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I was toying around with central simple algebras over a field $K$ today and thought that I should try to verify Noether-Skolem's theorem that any automorphism of such must be inner. So, let us take $K = \mathbb{Q}[\sqrt{5}]$, and the quaternion algebra $(1,-4)_K$, i.e. $i_1^2=1$,$j_1^2=-4$ and $i_1j_1=-j_1i_1$. Now, we can see that this is isomorphic to $M_2(K)$, the matrix ring, where the isomorphism is given by $$j_1 \rightarrow \pmatrix{0 & -4 \\ 1 & 0},$$ $$i_1 \rightarrow \pmatrix{1 & 0 \\ 0 & -1 }.$$ Call $\psi$ the isomorphism $(1,-4)_K \rightarrow M_2(K)$. Now, we can see that as a quaternion algebra, $(1,-4)_K \cong (1,1)$ ( call the generators here for $i_2$ , $j_2$) by the isomorphism $$i_1 \rightarrow i_2,$$ $$j_1 \rightarrow j_2(1+i_2)$$ and we call this isomorphism by $\phi$. Now, $(1,1)_K$ is isomorphic to $M_2(K)$ as well, let $\sigma:(1,1)_K \rightarrow M_2(K)$ be the isomorphism (which is of the same "form" as $\psi$). Now, this induces an automorphism: $\sigma \circ \phi \circ \psi^{-1}: M_2(K) \rightarrow M_2(K)$, which takes $$e= \pmatrix{1 & 0 \\ 0 & -1 }$$ onto itself and takes $$f= \pmatrix{0 & -4 \\ 1 & 0}$$ onto $\sigma(j(1+i))$. Now, I can not get this to be an inner automorphism! I get that all entries must be zero, and this is clearly nonsense. So please, mathstackexchange, before I lose my mind (that's an exaggeration, but it is annoying), how can this be resolved?

Update (A concrete description of $\sigma(j(i+1))$

In the comments I got asked whether I could write out what $\sigma(j(i+1))$ was. I left it out on purpose, since I believe this might be where the error lies. However, this is how my thinking goes: $$\sigma(j(i+1))= \sigma(j)\sigma(i+1) = \pmatrix{0 & 1 \\ 1 & 0 } [ \pmatrix{1 & 0 \\ 0 & -1 } + \pmatrix{1 & 0 \\ 0 & 1}].$$

Multiplying this, I get $$\pmatrix{ 0 & 0 \\ 2 & 0 }.$$

So if we assume that we have an inner automorphism: we should have $$\pmatrix{a & 0 \\ 0 & b } \pmatrix{0 & -4 \\ 1 & 0 } = \pmatrix{0 & 0 \\ 2 & 0} \pmatrix{a & 0 \\ 0 & b}$$

but this can not hold! So, some of my assumptions should be wrong, or some step, but I really can't see which one. I suspect that it lies in the step where I multiply by a norm. But it should hold, but maybe my isomorphism isn't an isomorphism. But there should be one, and I am quite sure that that one is correct!

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  • $\begingroup$ rschwieb: I can! I will edit in a moment (but note that I might have done some miscalculation here, so be a bit careful, and don't trust me too much). $\endgroup$ – Tedar Aug 27 '13 at 20:26
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    $\begingroup$ rschwieb: There you go! $\endgroup$ – Tedar Aug 27 '13 at 20:42
  • $\begingroup$ Hmm. Which of those $i,j$ are elements of $(1,-4)_K$ and which belong to $(1,1)_K$? Assuming that $\phi$ goes from the former to latter I calculate: $$\phi(-4)=\phi(j^2)=(j(1+i))^2=j(1+i)j(1+i)=j^2(1-i)(1+i)=j^2(1-i^2)=1\cdot0=0,$$ but isn't $\phi(-4)=-4$ by linearity? $\endgroup$ – Jyrki Lahtonen Aug 27 '13 at 20:54
  • $\begingroup$ @JyrkiLahtonen: I have marked them up now! $\endgroup$ – Tedar Aug 27 '13 at 20:57
  • $\begingroup$ Something wrong with your isomorphism as $(1+i_2)$ is a zero divisor, hence so is $j_2(1+i_2)$, but $j_1$ is a unit. No isomorphism can map a unit to a zero-divisor. $\endgroup$ – Jyrki Lahtonen Aug 27 '13 at 20:59
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As indicated in the comments something is wrong with the isomorphism from $(1,-4)_K$ to $(1,1)_K$.

Would $\phi:i_1\mapsto i_2$, $j_1\mapsto j_2(1+\sqrt5 i_2)$ work? Then you have $$ \phi(j_1i_1)=j_2(1+\sqrt 5 i_2)i_2=j_2i_2(1+\sqrt5 i_2)=-i_2j_2(1+\sqrt5 i_2)=\phi(-i_1j_1) $$ as well as $$ \phi(-4)=\phi(j_1^2)=j_2(1+\sqrt5 i_2)j_2(1+\sqrt5 i_2)=j_2^2(1-\sqrt5i_2)(1+\sqrt5i_2)=j_2^2(1-5i_2^2)=-4 $$ as you should.

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  • $\begingroup$ Past midnight here. Can't check the part about Noether-Skolem, sorry. $\endgroup$ – Jyrki Lahtonen Aug 27 '13 at 21:12
  • $\begingroup$ +++++++++++++++++++ A million if I could! Some fun story behind this: I have been toying with this on and on for some time, not seeing where the error went. I trusted someone elses isomorphism (posted here on stackexchange) which wasn't an isomorphism. Lesson? Never, ever, trust an isomorphism until' you've checked it really is. $\endgroup$ – Tedar Aug 27 '13 at 21:51
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    $\begingroup$ Thanks. Glad to hear the problem is solved and the logical structure of math saved :-). The only clue I had was that may be $\sqrt5$ was included in the center for a reason? $\endgroup$ – Jyrki Lahtonen Aug 28 '13 at 3:52
  • $\begingroup$ Nicely done! ${}{}$ $\endgroup$ – rschwieb Aug 28 '13 at 13:22
  • $\begingroup$ @Tedar: Your remark made me curious, so I did some searching. If you are referring to this answer you will see that I am using more or less the same recipe. There the algebra had $i^2=5$ as opposed $i^2=1$ here. $\endgroup$ – Jyrki Lahtonen Aug 28 '13 at 19:53

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