If K and L are extensions of Q, the rationals. why is it that if there is a prime p in Z such that p is unramified in K and totally ramified in L, then K and L are linearly disjoint. Any ideas? How does linearly disjointedness even come in?
linearly disjoint ramified primes
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number-theory
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1Let $F=K\cap L$. Then since $F$ is a subfield of $K$, $p$ is unramified in $F$. On the other hand, since $F$ is a subfield of $K$, $p$ must be totally ramified in $F$. The only way $p$ can be both unramified and totally ramified in $F$ is if $F=\mathbf{Q}$. If one of $K,L$ is Galois over $\mathbf{Q}$, then this implies $K$ and $L$ are linearly disjoint, but in general, I can't immediately see that this argument gives you linear disjointness. – 2012-11-11
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1Linear disjointness means $K\otimes_\mathbf{Q}L$ is a domain (equivalently since these extensions are finite, a field). If $K^g$ and $L^g$ are the Galois closures of $K$ and $L$, then $K\otimes_\mathbf{Q}L$ is a subring of $K^g\otimes_\mathbf{Q}L^g$, so $K$ and $L$ are linearly disjoint if $K^g$ and $L^g$ are. It's true that $p$ is unramified in $K^g$, but I don't think it has to be totally ramified in $L^g$. The problem is that (I think) $p$ need not be totally ramified in a composite of extensions in which it is totally ramified. So the argument I gave above can't be applied – 2012-11-11
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0to $K^g$ and $L^g$. – 2012-11-11
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0I'm sorry I'm not familiar with the definitely of your linear disjoint, I thought it means no basis vector in one field extension can appear in the other, or any subset of one basis remains independent over the other. – 2012-11-11
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0Keenan, This argument also shows that it suffices to show that $K^g$ and $L$ are linearly disjoint. But you have done that in your first comment. – 2012-11-11
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0@Jeff Ah, you're right. That never even occurred to me. Thanks! – 2012-11-11