I have been given the following question:
Suppose that $M$ and $L$ are fields and $\phi_1, \cdots , \phi_n$ are distinct embeddings of $M$ into $L$. Prove that there do not exist elements $\lambda_1, \cdots, \lambda_n$ of L, not all zero, such that $\lambda_1\phi_1(x) + \cdots + \lambda_n\phi_n(x) = 0$ for all $x \in M$. Deduce that if $K \subseteq M$ is a finite field extension and $\phi_1,\cdots,\phi_n$ are distinct $K$-automorphisms of $M$ then $n \leq [M : K ]$.
I have managed to do the first part of the question, but I don't understand how the second part follows from the first. I have seen a proof of this result but that relied on minimal polynomials etc.
I don't see where $K$ comes into the second part of the question. As $\phi_1,\cdots,\phi_n$ are distinct embeddings of $M$ into $M$ that preserve $K$ we know that $\phi_1,\cdots,\phi_n$ are linearly independent but I don't see how to proceed from there.
Any help is appreciated!
Never mind. I think I got it.
To show $n \leq [M:K]$ Consider the vector space of all $K$-linear maps from $M$ to $M$ (i.e. $f(x+y) = f(x)+f(y)$ and $f(kx) = kf(x)$ for all $k \in K$). It is easy to see that this forms a vector space, say $V$. Note that our $K$-homomorphisms are linearly independent members of $V$ from the first part. So it is enough to show the dimension of $V$ is $[M:K]$, then we have $n \leq [M:K]$.
But note that if $x_1,\cdots,x_m$ is a basis for $M$ over $K$ with $m = [M:K]$ then if we pick $\varepsilon_i \in V$ for $i \in \{1,\cdots,m\}$ such that $\varepsilon_i(x_j) = \delta_{ij}$ is a basis for $V$ (this specifies each $\varepsilon_i$ completely as we specify how it acts on a basis). So $\dim(V) = m = [M:K]$ and we are done (in fact $V$ is just the dual basis of $M$ over $K$).