \alpha_i$ for $i \in [6]$.
However, I don't really buy this argument, especially the part where I claim that each homomorphism corresponds to a root. I feel lthat some similar result might be true but I'm not sure where it's stated in the book.
Best Answer
To find all homomorphisms $\sigma: \mathbb{Q}(\alpha) \rightarrow \mathbb{C}$, we can follow these steps:
Find the minimal polynomial of $\alpha$. Find all roots of the minimal polynomial in $\mathbb{C}$. Each homomorphism is determined by where it sends $\alpha$. Step 1: Find the minimal polynomial of $\alpha$.
We are given that $\alpha$ is a complex root of the polynomial $X^6 + X^3 + 1$. To show that this polynomial is irreducible over $\mathbb{Q}$, consider the polynomial $g(X) = f(X-1) = (X-1)^6 + (X-1)^3 + 1$, where $f(X) = X^6 + X^3 + 1$. Expanding $g(X)$, we get:
$$ g(X) = X^6 - 6X^5 + 21X^4 - 42X^3 + 63X^2 - 54X + 27 $$
Applying the Eisenstein criterion to $g(X)$ with prime $p = 3$, we find that $g(X)$ is irreducible over $\mathbb{Q}$. Since $g(X)$ is irreducible, it follows that $f(X) = g(X+1)$ is also irreducible over $\mathbb{Q}$. Therefore, the minimal polynomial of $\alpha$ is $X^6 + X^3 + 1$.
Step 2: Find all roots of the minimal polynomial in $\mathbb{C}$.
Let $\omega = e^{\frac{2\pi i}{9}}$. Then, the roots of the minimal polynomial are:
$$ \alpha_1 = \omega^1, \quad \alpha_2 = \omega^2, \quad \alpha_3 = \omega^4, \quad \alpha_4 = \omega^5, \quad \alpha_5 = \omega^7, \quad \alpha_6 = \omega^8 $$
Step 3: Each homomorphism is determined by where it sends $\alpha$.
For each homomorphism $\sigma: \mathbb{Q}(\alpha) \rightarrow \mathbb{C}$, the image of $\alpha$ must be one of the roots of its minimal polynomial. So, we have 6 possible homomorphisms:
$$ \sigma_1(\alpha) = \alpha_1, \quad \sigma_2(\alpha) = \alpha_2, \quad \sigma_3(\alpha) = \alpha_3, \quad \sigma_4(\alpha) = \alpha_4, \quad \sigma_5(\alpha) = \alpha_5, \quad \sigma_6(\alpha) = \alpha_6 $$
Therefore, there are 6 homomorphisms $\sigma: \mathbb{Q}(\alpha) \rightarrow \mathbb{C}$, where each $\sigma_i(\alpha) = \alpha_i$ for $i = 1, 2, \ldots, 6$.