As I was reading about Algebra, the following came up:
It is often necessary, to write an Element $b \in K(a)$ of a simple extension $K(a)/K$ with $[K(a) : K]=n$, as linear combination $b=k_0+k_1a+...+k_{n-1}a^{n-1}$, with $k_0,...,k_{n-1} \in K$.
To find a method to do so, one just needs to consider the lemma:
Let $L/K$ be a field extension. If $a \in L$ is algebraic over $K$ with minimal polynomial $m(a,K)$, $deg (m(a,K))=n$, then $K(a)=K[a] \cong K[X]/(m(a,K))$ and $n=[K(a) : K]$.
Now I am trying to find this method:
The Lemma tells us that $K(a) \cong K[x]/(m(a,K))$, where $(m(a,K)):=\{p(x) m(a,K)(x) : p(x) \in K[x] \}$.
Now as far as I understand this if $p(a) \in K(a)$ the lemma syst that there exists some polynomial $R(x) \in K[x]/(m(a,K))$ such taht $ p(a)"="R(x)$.
If I take $P(a)$, look at $P(x)$ and divide $P(x)$ by $m(a,K)(x)$ I would get P(x)=m(a,K)(x)Q(x)+R(x).
Since $K[X]/(m(a,K))$ is an equivalence relation, I can choose a representative $v(x)$ for an equivalence class such that $deg(v(x))< deg(m(a,k)(x))$.
This means that $m(a,K)(x)Q(x)+R(x)$ is in the same equivalence class as $R(x)$, and I can choose $R(x)$ as my representative.
This means that $P(a)$ can be written as $R(x)$. But the statement above was that we would be able to write $R(x)$ als linear combination of $1,a,a^2,...,a^{n-1}$. So my attempt seems to be wrong. Could someone show me how this should work?