Problem: Let $\phi: R\to S$ be a surjective ring homomorphism (the rings are not necessarily commutative) such that for every $a\in \ker\phi$ there is a $n\in\mathbb{N},n\ge 1$ s.t. $a^n=0$.
i) The set $N:=1+\ker\phi$ is a normal subgroup of $R^*$, the group of units of $R$.
ii) $\phi$ induces an isomorphism $R^*/N\cong S^*$.
I am totally lost here.
i) I need to show that $gNg^{-1}=N,\forall g\in R^*$. We have $gNg^{-1}=1+g\cdot \ker\phi\cdot g^{-1}=N$, since $g\cdot \ker\phi\cdot g^{-1}\in \ker\phi$ because $\ker\phi$ is normal in $R$. But this means that $N$ is a normal subgroup of $R^*$. I'm confused that I didn't need the provided condition for this.
ii) Where do I go from here, what does "induced" mean in this context?
Best Answer
The homomorphism theorem says that from a surjective homomorphism $\phi\colon R\to S$ we can define an isomorphism $$ \tilde{\phi}\colon R/\ker\phi\to S $$ by mapping each coset $r+\ker\phi$ to $\phi(r)$.
If $x,y\in \ker\phi$, you have to show that $$ (1+x)(1+y)\in 1+\ker\phi $$ and that every element in $1+\ker\phi$ has an inverse which still belongs to $1+\ker\phi$. This ensures $1+\ker\phi$ is a subgroup of $R^*$.
For normality, you have to show that, for $x\in\ker\phi$ and $r\in R^*$, $r(1+x)r^{-1}\in 1+\ker\phi$.
Next, you have to show that $\phi^*\colon R^*\to S^*$ (the restriction of $\phi$) induces a morphism $R^*/N\to S^*$, that is, $N\subseteq \ker\phi^*$.
Just for completeness, here's a sketch.
If $x,y\in\ker\phi$, then $(1+x)(1+y)=1+(x+y+xy)\in 1+\ker\phi=N$, because $x+y+xy\in\ker\phi$.
If $x\in\ker\phi$, set $y=-x$; then we know that $y^n=0$, for some $n>0$. Then $$(1-y)(1+y+y^2+\dots+y^{n-1})=1-y^n=1$$ and so $1-y=1+x$ is invertible (the element is obviously also a left inverse).
If $x\in\ker\phi$ and $r\in R^*$, then $r(1+x)r^{-1}=1+rxr^{-1}\in N$.
Now $\phi$ induces a group morphism $\phi^*\colon R^*\to S^*$, because invertibles in $R$ are mapped to invertible elements of $S$ (because $\phi$ is surjective). It is obvious that $N\subset\ker\phi^*$, because $\phi^*(1+x)=\phi(1+x)=1+\phi(x)=1$.
Conversely, if $r\in\ker\phi^*$, then $\phi(r)=1$, so $r-1\in\ker\phi$ and $r=1+(r-1)\in N$.