linearly disjoint ramified primes 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?  
 A: I'm turning my comment into an answer (thanks to Jeff Tolliver for pointing out that it actually does provide an answer). 
If $K^g$ is the Galois closure of $K$ (over $\mathbf{Q}$), then since $p$ is unramified in each of the Galois conjugates of $K$, and $K^g$ is the composite of these conjugates, $p$ is unramified in $K^g$. Since $K\otimes_\mathbf{Q}L\rightarrow K^g\otimes_\mathbf{Q}L$ is is injective (since $L$ is flat over $\mathbf{Q}$), it suffices to prove that $K^g$ and $L$ are linearly disjoint (i.e. that $K^g\otimes_\mathbf{Q}L$ is a domain). Therefore I can replace $K$ with $K^g$ and reduce to the case where $K/\mathbf{Q}$ is Galois. 
Now I'm going to assume that $K$ and $L$ have been embedded into some common extension of $\mathbf{Q}$, so that it makes sense to take their composite and intersection. We then have a map $K\otimes_{\mathbf{Q}}L\rightarrow KL$, and I will prove it is injective. It is surjective because $K/\mathbf{Q}$ is finite.
Since $F:=K\cap L$ is contained in $K$, $p$ must be unramified in $F$. On the other hand, $F$ is contained in $L$, so $p$ must also be totally ramified in $F$. This forces $F=\mathbf{Q}$. A standard theorem in Galois theory says that the natural map $\mathrm{Gal}(KL/L)\rightarrow\mathrm{Gal}(K/\mathbf{Q})$ is injective with image $\mathrm{Gal}(K/F)$. Since $F=\mathbf{Q}$, the map is an isomorphism, and in particular, $[KL:L]=[K:\mathbf{Q}]=\dim_L(K\otimes_\mathbf{Q}L)$. This equality of dimensions implies that $K\otimes_\mathbf{Q}L\rightarrow KL$, which is $L$-linear and surjective, must be injective too. So $K\otimes_\mathbf{Q}L$ is a domain. 
EDIT: Here is a proof that the notion of linear disjointness mentioned by the OP in the comments is equivalent to the notion I'm using (in this situtation, when $K/\mathbf{Q}$ is finite and when both fields $K$ and $L$ are inside of some common field, so it makes sense to take their composite). Let $\alpha_1,\ldots,\alpha_n$ be a $\mathbf{Q}$-linearly independent set in $K$. I'm going to assume the elements form a $\mathbf{Q}$-basis (they are contained in one, so my argument goes through regardless). Then the tensors $\alpha_i\otimes 1$ form an $L$-basis for $K\otimes_\mathbf{Q}L$. So every element of $K\otimes_\mathbf{Q}L$ is of the form $\sum_{i=1}^n\alpha_i\otimes b_i$ for $b_i\in L$ (uniquely determined). The map $K\otimes_\mathbf{Q}L\rightarrow KL$ sends this element to $\sum_i\alpha_ib_i$. Thus the map is injective if and only if the $\alpha_i$ are $L$-linearly independent. 
So, if $K$ and $L$ are linearly disjoint in the sense of the OP, the map is injective, and so $K\otimes_\mathbf{Q}L$ is a domain. Conversely, if $K\otimes_\mathbf{Q}L$ is a domain, then, because it is finite over a field, it is a field, and therefore the map is injective, so the $\alpha_i$ are $L$-linearly independent, i.e., $K$ and $L$ are linearly disjoint in the sense of the OP. 
