Sub-fields generated by algebraically independent set Let $\{a,b,c,d\}$ be algebraically independent set over $\mathbb Q$. Is the following true? I think it is yes but my argument is not strong enough so I am asking what is the easiest way to show it. Here is my question  $$\mathbb Q(a,b)\cap \mathbb Q(c,d)=\mathbb Q$$ where $\mathbb Q(a,b)$ is the smallest sub field the contains $\mathbb Q\cup \{a,b\}$
Any help will be greatly appreciated
 A: I think this problem touches on some of the hidden difficulty of polynomial rings, in that we are all very familiar with them, but it is often difficult to justify the properties that they "obviously" have. As such, I'll give two solutions - an "easy" and a "hard" one, which involve different levels of rigour, and emphasise different aspects of the problem.
Note that the easy version is not really a solution at all; it is just a restatement of the problem in terms that might make you more comfortable with the answer.
Easy mode
Algebraic independence of $\{a, b, c, d\}$ tells us that $\mathbb{Q}(a, b, c, d)$ is isomorphic to the ring $\mathbb{Q}(x_1, x_2, x_3, x_4)$ of four-variable rational functions over $\mathbb{Q}$. Under this isomorphism, the subfields $\mathbb{Q}(a,b)$ and $\mathbb{Q}(c, d)$ correspond to $\mathbb{Q}(x_1, x_2)$ and $\mathbb{Q}(x_3, x_4)$ respectively.
Since $\mathbb{Q}(x_1, x_2)$ and $\mathbb{Q}(x_3, x_4)$ are rational functions in different variables, an element in their intersection cannot involve any of the variables, and hence it is rational.
Hard mode
Suppose that $\alpha \in \mathbb{Q}(a, b) \cap \mathbb{Q}(c, d)$.
Since $\alpha \in \mathbb{Q}(a, b)$, we may express it as
$$\alpha = \frac{f_1(a, b)}{ g_1(a, b)},
$$
where $f_1, g_1$ are polynomials with rational coefficients. Similarly, we have
$$
\alpha = \frac{f_2(c, d)}{g_2(c, d)},
$$
for polynomials $f_2, g_2$. Assume that $\alpha \neq 0$ so that the polynomials $f_i, g_i\neq 0$.
Then we have
$$\frac{f_1(a,b)}{g_1(a,b)} = \frac{f_2(c,d)}{g_2(c,d)},$$
and hence
$$
f_1(a,b)g_2(c,d) = f_2(c,d)g_1(a,b).
$$
Set $p(x_1, x_2, x_3, x_4) = f_1(x_1, x_2)g_2(x_3,x_4) - f_2(x_3,x_4)g_1(x_1,x_2)$ in the polynomial ring $R = \mathbb{Q}[x_1, x_2, x_3, x_4]$. Then $p(a, b, c, d) = 0$, so by independence of $a,b,c,d$ we have $p = 0$ in $R$. It follows that
$$
f_1(x_1, x_2)g_2(x_3,x_4) = f_2(x_3,x_4)g_1(x_1,x_2).
$$
View the equation $f_1(x_1, x_2)g_2(x_3,x_4) = f_2(x_3,x_4)g_1(x_1,x_2)$ in the ring $\mathbb{Q}(x_3,x_4)[x_1, x_2]$ of polynomials in $x_1, x_2$ over the field $\mathbb{Q}(x_3, x_4)$.
Write $f_1(x_1, x_2) = \sum_{i,j}a_{ij}x_1^ix_2^j$ and $g_1(x_1, x_2) = \sum_{i,j}b_{ij}x_1^ix_2^j$, where the $a_{ij}$ and $b_{ij}$ are nonzero finitely often. Note that $f_1, g_1 \in \mathbb{Q}[x_1, x_2]$ by definition, so the $a_{ij}, b_{ij}$ are all in $\mathbb{Q}$. We have
$$
\sum a_{ij}g_2(x_3,x_4) x_1^ix_2^j = \sum b_{ij}f_2(x_3,x_4)x_1^ix_2^j,
$$
so $a_{ij}g_2 = b_{ij}f_2$ for all $i, j$. In particular, $a_{ij} = 0 \iff b_{ij} = 0$, and when these are nonzero we have
$$
\alpha = \frac{f_2}{g_2} = \frac{a_{ij}}{b_{ij}} \in \mathbb{Q}.
$$
