Where are extension elements in a field extension taken from?

Question

For example, how can you conclude that a finite field extension of an algebraic field $K$ is algebraic if you don't know the type of the extension elements (the elements that you add to extend the field $K$).

Answer 1

On the assumption that you mean an extension $E$ of finite degree $n$, then $E$ is a vector space of dimension $n$ over $K$ - so every set on $n+1$ elements of $E$ is linearly dependent over $K$.

Suppose $e$ is an element of $E$ and consider the set $1, e, e^2 \dots e^n$ - this has $n+1$ elements and a linear dependence over $K$ gives a polynomial over $K$ which is satisfied by $e$. So $e$ is algebraic over $K$ - and $e$ was arbitrary, so every element of $E$ is algebraic over $K$.

Answer 2

Say we add on just one extension element. Then we are working with a simple extension. A concrete way to think about such field extensions $K \hookrightarrow F $ is to have the polynomial ring $K[x]$ in mind.

If the extension is algebraic ($\dim_{K}F < \infty$) then $F \cong K[x]/\langle f\rangle $ for some irreducible polynomial $f$ (unique up to a non-zero element of $K$) with degree equal to the dimension of the extension. In this case, you can think of the extension element as the indeterminate $x$ in $K[x]$ but with the condition that in the particular combination of scalars and powers of $x$ that gives $f(x)$, it is equal to zero.

If the extension is transcendental ($\dim_{K}F \geq \infty$) then $F \cong K(x)$ and so in this case you can think of the extension element as the indeterminate $x$, but this time with no condition.

To address your particular question, one either has or has not information regarding the extension. If one does, then it is always determinable whether it is a finite-dimensional extension (i.e., algebraic). If one does not have information regarding conditions on the extension elements, then we are just working with general field extensions, which of course may be either transcendental or algebraic.

Answer 3

Anything can "come from" anywhere - the actual elements are not what matters, the structure of the field (i.e., its operations) is what matters. The field $\mathbb{Q}(\sqrt{2})$ that is a subfield of $\mathbb{C}$ is isomorphic to, and just as good as, the field $K$ defined by $$K=\mathbb{Q}[x]/(x^2-2),$$ which in turn is isomorphic to and just as good as the field $L$ defined by $$L=\{(a,b)\mid a,b\in\mathbb{Q}\}\quad\mathsf{\text{with operations}}\quad \begin{align*} (a,b) +(c,d) &= (a+c,b+d)\\ (a,b)\cdot (c,d)\, &= (ac+2bd,ad+bc) \end{align*}$$ The field $\mathbb{Q}(\sqrt{2})$ consists of complex numbers, the field $K$ consists of equivalence classes of rational polynomials, and the field $L$ consists of ordered pairs of rational numbers. They're all isomorphic, and any statement you make about the structural properties of one will be true of all of them.

See the Wikipedia article on transport of structure.