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Proposition 1 $a,b$ are non-zero element of an integral domain $R$. $a$ divides $b$ and $b$ divides $a$ if and only if there exist a unit $u$ such that $a=bu$

$\Rightarrow :$
• $\mathrm{\exists }r,s\in R:sa=b$ and $rb=a$. Then, $a=rsa$ and $a(1-rs)=0$. Since $R$ is an integral domain and $a$ is non-zero, $1-rs=0$. Thus, $r,s$ are units and $a=rb$.
$\Leftarrow :$
• $a=bu⟹b\mid a$. Since $u$ is a unit, $\mathrm{\exists }t\in R:ut=1$. $at=b(ut)=b⟹a\mid b$.

Proposition 2 $a|b⟺⟨b⟩\subseteq ⟨a⟩$

Corollary 2.1 $a,b$ are associates $⟺⟨a⟩=⟨b⟩$

Proposition 3 $p$ is irreducible if and only if $⟨p⟩$ is maximal amongst all principal ideals that contain $⟨p⟩$.

$\Rightarrow :$
• For any principal ideal $⟨x⟩$ that contains $⟨p⟩$ and is not the unit ideal, $p\in ⟨x⟩$. So there exists $r\in R:p=xr$. Since $x$ divides $p$ and $p$ is irreducible, $x$ is either an unit or an associate of $p$. If $x$ is an unit, then $⟨x⟩$ is the unit ideal. Thus, $x$ is an associate of $p$. Based on the previous corollary, $⟨x⟩=⟨p⟩$ and $⟨p⟩$ is maximal amongst all principal ideals that contain $⟨p⟩$.
$\Leftarrow :$
• For any $r$ that divides $p$, $⟨p⟩\subseteq ⟨r⟩$. Since $⟨p⟩$ is maximal amongst all principal ideals, $⟨p⟩=⟨r⟩$ or $⟨r⟩$ is the unit ideal. So, $r$ is either an associate of $p$ or an unit. Thus, $p$ is irreducible.

Proposition 4 In an integral domain $R$, every prime is irreducible.

$p$ is a prime in an integral domain $R$. For any $a,b\in R$ such that $p=ab$. Since $p$ is a prime, WLOG, let’s assume $p|a$. Then there exist $t\in R$ such that $pt=abt=a$. So, $a(bt-1)=0$. Since $R$ is an integral domain and $a\ne 0$, $bt=1$ and $b$ is an unit. Therefore, $a$ is an associate of $p$ and $p$ is irreducible.

Definition 1 An ideal $I$ of $R$ is prime if the quotient $R/I$ is an integral domain. It is maximal if $R/I$ is a field.

Lemma 2 An ideal $I$ is prime if, and only if, for every pair of elements $r,s$ in $R$ such that $rs$ is in $I$, either $r$ is in $I$ or s is in $I$.

$\Rightarrow :$
• An ideal $I$ of $R$ is prime $⟹$ $R/I$ is an integral domain $⟹\mathrm{\forall }a,b\in R/I,ab=0+I⟹a=0+I\vee b=0+I$

  Since $\mathrm{\forall }a,b\in R/I,\mathrm{\exists }x,y\in R:a=x+I,b=y+I$. So, $ab=xy+I=0+I⟹x+I=0+I\vee y+I=0+I$. Thus, $\mathrm{\forall }xy-0\in I⟹x-0=x\in I\vee y\in I$.

$\Leftarrow :$
• $xy\in I⟹x\in I\vee y\in I$

  $\mathrm{\forall }x+I,y+I\in R/I,(x+I)(y+I)=xy+I=0+I⟹xy\in I⟹x+I=0+I\vee y+I=0+I$. Thus, $R/I$ is an integral domain and $I$ is prime.

Lemma 3 The only ideals of a field are the zero ideal and the unit ideal.

Assume $I$ is an ideal of the field $F$ that has a non-zero element $r$. By definition, $\mathrm{\exists }{r}^{-1}\in F:r{r}^{-1}=1\in I$. Since $1\in I$, unit ideal must be a subset of $I$. But every ideal is a subset of the unit ideal, so $I$ must be equivalent to the unit ideal.

Clearly, zero ideal is also an ideal of $F$.

Lemma 4 An ideal $I$ is maximal if, and only if, the only ideals of $R$ containing $I$ are $I$ and the unit ideal.

$\Rightarrow :$
• $I$ is maximal $⟹$ $R/I$ is a field

  Assume there exist an ideal $J$ of $R$ such that $I\subseteq J\subseteq R$. Note $J/I\subseteq R/I$. Since $J$ is an ideal of R, $J/I$ is an ideal of $R/I$. $R/I$ is a field. Based on the previous lemma, $J/I$ is zero ideal or unit ideal which means $J=I$ or $J=R.$

$\Leftarrow :$
• For any $r\in R\setminus I$, $I$ is an ideal of $I+⟨r⟩$. $r\notin I$, $I+⟨r⟩$ must be the unit ideal. Thus, $1\in (I+⟨r⟩)$. There exists $t\in R,i\in I:i+rt=1$. Thus, $rt\equiv 1modI$ and $r$ has an inverse in $R/I$. Note, $\mathrm{\forall }i\in I,i\equiv 0modI$. Therefore, $R/I$ is a field.

Proposition If $I+J=⟨1⟩$, then $I\cap J=IJ$

Third Isomorphism Theorem for Groups Let $G$ be a group and let $H$ and $K$ be normal subgroups of $G$, with $H\le K$. Then

1. $K/H⊴G/H$
2. $(G/H)/(K/H)\cong G/K$

1. $\mathrm{\forall }x\in K/H;y\in G/H,\mathrm{\exists }k\in K;g\in G:x=kH;y=gH$ and $yx{y}^{-1}=(gk{g}^{-1})H$. Since $K$ is a normal subgroup of $G$, $gk{g}^{-1}\in K$. Thus, $yx{y}^{-1}\in K/H$ and $K/H⊴G/H$
2. Consider a mapping $\varphi :G/H\to G/K$ by $\varphi (gH)=gK$

   The map is well-defined since if $aH=bH$, then $a^{-1}b\in H\subseteq K$ and

   $$\varphi (aH)=aK\circ (a^{-1}b)K=bK=\varphi (bH)$$

   The map is homomorphic since $\mathrm{\forall }aH,bH\in G/H$,

   $$\varphi (aH)\varphi (bH)=(aK)(bK)=(ab)K=\varphi ((ab)H)$$

   It is easy to tell $\varphi$ is surjective, so $\mathrm{im}(\varphi )=G/K$.

   $$\varphi (gH)=gK=K⟺g\in K⟺gH\in K/H⟹\ker (\varphi )=K/H$$

   Based on First Isomorphism Theorem, $(G/H)/(K/H)\cong G/K$

Second Isomorphism Theorem for Groups Let $G$ be a group, $H\le G$ and $N⊴G$. Then

1. $HN\le G$
2. $H\cap N⊴H$
3. $HN/N\cong H/(H\cap N)$

1. $N⊴G⟹HN=NH⟹HN\le G$
2. $\mathrm{\forall }x\in H\cap N,x\in H\wedge x\in N.$ Since $N⊴G$, $\mathrm{\forall }x\in H\cap N;h\in H,hx{h}^{-1}\in N$. Note, $x,h,h^{-1}\in H$. So, $hx{h}^{-1}\in H$. Thus, $hx{h}^{-1}\in H\cap N$ and $H\cap N⊴H$.
3. Consider now the canonical homomorphism $\phi :HN⟶HN/N$ by $\phi (h_{i}n)=h_{i}N$

   The restriction of $\phi$ to $H$ is a homomorphism of $H$ in $HN/N$ with kernel: $H\cap \ker (\phi )=H\cap N$. It is not difficult to see that $\mathrm{im}({\phi }_H)=HN/N$. Based on First Isomorphism Theorem, $H/(H\cap N)\cong HN/N$

First Isomorphism Theorem for Groups If $\varphi :G\to H$ is a homomorphism then

Define a mapping $f:G/\ker (\varphi )\to \mathrm{im}(\varphi )$ by $f(a\ker (\varphi ))=\varphi (a)$

The map is well-defined since if $a\ker (\varphi )=b\ker (\varphi )$, then $a^{-1}b\in \ker (\varphi )$ and

Let $h$ be an arbitrary element of $\mathrm{im}(\varphi )$, then $\mathrm{\exists }g\in G:\varphi (g)=h$. Since $f(g\ker (\varphi ))=\varphi (g)=h$, $f:G/\ker (\varphi )\to \mathrm{im}(\varphi )$ is surjective. It is also injective because $f(a\ker (\varphi ))=f(b\ker (\varphi ))⟹\varphi (a)=\varphi (b)⟹\varphi (a)\varphi (b^{-1})=e_H⟹a{b}^{-1}\in \ker (\varphi )⟹a\ker (\varphi )=b\ker (\varphi )$

$f:G/\ker (\varphi )\to \mathrm{im}(\varphi )$ is a homomorphism since

Proposition Suppose that $\alpha$ has an eventually periodic continued fraction expansion. Then $\alpha$ is a quadratic irrational.

We first show this when $\alpha$ has a periodic continued fraction expansion. We then have a $d$ such that

Since ${\alpha }_0,{\alpha }_1,\cdots ,{\alpha }_{d-1}$ are all integers

Since $\alpha$ is irrational, $z\ne 0$. Thus, $\alpha$ is a quadratic irrational.

If $\alpha =[{\alpha }_0,{\alpha }_1,\dots ,{\alpha }_m,{\alpha }_{m+1},\dots ,{\alpha }_{m+d-1},{\alpha }_{m+d},\dots ]$, then

Clearly $\beta$ has a periodic continued fraction expansion. So it is quadratic irrational.

Note,

Since $\alpha$ is irrational, $a^{\prime }\ne 0$. Thus, $\alpha$ is a quadratic irrational.

Liouville’s theorem on diophantine approximation Let $\alpha$ be an irrational number that is algebraic of degree $d$. Then for any real number $e>d$, there are at most finitely many rational numbers $\frac{p}{q}$ such that $|\frac{p}{q}-\alpha |<\frac{1}{q^e}$.

If there is no such rational number $\frac{p}{q}$ then the number of solution is clearly finite.

Now assume there exists at least one rational number $\frac{p}{q}$ such that $|\frac{p}{q}-\alpha |<\frac{1}{q^e}$.

Let $P(x)$ be a polynomial of degree $d$, with integers coefficients, such that $P(\alpha )=0$. Choose $ϵ$ such that $P(x)$ has no roots other that $\alpha$ on the interval $[\alpha -ϵ,\alpha +ϵ]$

Write $P(x)=(x-\alpha )\cdot Q(x)$ where $Q(x)$ is a monic polynomial with real coefficients of degree $d-1$. Since $Q(x)$ is continuous, there exists $K>0$ such that $|Q(x)|\le K$ on the interval $[\alpha -ϵ,\alpha +ϵ]$

For all rational number $\frac{p}{q}$ such that $|\frac{p}{q}-\alpha |<\frac{1}{q^e}$, $|\frac{p}{q}-\alpha |>ϵ$ or $|\frac{p}{q}-\alpha |\le ϵ$

For all rational number $\frac{p}{q}$ such that $ϵ<|\frac{p}{q}-\alpha |<\frac{1}{q^e}$. $q^e<1/ϵ$ and $p\in [q(\alpha -\frac{1}{q^e}),q(\alpha +\frac{1}{q^e})]$, it is not difficult to tell the number of such rational numbers will be finite.

For all rational number $\frac{p}{q}$ such that $|\frac{p}{q}-\alpha |<\frac{1}{q^e}\wedge |\frac{p}{q}-\alpha |\le ϵ$. Note

Since $P$ has degree $d$ and integer coefficients,

where both the nominator and the denominator are integers. Since $\alpha$ is irrational, $P(\frac{p}{q})\ne 0$. Then $\left|P\left(\frac{p}{q}\right)\right|\ge \frac{1}{q^d}$. Note

Similarly, the number of rational numbers $\frac{p}{q}$ such that $|\frac{p}{q}-\alpha |<\frac{1}{q^e}\wedge |\frac{p}{q}-\alpha |\le ϵ$ will be finite.

Therefore, there are at most finitely many rational numbers $\frac{p}{q}$ such that $|\frac{p}{q}-\alpha |<\frac{1}{q^e}$.

Theorem 1 Let $\alpha$ be an irrational number, and $Q>1$ an integer. Then there exist $p,q$ integers, with $1\le q<Q$, such that $|p-q\alpha |<\frac{1}{Q}$.

For $1\le k\le Q-1$: Set ${\alpha }_k=\lfloor k\alpha \rfloor$ such that $0<k\alpha -{\alpha }_k<1.$

Partition the interval $[0,1]$ into $[0,1/Q]\cup [1/Q,2/Q]\cup \cdots \cup [(Q-1)/Q,1]$. So there are in total $Q$ subintervals. Let $S=\{0,\alpha -{\alpha }_1,2\alpha -{\alpha }_2,\dots (Q-1)\alpha -{\alpha }_{Q-1},1\}$, $|S|=Q+1$. There there must be a subinterval from earlier which contains at least 2 elements of $S$ which are not 0 and 1. Set ${\alpha }_0=0,{\alpha }_Q=1$. Since $s_x$ and $s_y$ are in the same subinterval, $|s_x-s_y|=|m\alpha -({\alpha }_y-{\alpha }_x)|<1/Q$ where $1\le m<Q$.

Corollary 2 For any irrational $\alpha$ there are infinitely many $\frac{p}{q}$ such that

For any given $Q\in Z$ there exists integers $p,q$ that satisfies the above ineqaulity. Find $Q^{\prime }\in \mathbb{Z}:{\frac{1}{Q}}^{\prime }<|p-q\alpha |$. Then there exists integers $p^{\prime },q^{\prime }$ such that $|p^{\prime }-q^{\prime }\alpha |<{\frac{1}{Q}}^{\prime }<|p-q\alpha |$. Clearly $p\ne p^{\prime }\wedge q\ne q^{\prime }$. We can repeat this process to find infinitely many such $p,q$.

Theorem 3 For any squarefree d there is a nontrivial solution to $x^2-d{y}^2=1$

Since we are looking for nontrivial solution $y\ne 0$. $x^2-d{y}^2=1⟺\frac{x}{y}=\sqrt{d}$. Based on the earlier corollary, there are infinitely many $\frac{x}{y}$ such that

Note

Therefore,

Since $N(x+y\sqrt{d})\in \mathbb{Z}$ and is bounded in a finite interval, there exists an integer $M:|M|<3\sqrt{d}\wedge$there exists infinitely many pairs of $(x,y)$ such that $N(x+y\sqrt{d})=M$.

Since there are finitely many congruence classes mod $M$, there are infinitely many $x$ in at least one congruence classes $[x_0{]}_M$. And there are infinitely many $y$ that is paired up with such $x$ in at least one congruence classes $[y_0{]}_M$.

For any $(x_i,y_i)$ and $(x_j,y_j)$ that satisfy the above conditions,

Note,

Therefore, $\mathrm{\exists }a,b\in \mathbb{Z}$ s.t.

It is easy to tell that $N(a+b\sqrt{d})=M/M=1$. The solution is nontrivial when $(x_i,y_i)\ne (x_j,y_j)$.