Let $\varphi:K \rightarrow L$ be a homomorphism of fields and $N/K$ be a finite field extension. The continuation is definded as $\phi:N \rightarrow L$. 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 $\beta_i$
- $\varphi (\alpha)=\varphi^´(\alpha)\rightarrow \varphi = \varphi^´$
Unfortunately another exercise from an old algebra exam which I'm really struggling to solve. Maybe someone could give me a hint? Thanks in advance!
Further information:
- 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 $x \in A$.
- Definition of a finite field extension
$(N/K)$ 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 $K$. Finite meaning there are a finite amounts of elements
Edit: $N$ is given by $N=K(\alpha)$. For the first task you need to set and $f(\alpha)=0$, but I am unsure how to procced.
Let $\phi: K \to L$ be a fixed field morphism, let $n =$ deg$(f)$ and let $\phi(f)$ be the corresponding polynomial in $L[x]$. If $\psi: N \to L$ is a field morphism that extends $\phi$, then $\psi(\alpha)$ is a root of $\phi(f)$ since if $$ 0 \: = \: f(\alpha) \: = \: a_{n}\alpha^{n} \: + \: ... \: + \: a_{1}\alpha \: + \: a_{0} $$ then an application of $\psi$ shows that $$ \psi(0) \: = \: 0 \: = \: \phi(a_{n}) \cdot \psi(\alpha)^{n} \: + \: ... \: + \: \phi(a_{1}) \cdot \psi(\alpha) \: + \: \phi(a_{0}) \: = \: \phi(f)(\psi(\alpha)) $$ because $\psi$ is a field morphism that equals $\phi$ on $K$ (and therefore on the coefficients of $f$).
Next, since $N = k(\alpha)$ and $f$ is the minimal polynomial of $\alpha$ over $K$, the sequence $(\alpha_{i})_{i = 0}^{n - 1}$ forms a linear basis for $N$ over $K$. In other words, each element $b$ of $N$ may be written as $b_{n}\alpha^{0} + ... + b_{n}\alpha^{n}$ for coefficients $b_{i}$ in $K$. After applying $\psi$, we get $$ \psi(b) \: = \: \phi(b_{0}) \cdot \psi(\alpha)^{0} \: + \: ... \: + \: \phi(a_{n}) \cdot \psi(\alpha)^{n} $$ Hence every field morphism extension of $\phi$ to $N$ is completely determined by where it maps $\alpha$.