Good morning. I am trying to prove the following statement (I suspect it is true):
Consider a finite set $(a_{1},...,a_{n})$ of elements in an elementary extension $*\mathbb{Q}_{p}$ of $\mathbb{Q}_{p}$. Is $\mathbb{Q}(a_{1},...,a_{n})$ dense in $\mathbb{Q}_{p}(a_{1},...,a_{n})$? (The $a_{1},...,a_{n}$ are transcendental over $\mathbb{Q}_{p}$).
I attempted the proof as follows:
We can view polynomials from $\mathbb{Q}_{p}(a_{1},...,a_{n})$ as a quotient of two linear combinations of elements from $(a_{1},...,a_{n})$ raised to various powers and with coefficients from $\mathbb{Q}_{p}$. For each term $q_{i}a_{i}^{k}$, i used the fact that every $q_{i}$ in $\mathbb{Q}_{p}(a_{1},...,a_{n})$, if not an element of $\mathbb{Q}(a_{1},...,a_{n})$, is a limit point so there exists a net $(b_{\alpha})_{\alpha}$ in the latter which converges to $q_{i}$. Since $a_{i}^{k}$ is also in $\mathbb{Q}(a_{1},...,a_{n})$, so too is the net $ (b_{\alpha}a_{i}^{k})_{\alpha} \to q_{i}a_{i}^{k}$.
We can do this for every constituent term of the polynomial in the numerator, then form a net by adding the nets used for the individual terms. This net eventually converges to the entire polynomial in the numerator.
I then hoped to repeat this process for the denominator and divide the two nets, which would show that every element of $\mathbb{Q}_{p}(a_{1},...,a_{n})$ is a limit point of a net in $\mathbb{Q}(a_{1},...,a_{n})$, but I am concerned that the denominator of such a net might reach $0$ at some point prior to convergence, and can't think of a way to work around that.
Is there any convenient way to exclude this case, or will I need to revise my entire attempt?
Thank you in advance.