# Number of injective field homomorphism

Let $F$ and $L$ be two field extensions of $K$. The extension $F/K$ is finite of degree $n$ and the extension $L/K$ may be infinite also.

I want to count the number of injective homomorphism from $F$ to $L$ where K is fixed. Is it true that the number of injective homomorphism from $F$ to $L$ where K is fixed is less than or equal to $n$?

I know that the injective homomorphisms from $F$ to $L$ are lineraly independent over $L$. So, I'm trying to prove that if there are $n+1$ injective homomorphism, they must be linearly dependent. So,a contradiction. Is this approach correct?

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minor nitpick: you're looking at morphisms from $F$ to $L$ over $K$? (Edit: my bad, you said "where $K$ is fixed".) If $L=K$ ($n=1$) is a number field and $L=\mathbf{C}$ you get $[K:\mathbf{Q}]$ embeddings of $K$ into $\mathbf{C}$. This is bigger than $1$ iff $K\neq \mathbf{Q}$. Probably you want to assume that $K$ is the prime field? – Harry Apr 4 '12 at 15:24
You should also look at "separability" degree: en.wikipedia.org/wiki/Separable_extension – Harry Apr 4 '12 at 15:27
Harry is right, but since you mention linear independence of morphisms I'll give an answer along those lines rather than invoke separability. – Georges Elencwajg Apr 4 '12 at 16:11

Yes it is true (and keeep in mind that a morphism of fields is automatically injective).

The key point is that the set of $K$-linear maps $f:F\to L$ is a vector space over the field $L$.
Indeed, for $\lambda \in L$ the map $\lambda f: F\to L \;$ is defined (you guessed it!) by $(\lambda f)(x)=\lambda \cdot (f(x))$ where the dot $\cdot$ is the product in the field $L$.

Linear algebra then teaches us that the dimension of that $L$-vector spase $\mathcal L_{K-lin }(F,L)$ is $n=dim_K(F)$ [see proof in Edit below]

The theorem of linear independence of homomorphisms then states that the set of $K$-algebra morphisms $Hom_{K-alg }(F,L)$ is a linearly independent subset $Hom_{K-alg }(F,L)\subset \mathcal L_{K-lin }(F,L)$ of the aforementioned $L$-vector space $\mathcal L_{K-lin }(F,L)$, so that of course $$\operatorname {card} Hom_{K-alg }(F,L)\leq n$$ just as you wished.

Caveat
The set $Hom_{K-alg }(F,L)$ has no algebraic structure whatsoever: it is an unashamedly naked set, very possibly empty .

Edit
In order to answer user's question in the comments, here is a proof that $\mathcal L_{K-lin }(F,L)$ has dimension $n$ over $L$:

Choose a basis $a_1,...,a_n$ of $F$ over $K$.
Then the $K$-linear maps $f_i:F \to L$ defined by $f_i(\sum k_r a_r)= k_i$ are the required basis.
This is just a slight generalization of the usual concept of dual basis, which you recover if $L=K$ (which is a perfectly legitimate choice for $L$ in the question and in the answer!).

Be very wary of the confusing fact that all $f_i$'s have values in $K$ but that there exist linear maps $f\in \mathcal L_{K-lin }(F,L)$ capable of reaching any $\lambda\in L$: for example $(\lambda f_1)(a_1)=\lambda$ !

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why is the dimension of that L-vector spase $\mathcal L_{K-lin }(F,L)$ is equal to $n=dim_K(F)$? – Mohan Apr 4 '12 at 16:37
Dear @user: I have added a proof of that in an edit. – Georges Elencwajg Apr 4 '12 at 17:03