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Assume that $K,L$ are fields such that there is an isomorphism of groups $\mathrm{GL}_n(K) \cong \mathrm{GL}_n(L)$ for all $n \in \mathbb{N}$. Does it follow that $K \cong L$?

I am also interested in related results, for example if it is enough to test small values for $n$ and what happens for $\mathrm{PGL}_n$.

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Hmm... We know the group of units is isomorphic since $\mathrm{GL}_1(K)\cong \mathrm{GL}_1(L)$. Are there any examples of fields with isomorphic units that are not isomorphic? –  Alex Becker Dec 31 '13 at 23:21
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@Alex: If $R,S$ are factorial domains with $R^* \cong S^*$, and $R,S$ have the same number of prime elements (mod units), then $Q(R)^* \cong Q(S)^*$ (since $Q(R)^* \cong R^* \oplus \bigoplus\limits_{p} \mathbb{Z}$). For example we can take $R=\mathbb{Z}[x]$ and $S=\mathbb{Z}[x,y]$. Thus, $\mathbb{Q}(x)$ and $\mathbb{Q}(x,y)$ have isomorphic multiplicative groups. –  Martin Brandenburg Dec 31 '13 at 23:28
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Sorry I will delete this question within the next hours because mathoverflow.net/questions/106838 is the same. Actually it suffices to consider any given $n>1$ –  Martin Brandenburg Dec 31 '13 at 23:36
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This is a reasonable question, I suggest not deleting it. A quick summary (this is an old result of linear algebraic geometry, you can recover the field from the automorphism group of its projective geometry) would be nice. –  Jack Schmidt Jan 1 at 2:11
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@AlexBecker There are examples of non-isomorphic fields $K$ and $L$ with $(K,+)\simeq (L,+)$ and $(K^{\times},\cdot)\simeq (L^{\times},\cdot)$. –  user26857 Jan 1 at 10:40

1 Answer 1

Here is a proof when $\rm{char}(K)$ and $\rm{char}(L)$ are $\neq 2$.

Let $\phi : \rm{GL}_2(K) \xrightarrow{\sim} \rm{GL}_2(L)$ be an isomorphism. Then $\phi$ induces an isomorphism between the derived subgroups $\rm{SL_2}$.

Let's assume the following lemma (I may add the proof later) (we need $\rm{char}(K) \neq 2$ for this lemma).

Lemma: $\phi$ sends any transvection to a transvection.

Then $\phi(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $\phi(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$ are both transvections. Choose $e_1$ an eigen-vector of the former and $e_2$ an eigen-vector or the latter. Since the two matrices do not commute, $(e_1,e_2)$ is a basis. In this basis $\phi(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $\phi(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$ are of the form $(\begin{smallmatrix} 1 & * \\ 0 & 1 \end{smallmatrix})$ and $(\begin{smallmatrix} 1 & 0 \\ * & 1 \end{smallmatrix})$. So we can assume that $$\phi\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \quad \phi\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ * & 1 \end{pmatrix}.$$ The centralizer of $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ is $$C\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \left\{ \begin{pmatrix} a & x \\ 0 & a \end{pmatrix} ~|~ a,x \in K \text{ or } L \right\},$$ and the normalizer of $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ is $$N\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \left\{ \begin{pmatrix} a & x \\ 0 & b \end{pmatrix} ~|~ a,b,x \in K \text{ or } L \right\},$$ and same for $(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$.

The subgroup $\{ (\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}) \}$ is the set of transvections centralizing $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$, hence there is an additive isomorphism $\sigma : K \rightarrow L$, such that $\phi(\begin{smallmatrix} 1 & x \\ 0 & 1 \end{smallmatrix}) = (\begin{smallmatrix} 1 & \sigma(x) \\ 0 & 1 \end{smallmatrix})$ and $\sigma(1)=1$. If we show that $\sigma$ is multiplicative, then we are done.

The subgroup $\{ (\begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix})\}$ is the set of matrices in the derived subgroup normalizing both $(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix})$ and $(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix})$, hence there is multiplicative isomorphism $\tau : K^* \rightarrow L^*$ such that $\phi(\begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix}) = (\begin{smallmatrix} \tau(a) & 0 \\ 0 & \tau(a)^{-1} \end{smallmatrix})$. Note that $$\begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & 0 \\ 0 & a^{-1} \end{pmatrix}^{-1} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}.$$ Apply $\phi$ in both sides and get $$\begin{pmatrix} \tau(a) & 0 \\ 0 & \tau(a)^{-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \tau(a) & 0 \\ 0 & \tau(a)^{-1} \end{pmatrix}^{-1} = \begin{pmatrix} 1 & \sigma(a) \\ 0 & 1 \end{pmatrix}.$$ Hence $\sigma(a) = \tau(a)$ for all $a \in K^*$, and $\sigma : K \rightarrow L$ is an isomorphism of field.

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