I am looking at a problem and wondering something... here is the question
If $K \subseteq L$ are fields, show that for any $a$, that $[L(a):L] \leq [K(a):K]$
Now, one thing that bothers me; is this a simple extension? If it is, I may have solved it. Why I cannot be sure is because it is not stated that $a \in L$ which I think is a part of the definition of a simple extension.
My solution uses the following proposition
Let $K(a):K$ be a simple extension. If it is transcendental then $[K(a):K]= \infty$. If it is algebraic then $[K(a):K]=\partial{m}$ where $m$ is the minimal polynomial of $a$ over $K$.
Then, I can consider a few cases(my solution)
It is either that a minimal polynomial exists o not. If a minimal polynomial $m$ exists in $K$, then since $K \subseteq L$, it also exists in $L$. Then the degrees will be equal; $\partial{m}$.
If $m$ is in $L$, then either that $m \in K$ or $m \notin K$. So $[K(a):K] \geq \partial{m}$ and hence the inequality is satisfied.
If no such $m$ exists for both $K,L$, then $a$ is transcendental and the degree is $\infty$ and thus inequality is still satisfied.
It cannot be that $m$ exists in $K$ but not in $L$, so we have considered all possible cases and shown that the inequality is true for each. Hence proven.
If this isn't a simple extension, then I am out of ideas. Can someone either tell me if my attempt is wrong or correct? If wrong, what can I do to solve it?
$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\R}{\mathbb{R}}$You are wrong on one point. Suppose $a$ is algebraic over $K$. Consider the minimal polynomial $f \in K[x]$ of $a$ over $K$.
Since $K \subseteq L$, we have $0 \ne f \in K[x] \subseteq L[x]$, so $a$ is also algebraic over $L$.
But the minimal polynomial $g \in L[x]$ of $a$ over $L$ could be different from $f$. All you can say that $x - a \mid g \mid f$, where the first divisibility is meant in $L(a)[x]$.
As an example, let $K = \Q$, $L = \R$ and $a = \sqrt{2}$. Then the minimal polynomial of $a$ over $K$ is $x^{2} - 2$, but over $L$ is $x - \sqrt{2}$.