Value group of simple field extension: Are these value groups equal? Suppose that a field extension $L/K$ is finite,
K is a Henselian field with a exponential valuation $v$, 
and $w$ is an extension of $v$ to $L$.
(If it is necessary, we can also assume that $L/K$ is tamely ramified.)
Then, is the following claim true?

Claim.
  $$x \in L,\; w(x) \in v(K^\ast)
\implies w(K(x)^\ast) = v(K^\ast)$$

The following is my try.
Since $x$ is algebraic over $K$, when I write $n := [K(x) : K]$,
I can write $y \in K(x)$ as $y = \sum_{i = 0}^{n-1} a_ix^i$.
Because $w$ is nonarchimedean, I get
$$ w(y) \geq \min_i \{ a_i x^i \}. $$
If $w$ were discrete, I could see $w(y) \in w(K^\ast)$,
but now I can't get anything from this.
If it is true, I want to use it for Chapter II, Section 7, Exercise 3
of Neukirch, "Algebraic Number Theory".
This is the exercise.

If $L/K$ is a totally tamely ramified extension of Henselian fields,
  intermediate fields of $L/K$ correspond 1-1 to the subgroups of $w(L^\ast)$ which contains $v(K^\ast)$.

The following is my thought.
For a subgroup $v(K^\ast) \subset H \subset w(L^\ast)$ and an intermediate field$K \subset M \subset L$, I put
$$ \Phi(H) = \{ x \in L \mid x = 0 \lor w(x) \in H \} $$
$$ \Psi(M) = w(M^\ast) $$
I want to show $\Phi \circ \Psi = \text{id}$ i.e. $\Phi \circ \Psi(M) = M$.
($\supset$) is easy.
To show ($\subset$), suppose $x \in L^\ast$, $w(x) \in w(M^\ast)$.
Let $M' = M(x)$.
By using claim, I see $w(M'^\ast) = w(M^\ast)$ i.e. $e = (w(M'^\ast) : w(M^\ast)) = 1$.
On the other hand, since $M'/M$ is tamely ramified, I can use a fundamental identity, so I get
$$ [M':M] = e \cdot f = 1 $$
(inertia degree $f = 1$ by $L/K$ is totally tamely ramified).
Then I get $M' = M$, so $x \in M$.
 A: The claim is false. For $p \neq 2$, take $K = \mathbb{Q}_p$, $L = \mathbb{Q}_p(\zeta_p)$, and $x = \zeta_p$. For what it is worth, the extension $L/K$ is tamely totally ramified.
For the exercise, there is a concrete description of tamely totally ramified extensions $L/K$ with degree $n$: $L = K(\sqrt[n]{\pi})$ for some uniformizer $\pi$ of $K$. The intermediate fields are $K(\sqrt[d]{\pi})$ as $d$ runs over divisors of $n$. At the same time, $w(L^*)/v(K^*)$ is cyclic of order $n$, so its subgroups are in one-to-one correspondence with divisors of $n$. Of course, when we write "divisors," we mean positive divisors.

But wait Erin, can you explain/give a reference for why every tamely finite totally ramified extension is obtained by adjoining an $n$th root of a uniformizer? In general, we know that the generator element must satisfy  an Eisenstein polynomial. How can we deduce that we can find a generator which satisfies $X^n - \pi$?

The Eisenstein polynomial is just using the totally ramified condition. We have to use the tame part too. See pages 52-53 in Lang's Algebraic Number Theory or see the chapter on the Kronecker-Weber theorem in Washington's Introduction to Cyclotomic Fields. The proof uses Krasner's lemma. And the uniformizer depends on the extension, e.g. over $\mathbb{Q}_p$ such extensions are $\mathbb{Q}_p(\sqrt[n]{pu})$ for some $p$-adic unit $u$, but $u$ usually is not $1$.
