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I am trying to verify this fact that I am sure it is true (maybe not), but all the material online is just making me quite confused, and I would greatly appreciate some assistance.

I have a local field $K$ complete with respect to a discrete valuation $v$. Let $L$ be a finite extension of $K$. I want to say that there is a unique discrete valuation $w$ on $L$ such that $w|_K = v$. Is this true? Thank you.

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The answer is yes. If $K$ is a local field and $L/K$ is an algebraic extension (in particular, a finite extension), then the valuation $v_K$ can be extended uniquely to a valuation $v_L$ of $L$ such that $v_L$ restricted to $K$ is equal to $v_K$. This is one of the fundamental theorems about local fields. If $L/K$ is finite, $v_L$ is given by $$v_L (x)=\frac{1}{n}v_K(N_{L/K}(x)),$$ for every $x\in L$, where $n=[L:K]$ and $N_{L/K}$ is the norm.

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