Exercise 9.1.2

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Assume that $\gcd(n,m) =1$. By Lemma A.5.2, we have a ring isomorphism $\alpha : \Z/nm\Z \simeq \Z/n\Z 	imes \Z/m\Z$ that sends $[a]_{nm}$ to $([a_n],[a]_m)$. Prove that $\alpha$ induces a group isomorphism $(Z/nm\Z)^* \simeq (\Z/nZ)^* 	imes (\Z/m\Z)^*$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $A,B$ are commutative rings (with unity). Then $$(A	imes B)^* = A^*	imes B^*.$$

Indeed, let $(a,b)\in A	imes B$.

\begin{align*}
(a,b) \in (A	imes B)^* &\iff \exists (c,d) \in A	imes B, \ (a,b)(c,d) = (1,1)\
 &\iff \exists c \in A, \exists d \in B, \ ac=1,bd=1\
  &\iff a\in A^*, b\in B^* \
  &\iff (a,b)\in A^*	imes B^*.
 \end{align*}
Moreover, if $\varphi : A	o B$ is a ring isomorphism, then for all $a \in A^* \Rightarrow \varphi(a) \in A^*$, since $ab = 1_A \Rightarrow \varphi(a) \varphi(b) = \varphi(1_A) = 1_B$. So we can define
$\psi : A^* 	o B^*$ by restriction with $a\mapsto \psi(a) = \varphi(a)$.

$\psi$ is a group homomorphism: if $u,v \in A^*, \psi(uv) = \varphi(uv) = \varphi(u)\varphi(v) = \psi(u)\psi(v)$, and $\psi$ is bijective:

$\bullet\ \varphi$ is injective, so its restriction $\psi$ if also injective.

$\bullet$ If $b\in B^*$, then there exists $d \in B$ such that $bd=1$. If we write $a=\varphi^{-1}(b), c=\varphi^{-1}(d)$, then $b=\varphi(a), d=\varphi(c), 1 = bd = \varphi(ac)$, so $ac = 1, a \in A^*$, thus $b=\psi(a)$, so $\psi$ is surjective.
$$A\simeq B \Rightarrow A^* \simeq B^*.$$
If we apply these two results to the rings $\Z/n\Z, \Z/m\Z$, we obtain
$$(\Z/nm\Z)^* \simeq (\Z/n\Z 	imes \Z/m\Z)^* = (\Z/nZ)^* 	imes (\Z/m\Z)^*.$$
\end{proof}

Exercise 9.1.2

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Exercise 9.1.2

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