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I was studying field theory when a question came to my mind.

It is known that, for a simple extension $K(a)/K$, there exists a $K$-basis of $K(a)$ involving $a$: if $a$ is algebraic over $K$, then it is formed by finitely-many powers of $a$; otherwise, it is as stated here (thanks to a comment).

My question is if this insight can be generalised to other kinds of extensions. More precisely, given an extension $K(S)/K$, can one from a $K$-basis of $K(S)$ starting from $S$?

Amanda Wealth
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    If $a$ is not algebraic the powers do not form a basis. – Tina Mar 27 '25 at 10:40
  • I think this question is rather broad. Maybe fokus it a bit, for example to algebraic extensions, finite degree extension or else. – Tina Mar 27 '25 at 10:42
  • @Tina Thanks for your point, I hope the question is correct now. – Amanda Wealth Mar 27 '25 at 12:51
  • I think it is still not a basis, but only a generating set ? – Zoudelong Mar 27 '25 at 12:58
  • @Zoudelong Could you exhibit a non-trivial combination resulting in 0? So I can correct the question again, in case. – Amanda Wealth Mar 27 '25 at 13:01
  • For example, $\frac{s}{s+1}+\frac{1}{s+1}-\frac{1}{1}=0$? – Zoudelong Mar 27 '25 at 13:03
  • I don't know either, and I found the answer here: https://math.stackexchange.com/questions/126747/a-basis-for-kx-regarded-as-a-vector-space-over-k

    It uses the existence and uniqueness of a partial fraction decomposition for rational functions.

     – Zoudelong Mar 27 '25 at 13:07
  • @Zoudelong That's what I was thinking about, bounding the degree of the numerator. After all, one can use Euclidean division to obtain the other ones. Thank you. Maybe this insight can lead to a basis of $K(S)$ also in the general case. – Amanda Wealth Mar 27 '25 at 13:10
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    @AmandaWealth Yeah, by combining the above and tower theorem again... But it may need further assumptions for $S$, e.g. $S$ is a transcendental basis of $F(S)/F$... It is not easy to get a concise version as the algebraic case. – Zoudelong Mar 27 '25 at 13:22
  • If $F[X]$ is the ring of polynomials, $F(X)$ its field of fractions and $x$ transcendental over $F$, then $F(x) \cong F(X)$ is basically the same thing. I think a $F$-Basis is given by ${x^n/f(x)^k \mid k,n\in \mathbb N_0, n < \operatorname {deg} f, f \text{ normed and irreducible over } F}\cup { x^n \mid n\in \mathbb N_0}$, but maybe its just a generating set you should check this. This can be proven by some generalised partial fraction decomposition. – Tina Mar 27 '25 at 16:03

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As suggested by @Tina, let me focus on a simple case:

Let $F\leqslant E$ be a field extension, $S\subseteq E$ be a finite set whose elements are algebraic over $F$. Then the field extension $F\leqslant F(S)$ can be decomposed into a tower of simple extensions: $$ F\leqslant F(s_1)\leqslant F(s_1)(s_2)\leqslant \dots \leqslant F(s_1)\dots(s_n)$$

Then by tower theorem:

For (not necessarily finite) field extensions $F\leqslant K\leqslant L$, if $F\leqslant K$ has a basis $\{\alpha_i:i\in I\}$, $K\leqslant L$ has a basis $\{\beta_j:j\in J \}$, then $F\leqslant L$ has a basis $\{\alpha_i\beta_j: (i,j)\in I\times J\}$.

We know that $F\leqslant F(S)$ has a basis of the form: $$ \left\{s_1^{\alpha_1}\cdots s_n^{\alpha_n}: \alpha_i\leqslant N_i-1\right\}$$ where $N_i$ is the degree of the extension $F(s_1,\dots,s_{i-1})\leqslant F(s_1,\dots,s_i)$.

Warning: If $s$ is not algebraic over $F$, then $\{1,s,s^2,\dots,\}$ is a basis of $F[s]$, but not a basis of $F(s)$.

Zoudelong
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