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If a have a dominant morphism $\pi:X \rightarrow \mathbb{P}^{1}$ where $X$ is a projective, geometrically integral $k$-scheme. Then this gives rise to a pullback map \begin{align*} \pi^{*}:\text{Br}(k(\mathbb{P}^{1}))\rightarrow \text{Br}(k(X)) \end{align*} where $k(Y)$ denotes the function field a scheme $Y$.

I assume this arises from the fact we have a homomorphism of function fields $\pi_{1}^{*}:k(\mathbb{P^{1}})\rightarrow k(X)$ via composition $g \rightarrow g\circ \pi$, then we can just apply the Brauer group as a functor from the category of fields to the category of abelian groups, someone please correct me if I have this wrong.

My question is if I have quaternion algebra $A =(a,f(t))$ over $k(\mathbb{P}^{1})=k(t)$ what would the image of $A$ look like under the map $\pi^{*}$?

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    $\begingroup$ The quaternion algebra $(\pi ^*a,\pi ^*f(t))$. $\endgroup$
    – abx
    Commented Jun 5, 2021 at 19:35
  • $\begingroup$ @abx Where $\pi^{*}$ is the same as the pullback on the function field ? (In new edit $\pi_{1}^{*}$) $\endgroup$
    – H U
    Commented Jun 5, 2021 at 19:38
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    $\begingroup$ Yes. This is just standard functoriality. $\endgroup$
    – abx
    Commented Jun 6, 2021 at 4:26

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