I need to have really good picture in my mind about transcendental basis and What are the similarity between basis and transcendental basis. Let $\mathbb R$ be the set of all really number. Let $\mathbb B$ be a basis of $\mathbb R$ over $\mathbb Q$ so it is a maximal linear independent set and it spans all element of $\mathbb R.$ Also, each element of $\mathbb R$ can be written as linear combination of finitly many element of $\mathbb B.$
Let $T$ be a transcendental basis of $\mathbb R$ over $\mathbb Q$ so it is a maximal algebraically independent set. Also, $\mathbb Q(T)=\mathbb R.$
Is it true that each element of $\mathbb R$ can be spanned by finitely number algebraically independent elements of $T$? if so, why.
Can we choose for each element of $\mathbb R$ different element of $T$ that spans it? if so , why ?
what we can also say about $T$?
I hope from those that have a deep understanding for this subject to say more about transcendental basis and algebraically independent set?
Any help will be appreciated greatly
Best Answer
You're conflating transcendence basis and vector space (Hamel) basis.
If $K$ is an extension field of $F$, there exists a set $T$ consisting of transcendental elements of $K$ such that
Such a set is a transcendence basis of $K$ over $F$.
A set $U\subseteq K$ is algebraically independent if and only if, for every finite subset $S=\{a_1,a_2,\dots,a_n\}$ (elements pairwise distinct) of $U$, there is no nonzero polynomial $f(X_1,\dots,X_n)\in F[X_1,\dots,X_n]$ such that $f(a_1,\dots,a_n)=0$.
Using a technique very similar to the proof of existence of vector space basis, one can show that a transcendence basis exists and that any maximal algebraically independent set is such a basis; also, two transcendence bases have the same cardinality.
So let $T$ be a transcendence basis of $K$ over $F$. It is not possible to prove that $K=F(T)$, when $T$ is a transcendence basis. To see why, consider the simple case when $K=F(t)$, where $t$ is transcendental over $F$; then also $\{t^2\}$ is a transcendence basis, but clearly $K\ne F(t^2)$. On the other hand, $t$ is algebraic over $F(t^2)$, so $K$ is indeed algebraic over $F(t^2)$. Perhaps more simply, if $K$ is algebraic over $F$, then the empty set is a transcendence basis, but $K$ need not equal $F$.
In general, if $T$ is nonempty transcendence basis and you replace one of its elements, say $t$, by its square $t^2$, then the set $T'$ so obtained is still a transcendence basis, but definitely $F(T')\subsetneq F(T)$. So condition 2 above cannot be replaced by $K=F(T)$.
In the finite case $K=F(t)$ it is certainly false that a transcendence basis is also a vector space basis: indeed, $K=F(t)$ is infinite dimensional as vector space over $F$.
In the case of $F=\mathbb{Q}$ and $K=\mathbb{R}$ a transcendence basis must have the same cardinality as $\mathbb{R}$, by a cardinality argument: we have $$ \mathbb{Q}(T)=\bigcup_{\substack{S\subseteq T\\S\text{ finite}}}\mathbb{Q}(S) $$ and every subfield $\mathbb{Q}(S)$ is countable.
Also a vector space basis has the same cardinality as $\mathbb{R}$, but a transcendence basis cannot be a vector space basis. Let $T'=T\setminus\{t\}$, where $t\in T$ is fixed. Then $\mathbb{Q}(T)=F(t)$, where $F=\mathbb{Q}(T')$. Then $\mathbb{Q}(T)$ is infinite dimensional over $F$. If you remove one element from a vector space basis, the subspace spanned by the new set has codimension one.
What you can say is that every $r\in\mathbb{R}$ is algebraic over $\mathbb{Q}(S)$, where $S$ is some finite subset of $T$. Indeed, if $f$ is the minimal polynomial of $r$ over $\mathbb{Q}(T)$, it has finitely many coefficients, so it belongs to $\mathbb{Q}(S)[X]$, for some finite $S\subseteq T$.
For completeness, $\mathbb{R}$ is not a purely transcendental extension of $\mathbb{Q}$, that is, for every transcendence basis $T$ of $\mathbb{R}$ over $\mathbb{Q}$, $\mathbb{Q}(T)\subsetneq\mathbb{R}$. Indeed if equality holds, any permutation of $T$ would induce an automorphism of $\mathbb{R}$, but $\mathbb{R}$ has only the identity automorphism.