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The multiplicative Hilbert theorem 90 states that, for a finite Galois extension $L/K$ with Galois group $G$, every multiplicative cocycle $c:G\to L^{\times }$ satisfying $c_{\sigma \tau }=c_{\sigma }\sigma (c_{\tau })$ has the form $c_{\sigma }=\sigma (y)/y$ for some $y\in L^{\times }$. In the cyclic case $G=⟨\sigma ⟩$, it says

Here is the full existence argument. Distinct field homomorphisms $\sigma :L\to L$ are linearly independent as functions over $L$. To see this Artin independence theorem directly, assume a nonzero identically vanishing relation with the fewest nonzero coefficients. There are at least two terms. Choose two participating embeddings ${\sigma }_1,{\sigma }_j$ and an element $a$ with ${\sigma }_1(a)\ne {\sigma }_j(a)$. Evaluate the relation at $at$ and subtract ${\sigma }_j(a)$ times its value at $t$. The $j$th term disappears while the first remains nonzero, giving a shorter vanishing relation, a contradiction.

Apply this independence toThe coefficients are nonzero, so $A$ is not identically zero; choose $t$ with $y=A(t)\ne 0$. The cocycle identity gives $\tau (c_{\sigma }^{-1})=c_{\tau \sigma }^{-1}{c}_{\tau }$, and henceThis proves the general statement. In the cyclic case, given $N(x)=1$, set $c_1=1$ and $c_{{\sigma }^j}=x\sigma (x)\cdots {\sigma }^{j-1}(x)$ for $1\le j<n$. The norm-one identity makes this definition compatible with ${\sigma }^n=1$, and the cocycle identity follows by concatenating the products. The argument above gives $\sigma (y)/y=x$. Conversely that quotient has norm one because its conjugate product telescopes.

Finally let $L/K$ be unramified, and take its arithmetic Frobenius $\varphi$. For a unit $x$ of norm one, the cyclic theorem gives $b\in L^{\times }$ with $\varphi (b)/b=x$. Write $r=v_L(b)$ and choose the common uniformizer $\pi \in K$. Then $y={\pi }^{-r}b$ has valuation zero, and $\varphi (\pi )=\pi$ implies $\varphi (y)/y=\varphi (b)/b=x$. Thus the unit form of Hilbert theorem 90 isThe unramified hypothesis is what permits shifting an arbitrary integer valuation of $b$ by a power of a uniformizer belonging to $K$.