Degree of transcendental basis

[mathmari]

Hey!

Let $E_1, E_2$ be extensions of the field $F$, that are contained in a bigger field, so it makes sense to consider the field $E_1E_2$.

Without using the proposition:
$F \leq K \leq E \Rightarrow \text{tr.deg}(E/F) = \text{tr.deg}(E/K) + \text{tr.deg}(K/F)$

I want to show that $$\text{tr.deg}(E_1E_2/F) \leq \text{tr.deg}(E_1/F) + \text{tr.deg}(E_2/F)$$

In the book there is the following proposition:

If we have a field extension $E/F$, and the sets $S,T$ such that $S\subseteq T\subseteq E$, where $S$ is algebraically independent over $F$ and $T$ spans algebraically $E/F$, then there exists a transcendental basis $B$ of $E/F$, such that $S\subseteq B\subseteq T$.

From that it follow that each subset of $E$, that is algebraically independent over $F$ that is contained in a transcendental basis of $E/F$ and each subset of $E$ that spans algebraically $E/F$ contains a transcendental basis of $E/F$.

I have done the following:

Let $B_1$ be a transcendental basis of $E_1/F$ and let $B_2$ be a transcendental basis of $E_2/F$.

So, we have that the extensions $E_1/F(B_1)$ and $E_2/F(B_2)$ are algebraic.

We have that $B_1\subset B_1\cup B_2$ and $B_2\subset B_1\cup B_2$.

Do we get the extensions $F(B_1)\leq F(B_1\cup B_2)\leq E_1$ and $F(B_2)\leq F(B_1\cup B_2)\leq E_2$ ? Or do we not know if $B_2\subseteq E_1$ and $B_1\subseteq E_2$ ? (Wondering)

Since the extensions $E/F(B_1)$ and $E/F(B_2)$ are algebraic, we have that the extensions $E_1/F(B_1\cup B_2)$ and $E_2/F(B_1\cup B_2)$ are also algebraic, or not?

Is everything correct so far? (Wondering)

Do we have to show now that the extension $E_1E_2/F(B_1\cup B_2)$ is also algebraic?

Or how could we continue? (Wondering)

 

[jvicens]

Hello! Your approach so far is correct. To show that $E_1E_2/F(B_1\cup B_2)$ is algebraic, we can use the proposition you mentioned earlier. Since $B_1$ and $B_2$ are both transcendental bases of $E_1/F$ and $E_2/F$ respectively, they are also algebraically independent over $F$. Therefore, $B_1\cup B_2$ is also algebraically independent over $F$.

Now, to show that $B_1\cup B_2$ spans algebraically $E_1E_2/F$, we can use the fact that $B_1$ and $B_2$ are transcendental bases of $E_1/F$ and $E_2/F$ respectively. This means that any element in $E_1$ and $E_2$ can be written as a polynomial in elements of $B_1$ and $B_2$ respectively. Therefore, any element in $E_1E_2$ can be written as a polynomial in elements of $B_1\cup B_2$.

Thus, by the proposition, $B_1\cup B_2$ is a transcendental basis of $E_1E_2/F$. This means that $\text{tr.deg}(E_1E_2/F) = |B_1\cup B_2| = |B_1| + |B_2| = \text{tr.deg}(E_1/F) + \text{tr.deg}(E_2/F)$.

Therefore, we have shown that $\text{tr.deg}(E_1E_2/F) \leq \text{tr.deg}(E_1/F) + \text{tr.deg}(E_2/F)$ without using the proposition you mentioned. I hope this helps!