division rings $D,D'$ both of finite dimension over their center $F$ are isomorphic as rings iff isomorphic as $F$-algebras? Say I have two division rings, $D,D'$, both with center $F$ and both are of finite dimension over $F$, for some field $F$. Now suppose that $D\cong D'$ as rings, does if follow that $D\cong D'$ as $F$ algebras?
From this question we know that for general $F$ algebras $A,B$ this is not necessarily the case, but I think that in this instance it is true.
My argument goes as follows:
Say $\Sigma:D\to D'$ is a ring isomorphism. One can easily see that it induces an automorphism $\sigma:F\to F$. From the Skolem-Noether theorem there is a $0\neq u\in D' $
s.t. $\sigma^{-1}(f)=ufu^{-1}$ for all $f\in F$. Taking $u\Sigma u^{-1}:D\to D'$ yields a ring isomorphism $D\to D'$ that is trivial on $F$, hence $D\cong D'$ as $F$-algebras.
Is this argument valid? It seems "too good to be true" especially due to the linked question.
 A: $\newcommand{\md}[1]{\left\lvert #1 \right\rvert}\newcommand{\Q}{\mathbb{Q}}\DeclareMathOperator{\Aut}{Aut}$The comment already points out a flaw in the argument (namely, the use of Skolem-Noether). I now give an example to show that the result is false.

Setup.
Set $F := \Q(\sqrt{2})$, $K := \Q(\iota \sqrt[4]{2})$, where $\iota \in \Bbb C$ is as usual. (For later reference, put $\mu = \iota \sqrt[4]{2}$.)
Define the ring $D = K \oplus K \theta$ subject to the relations
$$\theta^2 = -1, \quad \theta \cdot z = \bar{z} \theta.$$
(The addition is componentwise. $\bar{z}$ denotes usual complex conjugation.)
Let $\eta \in \Aut(F)$ be the map determined by $\sqrt{2} \mapsto -\sqrt{2}$.

Example.
We now give $D$ the structure of an $F$-algebra in two different ways. Note that this amounts to giving maps $F \to Z(D)$.
We have the natural inclusion of rings $F \subset K \subset D$.
This gives us two $F$-algebra structures on $D$: the first is the inclusion $\alpha : F \hookrightarrow D$, and the second is $\beta = \alpha \circ \eta$.
That is, $\beta(a + b\sqrt{2}) = a - b \sqrt{2}$, for $a, b \in \Q$.
(Note that in both cases, the image of the ring map is the copy of $F$.)

Justification.
Part 1. General result.
It may not be clear that $D$ is a division ring or that $F$ is exactly the center of $D$.
Both these facts follow from a more general construction due to Dickson. One reference is T.Y. Lam's A First Course in Noncommutative Rings $\S 14$ Some classical constructions. (You can directly jump to this section since your question indicates that you have the necessary background.)
It reduces to showing that $-1$ is not in the image of the norm map $\mathrm{N}_{K/F}$, by Corollary 14.8. This follows by our choice of $K$. Indeed, the norm map is just the usual norm of a complex number and hence, nonnegative.
Alternatively, you could try proving it explicitly in this case.
Clearly, $\dim_{F}(D) = 4 < \infty$.
Part 2. Specific details.
$D$ is isomorphic to $D$ as a ring. We wish to show that the two $F$-algebra structures are not isomorphic. This amounts to showing that there is no ring automorphism $\sigma : D \to D$ such that $$\beta = \sigma \circ \alpha.$$
Assume that such a $\sigma$ existed. Evaluate both sides of the above equation at $-\sqrt{2}$ to get
$$\sqrt{2} = \sigma(-\sqrt{2}).$$
But $-\sqrt{2} = \mu^2$. Thus, we get
$$\sqrt{2} = \sigma(\mu^2) = (\sigma(\mu))^2.$$
We now show that there is no element in $D$ whose square is $\sqrt{2}$, giving us the desired contradiction.
Indeed, assume that $z + w \theta \in D$ squares to $\sqrt{2}$, where $z, w \in K$. Using the definition of the multiplication in $D$, we get the equation
$$(z^2 - \md{w}^2) + 2 w \Re(z) \theta = \sqrt{2}.$$
Thus, either $w = 0$ or $z \in \iota \Bbb R$. In the former case, we are left with $z^2 = \sqrt{2}$, which is a contradiction since $\sqrt{2}$ has no square root in $K$.
In the latter case, we see that $z^2 < 0$. This again gives a contradiction since then $$z^2 - \md{w}^2 < 0 < \sqrt{2}.$$

Remark. I think the following played a role here: We had the automorphism $\eta$ of $F$ that does not extend to an automorphism of $K$.
