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If $M=\mathbf{Z}[G]$ is the group considered as a module over itself, and with $\mathbf{Z}$-basis $\{1,x,\ldots,x^{n-1}\}$, is it possible to construct a module homomorphism $\varphi:M\to\mathbf{Z}$ such that $\varphi(x^k)=k$? If so, how do we show it is additive; for instance, how do we show $\varphi(x^k + x^l)=\varphi(x^k)+\varphi(x^l)$?

Alexander Gruber
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Sam Williams
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    It is not clear what $x^k$ should be (or what $M$ should be), considering you obviously use additive notation. – Pavel Čoupek Oct 25 '14 at 13:40
  • Thanks Pavel, hopefully my edit will clarify things somewhat. – Sam Williams Oct 25 '14 at 13:43
  • $R$ is the ring of integer? or is it something such that $\mathbb{Z}$ is a $R$-module? – Exodd Oct 25 '14 at 13:51
  • We can make it the ring of integers. – Sam Williams Oct 25 '14 at 13:56
  • Does this mean that $G=\langle x\rangle$ is cyclic of order $n$? A $\mathbb Z$-module homomorphism is just a homomorphism of abelian groups and $\mathbb Z[G]$ as abelian group is just a sum of $\mathbb Z$'s (i.e. is free), so I'm still not sure what you are asking – Hagen von Eitzen Oct 25 '14 at 14:03
  • @Hagen von Eitzen Yes, it does. – Sam Williams Oct 25 '14 at 14:05
  • You say ${\bf Z}[G]$ is considered a module over itself and speak of module homomorphisms ${\bf Z}[G]\to{\bf Z}$, but you did not give any ${\bf Z}[G]$-module structure to $\bf Z$. Or are you simply giving ${\bf Z}[G]$ the structure of a $\bf Z$-module? If so, that makes the problem utterly trivial... – anon Oct 25 '14 at 18:06

1 Answers1

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In general, if you define a $\mathbb{Z}$-module homomorphism from a base to anything in $\mathbb{Z}$, it extends to an homomorphism from the whole module.

In your case, the homomorphism extends to $$ \varphi(a_0+a_1x+a_2x^2+\dots+a_{n-1}x^{n-1})=a_1+2a_2+\dots+(n-1)a_{n-1} $$ This construction is forced, and now $\varphi:M\to \mathbb{Z}$ is a module homomorphism.

To show that $\varphi$ is linear, do this:

Every element of $M$ can be written in an unique way as a linear combination of the $x^i$. So define $\varphi$ with the relation above. You have that $$ x^i=0+0x+0x^2+\dots+1x^i+\dots+0x^{n-1} \implies\\ \varphi(x^i)=\varphi(0+0x+0x^2+\dots+1x^i+\dots+0x^{n-1})=i $$ So $\varphi$ respects $\varphi(x^k)=k$ for all $k$. Now take $v$,$w$ elements of $M$. You can write them as combination of the $x^i$, so $$ v=v_0+v_1x+v_2x^2+\dots+v_{n-1}x^{n-1}\quad w=w_0+w_1x+w_2x^2+\dots+w_{n-1}x^{n-1}\\ \varphi(v)=v_1+2v_2+\dots+(n-1)v_{n-1}\\ \varphi(w)=w_1+2w_2+\dots+(n-1)w_{n-1}\\ \varphi(v+w)=\varphi((v_0+w_0)+(v_1+w_1)x+(v_2+w_2)x^2+\dots+(v_{n-1}+w_{n-1})x^{n-1})=\\ =(v_1+w_1)x+2(v_2+w_2)+\dots+(n-1)(v_{n-1}+w_{n-1}) $$ And this means $\varphi(v)+\varphi(w)=\varphi(v+w)$

Let's suppose $M$ is an $R$-module with basis $m_1,m_2,\dots,m_n$.
If $a_1,a_2,\dots,a_n$ are arbitrary elements of $R$, then there exists only one module homomorphism $\varphi:M\to R$ such that $\varphi(m_i)=a_i$ for all $i$, that is $$ \varphi(b_0m_0+b_1m_1+b_2m_2+\dots+b_{n-1}m_{n-1})=b_0a_0+b_1a_1+b_2a_2+\dots+b_{n-1}a_{n-1} $$

The construction here is forced due to the linearity of $\varphi$:
in fact, if $b\in R$, and $m\in M$ you have $\varphi(bm)=b\varphi(m)$; Since $\varphi(x+y)=\varphi(x)+\varphi(y)$, $$ \varphi(b_0m_0+b_1m_1+b_2m_2+\dots+b_{n-1}m_{n-1})=\\ =\varphi(b_0m_0)+\varphi(b_1m_1+b_2m_2+\dots+b_{n-1}m_{n-1})=\\ =\varphi(b_0m_0)+\varphi(b_1m_1)+\varphi(b_2m_2+\dots+b_{n-1}m_{n-1})=\\ \dots\\ =\varphi(b_0m_0)+\varphi(b_1m_1)+\varphi(b_2m_2)+\dots+\varphi(b_{n-1}m_{n-1})=\\ =b_0\varphi(m_0)+b_1\varphi(m_1)+b_2\varphi(m_2)+\dots+b_{n-1}\varphi(m_{n-1})=\\ b_0a_0+b_1a_1+b_2a_2+\dots+b_{n-1}a_{n-1} $$

Exodd
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