MHB Algebraic element - Minimal polynomial

[mathmari]

Hey!

We suppose that $M/L/K$ are consecutive fields extensions and $a\in M$ is algebraic over $K$. I want to show that $a$ is algebraic also over $L$. I want to show also that the minimal polynomial of $a$ over $L$ divides the minimal polynomial of $a$ over $K$ (if we consider this polynomial as an element of $L[x]$). Then I want to conclude that the degree of $a$ over $L$ is, at most, equal to the degree of $a$ over $K$.

I have done the following:

Since $a\in M$ is algebraic over $K$, there is a non-zero $f(x)\in K[x]$ with $f(a)=0$.

Since $K\leq L$ we have that $f(x)\in L[x]$ and $f(a)=0$. Therefore, $a$ is algebraic also over $L$.

We have that $\text{Irr}(a,K)\mid f(x)$ and $\text{Irr}(a,K)\mid f(x)$, or not? (Wondering)

How could we continue to conclude that the minimal polynomial of $a$ over $L$ divides the minimal polynomial of $a$ over $K$ ? (Wondering) I haven't really understood the part "if we consider this polynomial as an element of $L[x]$"... (Thinking)

 

[Deveno]

Since both $K,L$ are fields, we have that both $K[x],L[x]$ are Euclidean domains.

Let $m_K(x)$ be the minimal polynomial of $a$ in $K[x]$, and let $m_L(x)$ be the same in $L[x]$.

Since $K \leq L$, we can consider $m_K$ as an element of $L[x]$.

Apply the division algorithm in $L[x]$, to obtain:

$m_K(x) = q(x)m_L(x) + r(x)$ where $\text{deg}(r) < \text{deg}(m_L)$, or $r = 0$.

Assume that $r \neq 0$. We then have:

$0 = m_K(a) = q(a)m_L(a) + r(a) = q(a)\cdot 0 + r(a) = r(a)$.

Then $r(x) \in L[x]$ is a polynomial of lesser degree (than $m_L$) for which $r(a) = 0$, contradicting the minimality of $m_L$ over $L[x]$.