Basis and transcendental basis

Question

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

Answer 1

Is it true that each element of $\mathbb R$ can be spanned by finitely number algebraically independent elements of $T$ ?

"Spanned" in the sense of linear combination, no. But any element of $r \in \mathbb R$ is a rational function of finitely many elements of $T$. That is what $\mathbb R = \mathbb Q(T)$ means. So linear combinations are not enough. Starting with $T$, you have to allow products and quotients as well as sums.

Can we choose for each element of $\mathbb R$ different element of $T$ that spans it?

This seems nonsensical to me. I do not know what "$r$ is spanned by $t$" means, where $r$ is a real number and $t$ is one element of $T$.

What can we say about $T$?

From $\mathbb R = \mathbb Q(T)$ it follows that $T$ is uncountable.

Answer 2

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

1. $T$ is algebraically independent;
2. $K$ is algebraic over $F(T)$.

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.