Isomorphism of algebras $E_v = \prod_{w\mid v} E_w$

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I am currently reading the article

Bayer–Fluckiger, Eva, B. Nivedita, and Raman Parimala. "Hasse principle for G–quadratic forms." Documenta Mathematica 18 (2013): 383-392.

in which there is a side remark which I would like to see a proof of. Unfortunatly I can neither find one in the literature nor have I been able to prove the statement myself so far. I'll briefly explain the setup.

Let $k$ be a global field of characteristic $\neq 2$ and $E$ a finite-dimensional simple $k$-algebra. If $v$ is a place of $k$ and $w$ a place of a finite extension field $K|k$ lying above $v$, we write $w\mid v$. Put $E_v := E \otimes_k k_v$. Now, the side remark I mentioned is "Note that $E_v = \prod_{w\mid v} E_w$."

What I have found so far is a similar statement for separable extions of valuated fields in Neurkich's Algebraic Number Theory, namely Theorem (8.3) of chapter II in the German edition.

If $L|K$ is separable, we have $L\otimes_K K_v \cong \prod_{w\mid v} L_w$.

Neukirch's proof of this theorem relies on the separability condition, in that it uses a primitive element $\alpha$, such that $L=K(\alpha)$. In the paper I am reading there is no mentioning of separability and clearly for a global field such as $\mathbb{F}_p(T)$ I cannot assume separability of an arbitrary extension.

Do you have any hints for me as to how I might be able to prove that there is an isomorphism $E_v \cong \prod_{w\mid v} E_w$ or does perhaps someone know of a suitable reference where I might find the proof of this statement?

PS: Note that the requirement $\mathrm{char}(k)\neq 2$ may be unrelated to the validity of the statement in question. I've just copied it from the setup in the article.