I was wondering if the following simple (pun unintended) proof of the quaternion algebra $A=\left(\frac{a,b}{F}\right)$ being simple is valid.
I saw many more complicated proofs online, eg: Proof that Quaternion Algebras are simple.
Note: This proof uses the fact that the norm $v(x)=0$ implies $x=0$, where $v(x)=xx^*$.
Let $I$ be a nonzero ideal of $A$. Let $x$ be a nonzero element in $I$.
Then $xx^*=v(x)\in I$.
Since $x\neq 0$, $v(x)\neq 0$. Since $v(x)\in F^\times$, so $\frac{1}{v(x)}\in F$. Thus $v(x)\times\frac{1}{v(x)}=1\in I$. If $1$ is in $I$, we can say that $I$ must be equal to $A$.
Is this valid?
See the notes referred to in the other question. Without some assumptions on $a$, $b$ and $F$, $A$ is not a division algebra and has non-zero elements whose norms are zero. E.g., take $a = b = -1$ and $F=\Bbb{C}$. and let me write $I$, $J$ and $K$ for the generators of $A$ (so that $i \in \Bbb{C}$ means what it usually does). Then $i$ commutes with $I$ and we have $\nu(i + I) = (i + I)(i - I) = i^2 - I^2 = -1 - a = 0$. If $A$ happens to be a division algebra, then your argument is correct, but it doesn't work in general. See the notes for lots more information.