Proof: Firstly, if $u$ is a unit of $\mathbb{Z}/m\mathbb{Z}$, then $(u,m)=1$, so that $(u,n)=1$, so that $u$ is a unit of $\mathbb{Z}/n\mathbb{Z}$, proving the mapping is as claimed. To show surjectivity, let $p_i$ be a prime not dividing $n$ but dividing $m$ for all applicable $i$. For a typical unit $x∈(\mathbb{Z}/n\mathbb{Z})^{\times}$, solve the simultaneous system of equations$$y ≡ x \mod{n}$$$$y ≡ 1 \mod{\prod p_i}$$which is always possible given that $n$ and $\prod p_i$ are coprime. For any prime $q$ dividing $m$, we have $q$ divides either $n$ or $\prod p_i$. In the former case, since $y$ is coprime to $n$, we have $y$ coprime to $q$. In the latter case, since $y$ is coprime to $\prod p_i$, we have $y$ coprime to $q$. Since $y$ is coprime to all the prime divisors of $m$, it is coprime to $m$ itself, and is thus a unit of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ which naturally projects to $x$ in $(\mathbb{Z}/n\mathbb{Z})^{\times}$.$~\square$
Saturday, May 4, 2013
Collapsing Units Between Modular Rings (7.6.7)
Dummit and Foote Abstract Algebra, section 7.6, exercise 7:
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Let $n \mid m$, and prove that the natural projection map $(\mathbb{Z}/m\mathbb{Z})^{\times} → (\mathbb{Z}/n\mathbb{Z})^{\times}$ is surjective.
Proof: Firstly, if $u$ is a unit of $\mathbb{Z}/m\mathbb{Z}$, then $(u,m)=1$, so that $(u,n)=1$, so that $u$ is a unit of $\mathbb{Z}/n\mathbb{Z}$, proving the mapping is as claimed. To show surjectivity, let $p_i$ be a prime not dividing $n$ but dividing $m$ for all applicable $i$. For a typical unit $x∈(\mathbb{Z}/n\mathbb{Z})^{\times}$, solve the simultaneous system of equations$$y ≡ x \mod{n}$$$$y ≡ 1 \mod{\prod p_i}$$which is always possible given that $n$ and $\prod p_i$ are coprime. For any prime $q$ dividing $m$, we have $q$ divides either $n$ or $\prod p_i$. In the former case, since $y$ is coprime to $n$, we have $y$ coprime to $q$. In the latter case, since $y$ is coprime to $\prod p_i$, we have $y$ coprime to $q$. Since $y$ is coprime to all the prime divisors of $m$, it is coprime to $m$ itself, and is thus a unit of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ which naturally projects to $x$ in $(\mathbb{Z}/n\mathbb{Z})^{\times}$.$~\square$
Proof: Firstly, if $u$ is a unit of $\mathbb{Z}/m\mathbb{Z}$, then $(u,m)=1$, so that $(u,n)=1$, so that $u$ is a unit of $\mathbb{Z}/n\mathbb{Z}$, proving the mapping is as claimed. To show surjectivity, let $p_i$ be a prime not dividing $n$ but dividing $m$ for all applicable $i$. For a typical unit $x∈(\mathbb{Z}/n\mathbb{Z})^{\times}$, solve the simultaneous system of equations$$y ≡ x \mod{n}$$$$y ≡ 1 \mod{\prod p_i}$$which is always possible given that $n$ and $\prod p_i$ are coprime. For any prime $q$ dividing $m$, we have $q$ divides either $n$ or $\prod p_i$. In the former case, since $y$ is coprime to $n$, we have $y$ coprime to $q$. In the latter case, since $y$ is coprime to $\prod p_i$, we have $y$ coprime to $q$. Since $y$ is coprime to all the prime divisors of $m$, it is coprime to $m$ itself, and is thus a unit of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ which naturally projects to $x$ in $(\mathbb{Z}/n\mathbb{Z})^{\times}$.$~\square$
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