Let $L/K$ be a cubic extension defined by an irreducible cubic polynomial $F(x)$, in other words, $L \simeq K[x]/{(F(x))}$. We assume that $F$ does not have a multiple root in an algebraic closure of $K$.
Let $v$ be a finite place of $K$ and suppose that $F(x)$ decomposes into $(x-a)g(x)$ for some $a\in K_v$ and $g(x)$ is irreducible over $K_v$. Let $M/{K_v}$ be a quadratic extension defined by $g(x)$, i.e., $M\simeq K_v[x]/{(g(x))}$. Then, we have an isomorphism $L_v:=K_v[x]/{(F(x))} \simeq K_v \times M$ from the decomposition $f(x)=(x-a)g(x)$. Let $v$ and $w$ be the normalized valuations of $K_v$ and $M$, respectively.
Now, I have a question. If we pick $\alpha \in L^\times$, and let $(\alpha_1, \alpha_2)$ be the image of $\alpha$ in $K_v \times M$ by the composition $$ L \hookrightarrow L_v=K_v[x]/{(F(x))} \simeq K_v \times M. $$ Then, by the definition of the normalized valuation, we have $$ v(N_{L/K}(\alpha))=v(\alpha_1)+f\cdot w(\alpha_2), $$ where $N_{L/K}$ is the norm map from $L^\times$ to $K^\times$, and $f$ is the inertia degree of $M$ over $K_v$. (Am I right? If then,) Is this always true that $$ v(\alpha_1) \equiv w(\alpha_2)\equiv 0 \pmod 2 $$ if $v(N_{L/K}(\alpha))$ is even?
For instance, if $f=2$ then we easily have $v(\alpha_1)$ is even. In this case, is it always true that $w(\alpha_2)$ is also even?
If $f=1$, then we just have $v(\alpha_1)+w(\alpha_2)$ is even.
I think there is some restriction on the possible values of $\alpha_1$ and $\alpha_2$ (from the fact that $F(x)$ is irreducible over K) but I do not know it at the moment. Is there any proof or a counterexample?