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$\Bbb R/ \Bbb Q$ has degree $\infty$ and is not algebraic

$\Bbb C/ \Bbb R$ has degree 2 and is algebraic

Is the degree of $\Bbb Q(x)$ over $\Bbb Q$ infinity or $1$?

The degree of an extension over a finite field is a positive integer, (if finite degree meant algebraic) does that mean $\Bbb F_{p^n}$ is algebraic over $\Bbb F_p$ for any $n\in \Bbb N$?

Now if an element $\alpha$ is algebraic over a field $F$ what can we say about $F(\alpha)$? The degree of that extension is certainly going to be finite (equal to the degree of the minimal polynomial $m_{\alpha,F}$). But does every element in $F(\alpha)$ have a corresponding polynomial in $F[x]$ that it is a root of?

If $\alpha$ was transcendental then the degree would be infinite and of course $F(\alpha)$ would not be algebraic since it would contain $\alpha$ itself...

Am I getting this right?

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Every element of $F(\alpha)$ would be algebraic over $F$.

To see this, suppose that $[F(\alpha):F]=n$ and suppose $a \in F(\alpha)$. Then, for each $i$ $a^i=a_{i,0}+a_{i,1}\alpha+\cdots+a_{i,n-1}\alpha^{n-1}$ for some $a_{i,0},\dots,a_{i,n-1} \in F$. By linear algebra, $a^0,\dots,a^n$ are linearly dependent (when $F(\alpha)$ is thought of as a vector space over $F$). This means that there exists $c_0,\dots,c_n$ such that $c_0+c_1a+\cdots+c_na^n=0$.

I have found that a lot of results about field extensions (and about Galois theory) are found by using linear algebra like this.

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