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Let $R=(R,+,\cdot)$ be a ring with $1_R$. We will write $U(R)=(U(R),\cdot)$ the multiplicative group of units of this ring and $R^+=(R,+)$ the additive group of this ring. The group $U(R)$ acts in the group $R^+$, if we define a group action $$*:U(R)\times R^+\longrightarrow R^+,\ (g,a)\mapsto g*a:=g\cdot a.$$ Then, we know that the function $$\phi: U(R)\longrightarrow \mathrm{Aut}(R^+),\ g\mapsto \phi(g):=\phi_g $$ is a group homomorphism. We can show that $\ker {\phi}=\{1_G\}$, so $\phi$ is injective and we have monomorphism.

My question: Is $\phi $ surjective?

Lets take an element $\psi \in \mathrm{Aut}(R^+)$. We want to find an element $g\in U(R)$ such that $\phi(g)=\psi$. Then $\phi _g=\psi \iff \phi_g(r)=\psi(r),\ \forall r\in R^+ \iff g\cdot r=\psi (r), \forall r\in R^+$.

But how can we solve the last equation, to find $g$? Any ideas please?

Thank you.

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    $\begingroup$ Have a look at the ring $\mathbb{Z}[x,y]$. What are it’s units and how do the corresponding automorphisms look like? Now look at the automorphism interchanging $x$ and $y$. $\endgroup$ – Maik Pickl Apr 24 '18 at 20:21
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    $\begingroup$ Let $Z$ be a ring with "few" units, which is a subring of a ring/field $C$ with "many" units. For instance $Z=\mathbb Z$ and $C=\mathbb C$. Consider inside the ring $C[X]$ the subring $R$ of polynomials with free coefficient in $Z$. (It is a $Z$-algebra.) Then $U(R)$ "is" $Z$, but each map $X\to cX$, $c$ unit in $R$ induces an automorphism of the abelian group obtained from $R$ by applying the forgetful functor. $\endgroup$ – dan_fulea Apr 24 '18 at 21:20
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    $\begingroup$ @Chris It was meant to show you that $\phi$ can’t be surjective. $\endgroup$ – Maik Pickl Apr 26 '18 at 20:06
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    $\begingroup$ @Chris Because there is an automorphism of $\mathbb{Z}[x,y]$ that isn’t in the image of $\phi$. Namely the automorphism that just exchanges $x$ and $y$. It leaves $\mathbb{Z}$ fixed and is therefore not $-Id$ but it’s also not $Id$ since it interchanges $x$ and $y$. $\endgroup$ – Maik Pickl Apr 26 '18 at 20:19
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    $\begingroup$ @Chris Exactly! $\endgroup$ – Maik Pickl Apr 26 '18 at 20:38

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