1.5 Embeddings
Let \(K\) be a number field. Then an embedding of \(K\) is a non-zero ring homomorphism \(\sigma :K \hookrightarrow \mathbb {C}\).
Note that since \(\sigma \) is a ring homomorphism, we must have that \(\sigma (x)=x\) for all \(x \in \mathbb {Q}\subset K\).
Prove that an embedding is injective.
Now, since by definition our number fields are subfields of \(\mathbb {C}\), 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.
Let \(K=\mathbb {Q}(\sqrt{2})\) then we have two embeddings:
\begin{align} & \sigma _1: \mathbb {Q}(\sqrt{2}) \hookrightarrow \mathbb {C}\qquad \text{ sending } \qquad a+b\sqrt{2} \mapsto a+b\sqrt{2}\\ & \sigma _2: \mathbb {Q}(\sqrt{2}) \hookrightarrow \mathbb {C}\qquad \text{ sending } \qquad a+b\sqrt{2} \mapsto a+b(-\sqrt{2}) \end{align}Here \(\sigma _1\) 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 \(L=\mathbb {Q}(\sqrt{2},\sqrt{3})\). This is a degree \(4\) extension of \(\mathbb {Q}\) and every element can be written as \(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\) with \(a,b,c,d \in \mathbb {Q}\). Now, the embeddings are:
\begin{align} & \nu _1: a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \longmapsto a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}\\ & \nu _2: a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \longmapsto a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}\\ & \nu _3: a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \longmapsto a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}\\ & \nu _4: a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \longmapsto a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6} \end{align}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 \(\nu _1,\nu _3\) both give the identity embedding \(\sigma _1:K \hookrightarrow \mathbb {C}\), while \(\nu _2,\nu _4\) restrict to \(\sigma _2\).
Let \(K/F\) be a finite extension of number fields and let \(\sigma :F \hookrightarrow \mathbb {C}\) and \(\nu : K \hookrightarrow \mathbb {C}\) be embeddings. Then we say \(\nu \) extends \(\sigma \) if \(\nu \) restricted to \(F\) agrees with \(\sigma \). Symbolically, we say \(\nu {\mid _F}=\sigma \).
Let \(K/F\) be an extension of number fields and let \(\sigma : K \hookrightarrow \mathbb {C}\) be an embedding such that \(\sigma \mid _F=\operatorname {id}\) where \(\operatorname {id}\) denotes the identity embedding. Then if \(f(x) \in F[x]\) is an irreducible polynomial and \(\alpha \) is one of its roots, then \(\sigma \) sends \(\alpha \) to a conjugate of \(\alpha \).
Let \(f(x)= a_0+a_1x+\cdots +a_nx^n\). Then since \(\sigma \) fixes \(F\), we see that since
it follows that
and therefore, \(\sigma (\alpha )\) also is a root of \(f\), which gives the result.
(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 \(\sigma :K \hookrightarrow \mathbb {C}\) is any embedding and \(f^\sigma (x)\) denotes the polynomial obtained by applying \(\sigma \) to each coefficient, then \(f^\sigma \) also has \(n\) distinct roots.
By the fundamental theorem of algebra, \(f(x)\) has \(n\) roots in \(\mathbb {C}\), 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 \(f^\sigma \) instead. So the result follows.
For every embedding \(\sigma : F \hookrightarrow \mathbb {C}\) there are \([K:F]\) embeddings of \(K\) that extend \(F\).
We will prove this by induction on \([K:F]\). If \(K=F\) there is nothing to prove. So assume \(K \neq F\) and let \(\sigma \) be an embedding of \(F\). Now take \(\alpha \in K \backslash F\) (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 \(m_{\alpha ,F}\). Now let \(\beta \) be a root of the polynomial you get from applying \(\sigma \) to \(m_{\alpha ,F}\), which is \(m_{\alpha ,F}^\sigma =m_{\alpha ,F^\sigma }\). Here \(F^\sigma \) denotes the image of \(F\) under \(\sigma \), which is a number field isomorphic to \(F\). Now, \(m_{\alpha ,F^\sigma }\) is again irreducible over \(F^\sigma \) since under an isomorphism an irreducible polynomial will stay irreducible.
Now, from Theorem 1.3.4 we have
therefore there is an isomorphism \(\sigma :F(\alpha ) \cong F^\sigma (\beta )\) which sends \(\alpha \) to \(\beta \) and restricts to \(\sigma \) on \(F\). Doing this for each root of \(m_{\alpha ,F}^\sigma \) we see that there are \(\deg (m_{\alpha ,F})=[F(\alpha ):F]\) extensions of \(\sigma \) to \(F(\alpha )\). Now, use the inductive hypothesis.
We say an embedding is real if its image is \(\mathbb {R}\subset \mathbb {C}\). Otherwise we say the embedding is complex. Note that if \(\sigma \) is a complex embedding, then so is its complex conjugate \(\overline{\sigma }\) (i.e. this is the embedding given by applying \(\sigma \) and then doing complex conjugation.) We call \(\sigma ,\overline{\sigma }\) a pair of complex conjugate embeddings.
Note that Proposition 1.5.8 applied to \(K/\mathbb {Q}\) 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 \(2r_2\) complex embeddings), then
Let \(K/F\) be a finite extension of number fields. Then there exists \(\alpha \in K\) such that \(K \cong F(\alpha )\).
We prove this by induction on \([K:F]\). If \(K=F\) there is nothing to prove. So assume that \(K \neq F\) and let \(\alpha \in K \backslash F\). Then by our inductive hypothesis we have \(K=F(\alpha ,\beta )\) for some \(\beta \). We claim that there are infinitely many \(c\) such that \(K=F(\alpha +c\beta )\). To do this we will show that there can only be finitely many \(c \in F\) such that \(K \neq F(\alpha +c\beta )\). Assume this is the case, then lets think about how many conjugates \(\alpha +c\beta \) has over \(F\). If \(K =F(\alpha +c\beta )\) then by Proposition 1.5.8 there would be \([K:F]\) conjugates of \((\alpha +c\beta )\), but since we are assuming we aren’t in this situation there must be fewer conjugates. In particular, we must have two distinct embeddings \(\eta ,\sigma : K \hookrightarrow \mathbb {C}\) which extend \(F\) and send \((\alpha +c\beta )\) to the same element. So
Now, note that \(\sigma (\beta ) \neq \eta (\beta )\) since otherwise, \(\eta (\alpha )=\sigma (\alpha )\) and thus, since \(K=F(\alpha ,\beta )\) we’d have \(\sigma =\eta \), which is a contradiction. Therefore
but by Proposition 1.5.6 there are only finitely many possibilities for \(\eta (a)\), \(\eta (\beta )\), \(\sigma (\alpha )\), \(\sigma (\beta )\).
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 \(\alpha \in K\) which is not contained in any of there intermediate fields, then \(F(\alpha )\) is an extension of \(F\) not contained in any proper subfield of \(K\), so \(K=F(\alpha )\).