Question 1199314: How to see that a complete subfield F in Q_p with absolute value | |_p, is actually Q_p itself?
We have one inclusion: F\subset Q_p.
Trying to show that Q_p\subset F. Q_p is complete with respect to | |_p. take an element x in F\subset Q_p, so there exists a cauchy sequence x_n in Q_p such that x_n—>x.
But F id also complete so there exists y_n in F such that
y_n—>x, but then x_n=y_n, so can we say that Q_p\subset F and we're done?
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Key Idea**
* The crux of the argument lies in the uniqueness of limits in complete metric spaces.
**2. Proof Outline**
* **Assume:** We have a complete subfield F of Q_p. This means F is a field itself, and it's complete with respect to the same p-adic absolute value |.|_p as Q_p.
* **Show Q_p ⊆ F:**
* Take any x ∈ Q_p.
* Since Q_p is complete, there exists a Cauchy sequence (x_n) in Q ⊆ F such that x_n → x in Q_p.
* Since F is a subfield of Q_p, it contains all rational numbers (Q). Therefore, (x_n) is also a Cauchy sequence in F.
* Since F is complete, this Cauchy sequence (x_n) must converge to a limit y in F.
* **Uniqueness of Limits:** In any metric space (and hence in Q_p and F), limits are unique. Since x_n → x in Q_p and x_n → y in F, we must have x = y.
* This shows that for any x ∈ Q_p, there exists a corresponding y ∈ F (namely, y = x).
* Therefore, Q_p ⊆ F.
* **Conclusion:**
* We started with F ⊆ Q_p and proved Q_p ⊆ F.
* Combining these, we conclude that F = Q_p.
**In essence:**
* The completeness of both F and Q_p, along with the uniqueness of limits in complete metric spaces, forces any complete subfield of Q_p to be equal to Q_p itself.
**Note:** This proof relies heavily on the uniqueness of limits in complete metric spaces.
Let me know if you have any other questions or would like to explore related concepts!
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