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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?

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    $\begingroup$ How is that thing $\;A=\left(\frac{a,b}F\right)\;$ related to quaternions? Do you mean Hamilton quaternions $\;\Bbb H\;$ or something else? $\endgroup$ – DonAntonio Feb 25 '16 at 11:42
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    $\begingroup$ @Joanpemo: follow the links in the question. $\endgroup$ – Rob Arthan Feb 25 '16 at 11:43
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    $\begingroup$ @RobArthan Thank you very much. Then it is the usual quaternion algebra. $\endgroup$ – DonAntonio Feb 25 '16 at 11:44
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    $\begingroup$ Your argument would imply that $A$ is a division algebra, but that is not the case in general according to the paper linked to in the other question. $\endgroup$ – Rob Arthan Feb 25 '16 at 11:48
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    $\begingroup$ Does the link you gave not already show the problem with your argument? Look below the horizontal line at that link. They begin by saying that the norm is in $I$, and then they say it would be nice if it were nonzero. In your argument, you just assume that it is, but in the comments at the other page, you can see that $x \neq 0$ does not necessarily imply that $v(x) \neq 0$. $\endgroup$ – Barry Smith Feb 25 '16 at 11:49
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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.

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  • $\begingroup$ I see the problem now. Thanks. $\endgroup$ – yoyostein Feb 25 '16 at 12:50

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