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Let $I$ be an ideal of a ring $R$. It is mentioned in the book An Introduction to Group Rings (by Sehgal and Milies) that the canonical homomorphism $RG \rightarrow (R/I)G$ maps $Z(RG)$, center of group ring $RG$, onto $Z((R/I)G)$.

It can't see why this restriction is actually onto.

Thanks for any help.

Update: According to the below counterexample the statement is not true in general. it can be corrected as

Proposition: Let $I$ be an ideal of a ring $R$. Then for any group $G$, the canonical homomorphism $RG \rightarrow (R/I)G$ maps $Z(RG)$ onto $Z((R/I)G)$ iff it maps $Z(R)$ onto $Z((R/I))$.

which its proof is given by Alex W.

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Below for any finite subset $S$ of a ring we denote $\widetilde S:=\sum_{a\in S}a$. We will use the following well known fact (which is easy to check).

Theorem 1. Let A be any ring, $G$ - finite group, and $C_1,\ldots,C_n$ - all distinct conjugacy classes of $G$. Then $Z(AG)$ is a free $Z(A)$-module with a basis $\widetilde{C}_1,\ldots,\widetilde{C}_n$.

We now prove the following statement.

Theorem 2. Let $\phi:A\to B$ be a rings' homomorphism, $G$ - a finite group, $\phi:AG\to BG$ - standard continuation of $\phi$. Then $\phi(Z(AG))=Z(BG)$ iff $\phi(Z(A))=Z(B)$.

Proof. By the theorem 1, $Z(AG)=\bigoplus_{i=1}^n Z(A)\widetilde{C}_i$, hence $$ \phi(Z(AG))=\phi(\sum_{i=1}^n Z(A)\widetilde{C}_i)= \sum_{i=1}^n\phi(Z(A)\widetilde{C}_i)= \sum_{i=1}^n \phi(Z(A))\widetilde{C}_i=\bigoplus_{i=1}^n\phi(Z(A))\widetilde{C}_i. $$ Besides that, $$ Z(BG)=\bigoplus_{i=1}^n Z(B)\widetilde{C}_i. $$ In such a way, $$ \phi(Z(AG))=Z(BG)\Leftrightarrow \bigoplus_{i=1}^n\phi(Z(A))\widetilde{C}_i= \bigoplus_{i=1}^n Z(B)\widetilde{C}_i\Leftrightarrow\phi(Z(A))=Z(B). $$

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  • $\begingroup$ Did you mean ϕ(Z(AG))=Z(BG) in Theorem 2? If so, I think Theorem 2 is exactly equivalent to what we want to prove (in the case of commutative rings). So we can't use it here. Besides, R is not supposed to be commutative in the question. $\endgroup$ – user97635 May 15 '15 at 13:52
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    $\begingroup$ @user97635 I prove theorem 2 ) $\endgroup$ – Alex W May 15 '15 at 13:57
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    $\begingroup$ I think it must be $ϕ(Z(AG))=Z(BG)$ in the statement of Theorem 2, Isn't it? Besides, according to your last argument, here in the question we must have $ϕ(Z(R))=Z(R/I)$. I wonder whether or not we always could have $ϕ(Z(R))=Z(R/I)$, where ϕ is canonical homomorphism $R \rightarrow R/I$. $\endgroup$ – user97635 May 15 '15 at 14:15
  • $\begingroup$ @user97635 Yes ) I correct. $\endgroup$ – Alex W May 15 '15 at 14:26
  • $\begingroup$ @user97635 I agree, it is very unlikely that always $\phi(Z(A))=Z(B)$ for epimorphisms. $\endgroup$ – Alex W May 15 '15 at 14:36
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Below we will find a rings' epimorphism $\phi:A\to B$ such, that $\phi(Z(A))\neq Z(B)$. By theorem 2, it will give a counterexample to the statement of interest to us.

Let $G=\mathbb{Q}_8=\{\pm 1,\pm i,\pm j,\pm k\}$ be a quaternion group, $Z=Z(G)= \{\pm 1\}$, $\alpha:G\to G/Z$ - natural homomorphism. Further, let $F={\rm GF}(2)$ - field of order $2$, $A=FG$, $B=F(G/Z)$ and $\phi:A\to B$ - a standard continuation of $\alpha$ by linearity. Let's prove, that $\phi(ZA)\neq ZB$. Conjugacy classes of $G$ are $$ \{1\},\{-1\},\{\pm i\},\{\pm j\},\{\pm k\}. $$ Image of $g\in G$ in $A$ and $B$ denote $[g]$. By our description of $Z(A)$, $Z(A)$ is a vector space over $F$ with a basis $$ [1],[-1],[i]+[-i],[j]+[-j],[k]+[-k]. $$ Their images under $\phi$ is $$ \phi([1])=[1],\phi([-1])=[1], $$ and for $g\in\{i,j,k\}$ $$ \phi([g]+[-g])=2[g]=0. $$

Since $\phi$ is a $F$-linear, then $\phi(Z(A))$ is a vector space over $F$, spanned by $[1]$. In such a way, $\phi(Z(A))=\{0,[1]\}$. But $G/Z$ is abelian group, hence $B$ - commutative ring and $Z(B)=B$. Now it is obvious that $\phi(Z(A))\neq Z(B)$.

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