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I am aware that for a non-archimedean field $K$ and its completion $\hat{K}$ with respect to a valuation $v$ (and corresponding absolute value $|\cdot|$), with $v$ extending to valuation $\hat{v}$ on $\hat{K}$, the value groups of $v$ and $\hat{v}$ coincide: it is clear $v(K) \subseteq \hat{v}(\hat{K})$, but by considering a Cauchy sequence $x_n \to x$ in $\hat{K}$, eventually we must have $|x_n - x| \leq |x|$ for $n$ sufficiently large, and therefore $|x_n| = |x + (x_n - x)| = |x|$ by the non-archimedean property, so any valuation obtained in $\hat{K}$ is also obtained in $K$.

However, is this property also true if the field is archimedean: the value groups coincide? I'd be very grateful for a proof or counterexample, as I couldn't come up with one myself. Many thanks.

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It depends upon what you allow as a valuation. Does $\mathbf Q$ with its usual archimedean absolute value count as a $K$? You could define $v$ by setting $v(a) = \log |a|$. –  Dylan Moreland Jan 30 '12 at 6:17
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