# (The number of) embeddings of an algebraic extension of $\mathbb{Q}$ into $\mathbb{C}$

While reading a book (Neukirch) I came upon the claim that for an algebraic extension $K/\mathbb{Q}$ of degree $n$ there are $n$ embeddings of $K$ into $\mathbb{C}$. Why is it so, and in general how do such embeddings "look"?

## 2 Answers

Here's a short anwser. The Primitive element theorem tells you that $K/\mathbb{Q}$ is generated by a single element: $K = \mathbb{Q}(\alpha) = \mathbb{Q}[X]/(P)$ where $P$ is the minimal polynomial of $\alpha$.

Now like any irreductible polynomial over $\mathbb{Q}$, $P$ has $n = [K:\mathbb{Q}]$ distinct roots in $\mathbb{C}$ (because it is prime to its derivative $P'$). Let $\alpha_1,\ldots,\alpha_n$ be these roots. Then you get $n$ distinct embeddings $\sigma_i: K = \mathbb{Q}[X]/(P) \hookrightarrow \mathbb{C}$, $X \mapsto \alpha_i$. And obviously there can't be more since the image of $X$ has to be a root of $P$.

I should mention that all of this remains true for any finite extension $K/F$ that is separable (meaning that the minimal polynomials of elements $x\in K$ have distinct roots in $\bar{F}$.)

If $K/\mathbb{Q}$ is of degree $n$, then by the Primitive Element Theorem there exists $\alpha\in K$ such that $K=\mathbb{Q}(\alpha)$; the irreducible polynomial of $\alpha$ over $\mathbb{Q}$ has degree $[K:\mathbb{Q}]=n$. Thus, it is of the form $$f(x) = x^n + a_{n-1}x^{n-1}+\cdots + a_1x + a_0, \qquad a_i\in\mathbb{Q}$$ and every element of $K$ can be written uniquely as $$\beta_0 + \beta_1\alpha + \beta_2\alpha^2+\cdots + \beta_{n-1}\alpha^{n-1}$$ for some $\beta_0,\ldots,\beta_{n-1}\in\mathbb{Q}$.

Let $\rho_1,\rho_2,\ldots,\rho_n$ be the $n$ distinct complex roots of $f(x)$ in $\mathbb{C}$ (distinct since irreducible polynomials over $\mathbb{Q}$ are always square free). Then for each $j$, you have an embedding $\iota_j\colon K\to\mathbb{C}$ obtained by mapping $\mathbb{Q}$ to itself, and $\alpha$ to $\rho_j$; i.e., $$\iota_j(\beta_0 + \beta_1\alpha + \beta_2\alpha^2+\cdots + \beta_{n-1}\alpha^{n-1}) = \beta_0 + \beta_1\rho_j + \beta_2\rho_j^2+\cdots + \beta_{n-1}\rho_j^{n-1}.$$

For example, the two embeddings of $\mathbb{Q}(\sqrt{2})$ into $\mathbb{C}$ are given by (i) sending $a+b\sqrt{2}$ to $a+b\sqrt{2}\in\mathbb{C}$; and (ii) sending $a+b\sqrt{2}$ to $a-b\sqrt{2}\in\mathbb{C}$; since the two roots of $x^2-2$ are $\sqrt{2}$ and $-\sqrt{2}$.

Since an embedding from $K$ to $\mathbb{C}$ is completely determined by what happens to $\alpha$, and the image of $\alpha$ must be a complex root of $f(x)$, the $n$ embeddings described above are all the embeddings of $K$ into $\mathbb{C}$.