Clearly, we cannot construct any such DVR's when $K=\mathbb{Q}$, or when $K=\mathbb{F}_q$, so we must look to more intricate examples. Is it possible to construct such DVR's when $\mathrm{char\,} K = 0$? My intuition tells me that we should consider positive characteristic, but I can't figure out how to construct anything like this.
Given a field, $K$, are there two DVRs $A,B$ for the same field $K$, such that $A\cap B$ is also a field?
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abstract-algebra
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0$char \mathbb Q=0$. – 2017-01-23
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0What is a D.V.R? – 2017-01-23
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0But $\mathbb{Q}$ is far from the only field with characteristic $0$. What's your point? – 2017-01-23
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0https://en.wikipedia.org/wiki/Discrete_valuation_ring (there are many equivalent definitions) – 2017-01-23
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0You are asking if this can be done in characteristic zero, which typically means if this is always doable in characteristic zero. If you are asking for an example, you should start your question by "does there exists a $K$ such that..." – 2017-01-23
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0I realized that the title was a little confusing, so I corrected it. – 2017-01-23
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4The answer is no. If you pick two DVRs with the same field of quotients, then their intersection is a ring with exactly two maximal ideals: in particular it can never be a field. – 2017-01-23
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3See Theorem 12.2 of [Matsumura , Commutative Ring Theory](http://www.math.hawaii.edu/~pavel/cmi/References/Matsumura_Commutative_Theory.pdf) for a proof of the fact I said in the above comment. – 2017-01-23