# Involutions of full matrix ring $M_n(R)$

Hellow, I want to describe all involutions of full matrix ring over field and all involutions of matrix polynomial ring. Is it true or false that every involution of the full matrix ring $T = M_n(R)$ over field $R$ has the follwing form $$A \to C^{-1}A^TC,$$ for all $A\in M_n(R)$ and some fixed matrix $C$? What can we say about involutions of the matrix-polynomial ring $T[x]$?

• The matrix $C$ should be symmetric. – egreg Jan 19 '16 at 14:56
• Thank you, I agree with you. But is this an explicit descripton of all involutions? – Mikhail Goltvanitsa Jan 19 '16 at 15:06
• I assume that your involutions are linear. What about $A\rightarrow PA^TP^{-T}$ or $A\rightarrow -PA^TP^{-T}$ where $P$ is invertible or $A\rightarrow PAR$ where $P^2=R^2=I$. – loup blanc Jan 19 '16 at 15:53
• I think that for the third type of yor map the multiplicative property $\phi(AB) = \phi(B)\phi(A)$ of involution is not valid. – Mikhail Goltvanitsa Jan 19 '16 at 16:06
• "My definition" has the following form. Let $R$ be an arbitrary ring and $\phi: R\to R$ be a bijection such that $\forall r_1,r_2\in R$ $\ \phi(\phi(r_1)) = r_1, \ \phi(r_1+r_2) = \phi(r_1)+\phi(r_2),\ \phi(r_1r_2) = \phi(r_2)\phi(r_1)$. The most interesting problem for me is to describe maps with the third property in the ring $M_n(R)[x]$, where $R$ is Galois ring. Obviously, every involution of $M_n(R)$ can be extended to involution of $M_n(R$ by natural way. But i don't know are there exist other involutions... – Mikhail Goltvanitsa Jan 20 '16 at 11:02

Every involution of $T$ is of the form $\phi(A)=C^{-1}\sigma(A^T)C$ for an invertible matrix $C$, and a field automorphism $\sigma\colon R\to R$ satisfying $\sigma^2=\iota$.
Note that, for an involution $\phi$, then $\theta\colon T\to T$ defined by $\theta(A)=\phi(A^T)$ is a ring automorphism. So $\theta$ preserves the center of $T$. As the center of $T$ is the diagonal elements $\lambda I$ for $\lambda\in R$, we have $\theta(\lambda I)=\sigma(\lambda)I$ for a field automorphism $\sigma$. As $\phi\phi(\lambda I)=\lambda I$, we have $\sigma^2(\lambda)=\lambda$. Then, $\chi(A)=\phi(\sigma(A^T))$ is an automorphism of $T$, which preserves $\lambda I$ for each $\lambda\in R$, so is an automorphism of $R$-algebras. So, $\chi$ is an inner automorphism, meaning it is of the form $\chi(A)=C^{-1}AC$ for an invertible $C$. Therefore, $\phi(A)=C^{-1}\sigma(A^T)C$.
In fact, $\phi\phi(A)=A$, so $C^{-1}\sigma(C)^TA\sigma(C)^{-T}C=A$, showing that $C^{-1}\sigma(C)^T$ is in the center of $T$, so is equal to $\lambda I$ for some $\lambda\in R$. So, $C^T=\lambda \sigma(C)$. Taking the transpose $C=C^{TT}=\lambda\sigma(C^T)=\lambda^2 C$. So, $\lambda=\pm1$ and either $C^T=\sigma(C)$ or $C^T=-\sigma(C)$.
• Thank you very much! George, may be you know something about maps $\phi: M_n(R)[x]\to M_n(R)[x]$ such that $$\phi(A\cdot B) = \phi(B)\cdot \phi(A)$$ or the maps $\psi: M_n(R)[x]\to M_n(R)[x]$ such that $$\psi(A\cdot B) = \psi(A)\cdot \psi(B)$$. Obviously, every involution and every automorphism of $M_n(R)$ can be extended to such a map $\phi$ and $\psi$ correspondingly. But are there exist another types of these maps? – Mikhail Goltvanitsa Jan 26 '16 at 7:32
• @Michel: Note that $M_n(R)[x]$ is isomorphic to $M_n(R[x])$. Using the fact that $R[x]$ is a UFD it can be shown that involutions of this ring are of the form $A\mapsto C^{-1}\sigma(A^T)C$ where $\sigma$ is an automorphism of $R[x]$ and $C$ is invertible in $M_n(R[x])$. – George Lowther Jan 28 '16 at 2:29
• @ George: thank you. And what about automorphisms of $M_n(R)[x]$... Are all of them are inner? – Mikhail Goltvanitsa Jan 28 '16 at 16:17
• Automorphisms are of the form $A\mapsto C^{-1}\sigma(A^T)C$ where $\sigma$ is an automorphism of $R[x]$ and $C$ is an invertible element of $M_n(R)[x]$. This works because $R[x]$ is a UFD. I worked out a proof of this, then found the result in the paper dx.doi.org/10.1016/0024-3795(80)90221-9 (part a of second theorem stated). – George Lowther Jan 29 '16 at 20:47