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Algebra-II: Companion notes

Abhay Soman

Section 4.1 Arithmetic in \(\Z\)

Recall that every ideal in \(\Z\) is generated by a single integer, i.e., every ideal in \(\Z\) is principal (see Example 1.6.6). For \(a\in\Z\) it is easy to see that \(a\Z=(-a)\Z\text{.}\) We use this fact without explicitly mentioning it.
  1. Suppose that \(n\mid m\text{,}\) i.e., there exists \(k\in\Z\) such that \(m=nk\text{.}\) Hence, \(m\in n\Z\) and therefore any multiple of \(m\) is also in the ideal \(n\Z\text{.}\) Thus \(m\Z\subseteq n\Z\text{.}\) Conversely, if \(m\Z\subseteq n\Z\) then there exists \(\ell\in\Z\) such that \(m=n\ell\text{,}\) i.e., \(n\) divides \(m\text{.}\)
  2. Since every ideal of \(\Z\) is principal and the intersection of ideals is an ideal, there exists \(e\in\N\) such that \(m\Z\cap n\Z=e\Z\text{.}\) In particular, \(e\Z\subseteq m\Z\) and \(e\Z\subseteq n\Z\text{.}\) By Lemma 4.1.1(1), we have \(m\mid e\) and \(n\mid e\text{.}\) Therefore, \(\ell\mid e\) and hence \(e\Z=m\Z\cap n\Z\subseteq \ell\Z\text{.}\) Conversely, since \(\ell=lcm(m,n)\text{,}\) \(m\mid \ell\) and \(n\mid \ell\) we get that \(\ell\Z\subseteq m\Z\cap n\Z\text{.}\) We thus proved the result.
  3. This is left to the reader. Recall that \(m\Z+n\Z\) is the smallest ideal containing both \(m\Z\) and \(n\Z\text{.}\)
We next consider quotient rings.
  1. Suppose that \(n\) is not a prime. Thus there exists natural numbers \(1\lt k\lt n\) and \(1\lt\ell\lt n\) such that \(n=k\ell\text{.}\) Hence \([n]=[0]=[k][\ell]\) where, \([k]\neq[0]\) and \([\ell]\neq[0]\text{.}\)
  2. Suppose that \(n\) is a prime number. For any \(1\leq i\lt n\text{,}\) \(\gcd(i,n)=1\) and hence \(\Z=i\Z+n\Z\) (see Lemma 4.1.1(3)), i.e., there exists \(x,y\in\Z\) with \(1=ix+ny\text{.}\) Therefore, \([1]=[i][x]\in\Z/n\Z\text{,}\) i.e., \([i]\) is invertible. Converse follows from the part (1) of this proposition.
  3. Suppose that \([i]\in\Z/n\Z\) is a unit, i.e., there exists a unique \([j]\in\Z/n\Z\) such that \([i][j]=[1]\text{,}\) i.e., \(ij=1+nk\) for some \(k\in\Z\text{.}\) If \(d\) is a (positive) divisor of both \(i\) and \(n\) then \(d\) must divide \(1=ij-nk\text{.}\) Hence \(d=1\text{,}\) i.e., \(\gcd(i,n)=1\text{.}\) Converse is left to the reader.

Convention 4.1.3.

Let \(p\) be a prime number. The field \(\Z/p\Z\) is denoted by \(\mathbb{F}_p\text{.}\)
Let \(\varphi(n)\) be the Euler’s phi function, i.e., \(\varphi(n)\) is the number of positive integers less equal \(n\) and coprime to \(n\text{.}\) Use a result from Group theory to show that if \(i\) is a positive integer such that \(\gcd(i,n)=1\) then
\begin{equation*} i^{\varphi(n)}\equiv 1\mod n. \end{equation*}
For a nonzero element \(r\in\Q\) (resp., \(\alpha\in F\)) its inverse is denoted by \(1/r\) (resp., \(1/\alpha\in F\)). For any \(0\neq m\in\Z\text{,}\) \(\varphi(m)\neq 0\in F\) because \(\varphi\) is a monomorphism of rings. We define \(\widetilde{\varphi}\) using \(\varphi\) as follows.
\begin{equation*} \widetilde{\varphi}\left(m/n\right)=\varphi(m)/\varphi(n). \end{equation*}
It is left to the reader to check that \(\widetilde{\varphi}\) is a well-defined ring monomorphism, and that \(\widetilde{\varphi}\circ i_{\Z}=\varphi\text{.}\)
We show the uniqueness. Suppose that \(\psi\colon\Q\to F\) be a ring monomorphism such that \(\psi\circ i_{\Z}=\varphi\text{.}\) Since \(\psi\) is a ring homomorphism we must have
\begin{equation*} \psi(1)=1\quad\text{and}\quad\psi(m)=\psi(\underbrace{1+\cdots+1}_{m\text{-times}})=\underbrace{\psi(1)+\psi(1)+\cdots\psi(1)}_{m\text{-times}}=m. \end{equation*}
Thus we get \(\varphi(m)=\psi(m)=m\text{,}\) for any \(m\in\Z\text{.}\) By ring homomorphism properties we have \(\psi(m/n)=\psi(m)/\psi(n)=m/n=\varphi(m)/\varphi(n)\text{.}\) Hence uniqueness is proved.
  1. [Well-definedness].
    Suppose that \(x+n\Z=y+n\Z\text{,}\) i.e., \(x-y\in n\Z\text{.}\) Hence \(x-y=nk\) for some \(k\in\Z\text{.}\) For each \(i\) we therefore have \(p_i^{a_i}\mid x-y\) (see Fundamental theorem of arithmetic 1 ). Hence, for each \(i\text{,}\) \(x+p_i^{a_i}=y+p_i^{a_i}\) and thus \(f\) is well-defined.
  2. To check \(f\) is a ring homomorphism is left to the reader.
  3. [Injectivity].
    Suppose that \(f(x+n\Z)=0\text{,}\) i.e., \(x\in p_i^{a_i}\Z\) for each \(i\text{.}\) In other words, \(x\in\bigcap_{i=1}^{r}p_i^{a_i}\Z\text{.}\) Since \(lcm(p_1^{a_1},\ldots,p_r^{a_r})=n\text{,}\) by Lemma 4.1.1, we get that \(x\in n\Z\text{.}\) Hence \(f\) is injective.
  4. [Surjectivity].
    We only show that \((1+p_1^{a_1},0+p_2^{a_2},\ldots,0+p_r^{a_r})\) is in the image of \(f\text{.}\) By Lemma 4.1.1, we have \(p_1^{a_1}+p_i^{a_i}=\Z\) for any \(1\lt i\leq r\text{.}\) Therefore, for every \(1\lt i\leq r\text{,}\) there are \(x_i\in p_1^{a_1}\Z\) and \(y_i\in p_i^{a_i}\Z\) such that
    \begin{equation*} x_i+y_i=1. \end{equation*}
    Put \(y=y_2y_3\cdots y_r\text{.}\) Then
    \begin{equation*} y+p_1^{a_1}\Z=\prod_{i=2}^r (1-x_i)+p_1^{a_1}\Z=1+p_1^{a_1}\Z \end{equation*}
    and
    \begin{equation*} y+p_i^{a_i}=0+p_i^{a_i}\Z\quad\text{for all}\quad i\in\{2,3,\ldots,r\}. \end{equation*}
    Hence \(f(y)=(1+p_1^{a_1},0+p_2^{a_2},\ldots,0+p_r^{a_r})\text{.}\)

Remark 4.1.7.

Since the isomorphism given in Theorem 4.1.6 is a ring homomorphism the group of units on both sides are also isomorphic. Thus
\begin{equation*} (\Z/n\Z)^\times\simeq(\Z/p_1^{a_1}\Z)^\times\times(\Z/p_2^{a_2}\Z)^\times\times\cdots\times(\Z/p_r^{a_r}\Z)^\times. \end{equation*}
en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
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