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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?

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    $\begingroup$ With regard to the meaning of "induced": An element of a quotient group is a coset; so given some element of $R^*/N$ we have a coset, but we can also consider a coset representative. This representative will be in $R^* \subset R$, so that we can apply $\phi$ to it and end up in $S^*$; after all, $\phi: R \rightarrow S$, so we must ensure this map lands in the group of units in $S$. A key component, though, is that the map is "well-defined," i.e., that if we had chosen a different representative in $R^*$ we still would have gotten the same element of $S^*$. (Note $\phi$'s role in our new map.) $\endgroup$ – Benjamin Dickman Nov 23 '13 at 14:09
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    $\begingroup$ In particular, $\phi: R \rightarrow S$ can be used in defining a new map $\phi^*: R^*/N \rightarrow S$. This idea of somehow naturally using map 1 to define map 2 is generally what is meant by saying "map 1 induces map 2." $\endgroup$ – Benjamin Dickman Nov 23 '13 at 14:13
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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.

  1. If $x,y\in\ker\phi$, then $(1+x)(1+y)=1+(x+y+xy)\in 1+\ker\phi=N$, because $x+y+xy\in\ker\phi$.

  2. 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).

  3. 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 $r=1+(r-1)\in N$.

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  • $\begingroup$ I already showed that $N$ is a subgroup but left it out since I could do that part. I did show the normality of N above. The homomorphism theorem is clear to me, but don't we have to make sure that $N=ker\phi^*$ to ensure that we don't "cut out too little" of $R^*$ and thus won't get an isomorphism? $\endgroup$ – blst Nov 23 '13 at 14:10
  • $\begingroup$ @blst No; if $f\colon G\to G'$ is a group homomorphism, it's sufficient that $N$ is a normal subgroup of $G$ with $N\subseteq\ker f$ to ensure that $f$ induces a homomorphism $G/N\to G'$ with $xN\mapsto f(x)$. $\endgroup$ – egreg Nov 23 '13 at 14:50
  • $\begingroup$ I don't understand why. By the homomorphism theorem we get an isomorphism if we take $G/\ker f$. How can we also get an isomorphism if we take a subgroup of $\ker f$? $\endgroup$ – blst Nov 23 '13 at 14:54
  • $\begingroup$ @blst I didn't say isomorphism. $\endgroup$ – egreg Nov 23 '13 at 14:54
  • $\begingroup$ Okay I thought that was implicit because that's what the problem states. How does the step from $\phi^*$ morphism to isomorphism work? $\endgroup$ – blst Nov 23 '13 at 15:17

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