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Let $\varphi:K \rightarrow L$ be a homomorphism of fields and $N/K$ be a finite field extension. Show that there are at most $(N/K)$ continuations $\phi:N \rightarrow L$ of $\varphi$.

As I'm currently studying for my upcoming algebra exam, I found this rather frustrating exercise from an old algebra exam. I'm not sure how to begin solving this task so maybe someone could give me hint? Thanks in advance!

Edit: For clarification

  • $(N/K)$ is a finite field extension of the fields $N$ and $K$

A field extension is a pair of fields $K \subseteq N$, such that the operations of $K$ are those of $N$ restricted to $K$. In this case, $N$ is an extension field of $K$ and $K$ is a subfield of $N$. For example the complex numbers are an extension field of the real numbers and the real numbers are a subfield of the complex numbers.

If written like above $N/K$ then $K$ is a subfield of $N$ and $N$ is the extension field/extension of $K$.

  • Definition of a continuation:

A continuation of a function is a another function. The domain of the other function includes a subset which is equal to the original function. So the official definition is:

Let $X, Y$ and $A$ sets. A function $f:X \rightarrow Y$ is called a continuation of the function $g: A \rightarrow Y$, if $A$ is a Subset of $X$ and $g(x)=f(x)$ for all $x \in A$.

$2$nd Edit:

I have a rough idea on how to solve this maybe.

Let $\alpha\in N$ and $f$ the minimal polynom of $\alpha$ over $K$.

Now consider the zero points $\beta_1,...,\beta_k$ of $f$ in $L$ and show that:

  • $\varphi$ maps $\alpha$ to a $\beta_i$
  • $\varphi (\alpha)=\varphi^´(\alpha)\rightarrow \varphi = \varphi^´$

It follows that:

$\# \{$field extensions over $K(\alpha) \} \leq k \leq \text{deg}(f) = (K(\alpha)/K)$

Now write $N=K(\alpha_1,...,\alpha_m)$ and first extend $\varphi$ to $K(\alpha_1)$, then extend it to $K(\alpha_1, \alpha_2)$ and so forth...

Using the relation $[M:K]=[M:L] \cdot [L:K]$ we can see that $(N/K)=(K(\alpha_1)/K) \cdot (K(\alpha_1, \alpha_2)/K(\alpha_1)) \cdot...\cdot(N/K(\alpha_1,...,\alpha_{m-1}))$

Now you can to extend this to the extension $N/K$.

Is this correct?

Isaac16726
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    Would you mind defining what you mean by $(N/K)$ and continuation? – qualcuno Jan 27 '21 at 22:17
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    Are you sure there are no hypothesis on the characteristic? For example, here it is exhibited a finite extension $N/K$ with infinitely many quadratic subextensions $K(\alpha)$ with $\alpha^2 \in K, \alpha\in N \setminus K$. Now, we can consider the automorphism that sends $\alpha$ to its conjugate and lift it to $N$, since $N/K(a)$ is algebraic, giving rise to infinitely many morphisms. If $N/K$ has a finite amount of subextensions (for example when $N/K$ is separable, which is implied by $\mathsf{char} K = 0$) – qualcuno Jan 27 '21 at 23:52
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    (cont.) then it is cyclic, hence a map $N=K(\xi) \to L$ corresponds to a certain choice of a root of $m(\xi,K)$, there are $\deg m(\xi,K) = [N:K]$ of them – qualcuno Jan 27 '21 at 23:52

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