Math4302 Modern Algebra (Lecture 27)

For the forward direction, we proved if $ax\equiv b\mod n$ then $ax-b=ny$, $y\in\mathbb{Z}$.

then $b=ax-ny$, $d|(ax-ny)$ implies that $d|b$.

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For the backward direction, assume $d=\operatorname{gcd}(a,n)=1$. Then we need to show, there is exactly $1$ solution between $0$ and $n-1$.

If $ax\equiv b\mod n$, then in $\mathbb{Z}_n$, $[a][x]=[b]$. (where $[a]$ denotes the remainder of $a$ by $n$ and $[b]$ denotes the remainder of $b$ by $n$)

Since $\operatorname{gcd}(a,n)=1$, then $[a]$ is a unit in $\mathbb{Z}_n$, so we can multiply the above equation by the inverse of $[a]$. and get $[x]=[a]^{-1}[b]$.

Now assume $d=\operatorname{gcd}(a,n)$ where $n$ is arbitrary. Then $a=a'd$, then $n=n'd$, with $\operatorname{gcd}(a',n')=1$.

Also $d|b$ so $b=b'd$. So

\begin{aligned} ax\equiv b \mod n&\iff n|(ax-b)\\ &\iff n'd|(a'dx-b'd)\\ &\iff n'|(a'x-b')\\ &\iff a'x\equiv b'\mod n' \end{aligned} $$. Since $\operatorname{gcd}(a',n')=1$, there is a unique solution $x_0\in \mathbb{Z}_{n'}$. $0\leq x_0\leq n'+1$. Other solution in $\mathbb{Z}$ are of the form $x_0+kn'$ for $k\in \mathbb{Z}$. And there will be $d$ solutions in $\mathbb{Z}_n$, </details> <details> <summary>Examples</summary> Solve $12x\equiv 25\mod 7$. $12\equiv 5\mod 7$, $25\equiv 4\mod 7$. So the equation becomes $5x\equiv 4\mod 7$. $[5]^{-1}=3\in \mathbb{Z}_7$, so $[5][x]\equiv [4]$ implies $[x]\equiv [3][4]\equiv [5]\mod 7$. So solution in $\mathbb{Z}$ is $\{5+7k:k\in \mathbb{Z}\}$. --- Solve $6x\equiv 32\mod 20$. $\operatorname{gcd}(6,20)=2$, so $6x\equiv 12\mod 20$ if and only if $3x\equiv 6\mod 10$. $[3]^{-1}=[7]\in \mathbb{Z}_{10}$, so $[3][x]\equiv [6]$ implies $[x]\equiv [7][6]\equiv [2]\mod 10$. So solution in $\mathbb{Z}_{20}$ is $[2]$ and $[12]$ So solution in $\mathbb{Z}$ is $\{2+10k:k\in \mathbb{Z}\}$ </details> ### Ring homomorphisms #### Definition of ring homomorphism Let $R,S$ be two rings, $f:R\to S$ is a ring homomorphism if $\forall a,b\in R$, - $f(a+b)=f(a)+f(b)\implies f(0)=0, f(-a)=-f(a)$ - $f(ab)=f(a)f(b)$ #### Definition of ring isomorphism If $f$ is a ring homomorphism and a bijection, then $f$ is called a ring isomorphism. <details> <summary>Example</summary> Let $f:(\mathbb{Z},+,\times)\to(2\mathbb{Z},+,\times)$ by $f(a)=2a$. Is not a ring homomorphism since $f(ab)\neq f(a)f(b)$ in general. --- Let $f:(\mathbb{Z},+,\times)\to(\mathbb{Z}_n,+,\times)$ by $f(a)=a\mod n$ Is a ring homomorphism. </details> ### Integral domains and their file fo fractions. Let $R$ be an integral domain: (i.e. $R$ is commutative with unity and no zero divisors). #### Definition of field of fractions If $R$ is an integral domain, we can construct a field containing $R$ called the field of fractions (or called field of quotients) of $R$.

S={(a,b)|a,b\in R, b\neq 0}

a relation on $S$ is defined as follows: $(a,b)\sim (c,d)$ if and only if $ad=bc$. <details> <summary>This equivalence relation is well defined</summary> - Reflectivity: $(a,b)\sim (a,b)$ $ab=ab$ - Symmetry: $(a,b)\sim (c,d)\Rightarrow (c,d)\sim (a,b)$ - Transitivity: $(a,b)\sim (c,d)$ and $(c,d)\sim (e,f)\Rightarrow (a,b)\sim (e,f)$ - $ad=bc$, and $cf=ed$, we want to conclude that $af=be$. since $ad=bc$, then $adf=bcf$, since $cf=ed$, then $cfb=edb$, therefore $adf=edb$. - Then $d(af-be)=0$ since $d\neq 0$ then $af=be$. </details> Then $S/\sim$ is a field.