1.5 Embeddings

Definition 1.5.1

Let $K$ be a number field. Then an embedding of $K$ is a non-zero ring homomorphism $σ : K ↪ C$ .

Remark 1.5.2

Note that since $σ$ is a ring homomorphism, we must have that $σ (x) = x$ for all $x \in Q \subset K$ .

Exercise 1.5.3

Prove that an embedding is injective.

Now, since by definition our number fields are subfields of $C$ , then we have at least one embedding, which is just the identity embedding (i.e, send $x$ to $x$ ), sometimes called the standard embedding. But there can be others.

Example 1.5.4

Let $K = Q (\sqrt{2})$ then we have two embeddings:

\begin{aligned} σ_{1} : Q (\sqrt{2}) ↪ C sending a + b \sqrt{2} \mapsto a + b \sqrt{2} \\ σ_{2} : Q (\sqrt{2}) ↪ C sending a + b \sqrt{2} \mapsto a + b (- \sqrt{2}) \end{aligned}

Here $σ_{1}$ is just the identity embedding. Note that since the embedding has to keep rational numbers fixed, the $a, b$ above stay the same, what the different embeddings change is where $\sqrt{2}$ maps to, but $\sqrt{2}$ cant just map to anything, we will see later that in fact it has to map to a conjugate root.

Lets take this as a given and now consider the embeddings of $L = Q (\sqrt{2}, \sqrt{3})$ . This is a degree $4$ extension of $Q$ and every element can be written as $a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$ with $a, b, c, d \in Q$ . Now, the embeddings are:

\begin{aligned} ν_{1} : a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} ⟼ a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \\ ν_{2} : a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} ⟼ a - b \sqrt{2} + c \sqrt{3} - d \sqrt{6} \\ ν_{3} : a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} ⟼ a + b \sqrt{2} - c \sqrt{3} - d \sqrt{6} \\ ν_{4} : a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} ⟼ a - b \sqrt{2} - c \sqrt{3} + d \sqrt{6} \end{aligned}

Notice that since $K \subset L$ (just take elements with $c = d = 0$ ), then it makes sense to restrict the embeddings of $L$ to $K$ . If you do this, then we see that $ν_{1}, ν_{3}$ both give the identity embedding $σ_{1} : K ↪ C$ , while $ν_{2}, ν_{4}$ restrict to $σ_{2}$ .

Definition 1.5.5

Let $K / F$ be a finite extension of number fields and let $σ : F ↪ C$ and $ν : K ↪ C$ be embeddings. Then we say $ν$ extends $σ$ if $ν$ restricted to $F$ agrees with $σ$ . Symbolically, we say $ν ∣_{F} = σ$ .

Proposition 1.5.6

Let $K / F$ be an extension of number fields and let $σ : K ↪ C$ be an embedding such that $σ ∣_{F} = id$ where $id$ denotes the identity embedding. Then if $f (x) \in F [x]$ is an irreducible polynomial and $α$ is one of its roots, then $σ$ sends $α$ to a conjugate of $α$ .

Proof ▼

Let $f (x) = a_{0} + a_{1} x + \dots + a_{n} x^{n}$ . Then since $σ$ fixes $F$ , we see that since

a_{0} + a_{1} α + \dots + a_{n} α^{n} = 0

it follows that

a_{0} + a_{1} σ (α) + \dots + a_{n} σ (α)^{n} = 0

and therefore, $σ (α)$ also is a root of $f$ , which gives the result.

Lemma 1.5.7

(Separability Lemma) Let $K$ be a number field and let $f (x) \in K [x]$ be an irreducible polynomial of degree $n \geq 1$ . Then $f (x)$ has $n$ distinct roots.

Moreover, if $σ : K ↪ C$ is any embedding and $f^{σ} (x)$ denotes the polynomial obtained by applying $σ$ to each coefficient, then $f^{σ}$ also has $n$ distinct roots.

Proof ▼

By the fundamental theorem of algebra, $f (x)$ has $n$ roots in $C$ , so we only need to show that there are no repeated roots. For this, consider the derivative $f^{'} (x)$ of $f (x)$ . $f^{'}$ is also in $K [x]$ and is non-zero. If $f (x)$ has a repeated root, then $f$ and $f^{'}$ would share a common factor, call it $h$ . So let $g$ denote the greatest common divisor of $f$ and $f^{'}$ . Then $\deg (g) \leq n - 1$ and divides $f$ , but $f$ is irreducible, so $g$ must be a constant (i.e. $\deg (g) = 0$ ), but then as $h | g$ this means $h$ is also a constant and thus, $f, f^{'}$ do not share a common factor.

The same proof works using $f^{σ}$ instead. So the result follows.

Proposition 1.5.8

For every embedding $σ : F ↪ C$ there are $[K : F]$ embeddings of $K$ that extend $F$ .

Proof ▼

We will prove this by induction on $[K : F]$ . If $K = F$ there is nothing to prove. So assume $K \neq F$ and let $σ$ be an embedding of $F$ . Now take $α \in K ∖ F$ (i.e. in $K$ but not $F$ , which we can do since we are assuming $K \neq F$ ). Consider its minimal polynomial over $F$ , denoted $m_{α, F}$ . Now let $β$ be a root of the polynomial you get from applying $σ$ to $m_{α, F}$ , which is $m_{α, F}^{σ} = m_{α, F^{σ}}$ . Here $F^{σ}$ denotes the image of $F$ under $σ$ , which is a number field isomorphic to $F$ . Now, $m_{α, F^{σ}}$ is again irreducible over $F^{σ}$ since under an isomorphism an irreducible polynomial will stay irreducible.

Now, from Theorem 1.3.4 we have

F (α) ≅ F [x] / (m_{α, F}) ≅ F^{σ} [x] / (m_{α, F}^{σ}) ≅ F^{σ} [x] / (m_{β, F^{σ}}) ≅ F^{σ} (β)

therefore there is an isomorphism $σ : F (α) ≅ F^{σ} (β)$ which sends $α$ to $β$ and restricts to $σ$ on $F$ . Doing this for each root of $m_{α, F}^{σ}$ we see that there are $\deg (m_{α, F}) = [F (α) : F]$ extensions of $σ$ to $F (α)$ . Now, use the inductive hypothesis.

Definition 1.5.9

We say an embedding is real if its image is $R \subset C$ . Otherwise we say the embedding is complex. Note that if $σ$ is a complex embedding, then so is its complex conjugate $\overset{―}{σ}$ (i.e. this is the embedding given by applying $σ$ and then doing complex conjugation.) We call $σ, \overset{―}{σ}$ a pair of complex conjugate embeddings.

Remark 1.5.10

Note that Proposition 1.5.8 applied to $K / Q$ tells us that if $r_{1}$ is the number of real embedding and $r_{2}$ is the number of complex conjugate embeddings (i.e. there are $2 r_{2}$ complex embeddings), then

[K : Q] = r_{1} + 2 r_{2} .

Theorem 1.5.11 Primitive element theorem

Let $K / F$ be a finite extension of number fields. Then there exists $α \in K$ such that $K ≅ F (α)$ .

Proof ▼

We prove this by induction on $[K : F]$ . If $K = F$ there is nothing to prove. So assume that $K \neq F$ and let $α \in K ∖ F$ . Then by our inductive hypothesis we have $K = F (α, β)$ for some $β$ . We claim that there are infinitely many $c$ such that $K = F (α + c β)$ . To do this we will show that there can only be finitely many $c \in F$ such that $K \neq F (α + c β)$ . Assume this is the case, then lets think about how many conjugates $α + c β$ has over $F$ . If $K = F (α + c β)$ then by Proposition 1.5.8 there would be $[K : F]$ conjugates of $(α + c β)$ , but since we are assuming we aren’t in this situation there must be fewer conjugates. In particular, we must have two distinct embeddings $η, σ : K ↪ C$ which extend $F$ and send $(α + c β)$ to the same element. So

η (α) + c η (β) = σ (α) + c σ (β) .

Now, note that $σ (β) \neq η (β)$ since otherwise, $η (α) = σ (α)$ and thus, since $K = F (α, β)$ we’d have $σ = η$ , which is a contradiction. Therefore

c = \frac{η (α) - σ (α)}{σ (β) - η (β)},

but by Proposition 1.5.6 there are only finitely many possibilities for $η (a)$ , $η (β)$ , $σ (α)$ , $σ (β)$ .

If you use Galois theory then there is a much quicker proof: Since $[K : F]$ is finite, there are only finitely many intermediate fields. Now just pick $α \in K$ which is not contained in any of there intermediate fields, then $F (α)$ is an extension of $F$ not contained in any proper subfield of $K$ , so $K = F (α)$ . □