$\DeclareMathOperator\Cl{Cl}$
- Find $\Cl(K)$, where $K=๐(\sqrt{-6})$.
Since $dโก2\pmod4$, $๐ช_K=๐[\sqrt{-6}]$ and $โ_K=-6ร4$. Therefore $M_K=\frac4ฯ\sqrt{6}โ3.1$. Thus generators of $\Cl(K)$ may be found amongst the ideal prime factors of (2) and (3).
The minimal polynomial $m(X)$ for $\sqrt{-6}$ is $X^2+6$.
Over $๐ _2$ this factors as $X^2$, so by Dedekind $(2) = ๐ญ_2^2$ where $๐ญ_2=(2,\sqrt{-6})$ has norm 2 so not principal.
Over $๐ _3$ this factors as $X^2$, so by Dedekind $(3) =๐ญ_3^2$ where $๐ญ_3=(3,\sqrt{-6})$ has norm 3 so not principal.
$๐ญ_2๐ญ_3=(6,\sqrt{-6})=(\sqrt{-6})$ is principal, so $[๐ญ_2]=[๐ญ_3]$. So $\Cl(K)โ C_2$ is generated by $[๐ญ_2]$. - Let $K$ be a number field, other than ๐. Show that ${|ฮ_K|}>1$.
Let $n$ be the degree of $K$ over ๐, and $2r_{2}=n-r_{1}$ be the number of complex embeddings. Then every class in the ideal class group of $K$ contains an ideal of norm not exceeding Minkowski's bound \[M_K=\left(\frac{4}{ฯ}\right)^{r_2}\frac{n!}{n^n}\sqrt{|ฮ_K|}\] Since an ideal has norm at least one, we have $1โคM_K$, so that \[ \sqrt{|ฮ_K|}โฅ \left(\fracฯ4\right)^{r_{2}}{\frac {n^{n}}{n!}}โฅ \left({\frac {ฯ }{4}}\right)^{n/2}{\frac {n^n}{n!}}=\fracฯ2\prod_{j=2}^{n-1}\frac{\left(1+\frac{1}{j}\right)^j}{\frac2{\sqrtฯ}}>1โ\text{since }\left(1+\frac{1}{j}\right)^j>\frac2{\sqrtฯ}\text{for }j>1 \] so ${|ฮ_K|}>1$. - Find $\Cl(K)$, where $K=๐(\sqrt{-34})$.
Since $dโก2\pmod4$, $๐ช_K=๐[\sqrt{-34}]$ and $โ_K=-34ร4$. Therefore $M_K=\frac4ฯ\sqrt{34}โ7.4$. Thus generators of $\Cl(K)$ may be found amongst the ideal prime factors of (2) (3) (5) and (7).
The minimal polynomial $m(X)$ for $\sqrt{-34}$ is $X^2+34$.
Over $๐ _2$ this factors as $X^2$, so by Dedekind $(2) = ๐ญ_2^2$ where $๐ญ_2=(2,\sqrt{-34})$ has norm 2 so not principal.
Over $๐ _3$ this is irreducible, so by Dedekind (3) is inert.
Over $๐ _5$ this factors as $(X+1)(X-1)$, so by Dedekind $(5)=๐ญ_5\overline{๐ญ_5}$ where $๐ญ_5=(5,1+\sqrt{-34})$ has norm 5 so not principal.
Over $๐ _7$ this factors as $(X+1)(X-1)$, so by Dedekind $(7)=๐ญ_7\overline{๐ญ_7}$ where $๐ญ_7=(7,1+\sqrt{-34})$ has norm 7 so not principal.
Since $[๐ญ_5]^{-1}=[\overline{๐ญ_5}],[๐ญ_7]^{-1}=[\overline{๐ญ_7}]$, the class group is generated by $[๐ญ_2],[๐ญ_5],[๐ญ_7]$.
Note that $(1+\sqrt{-34})(1-\sqrt{-34})=35=5ร7=๐ญ_5\overline{๐ญ_5}๐ญ_7\overline{๐ญ_7}$, considering factorisation of principal ideal $(1ยฑ\sqrt{-34})$ we have either $๐ญ_5๐ญ_7$ or $๐ญ_5\overline{๐ญ_7}$ is principal, so either $[๐ญ_7]=[๐ญ_5]^{-1}$ or $[๐ญ_7]=[๐ญ_5]$, so $\Cl(K)$ is generated by $๐ญ_2,๐ญ_5$.
(Or check directly:\begin{align*} ๐ญ_5๐ญ_7&=(5,1+\sqrt{-34})(7,1+\sqrt{-34}) \\&=(35,5+5\sqrt{-34},7+7\sqrt{-34},-33+2\sqrt{-34}) \\&=(35,1+\sqrt{-34},-33+2\sqrt{-34})&1+\sqrt{-34}&=3ร(5+5\sqrt{-34})-2ร(7+7\sqrt{-34}) \\&=(35,1+\sqrt{-34})&-33+2\sqrt{-34}&=2ร(1+\sqrt{-34})-35 \\&=(1+\sqrt{-34})&35&=(1-\sqrt{-34})ร(1+\sqrt{-34}) \end{align*} so $[๐ญ_7]=[๐ญ_5]^{-1}$.)
Note that $(4+\sqrt{-34})(4-\sqrt{-34})=50=5^2ร2=(๐ญ_5\overline{๐ญ_5})^2๐ญ_2^2$, considering factorisation of principal ideal $(4ยฑ\sqrt{-34})$ we have either $๐ญ_2๐ญ_5^2$ or $๐ญ_2\overline{๐ญ_5}^2$ is principal, so either $[๐ญ_2]=[๐ญ_5]^{-2}$ or $[๐ญ_2]=[๐ญ_5]^2$, so $\Cl(K)$ is generated by $๐ญ_5$.
(Or check directly: \begin{align*} ๐ญ_5^2&=(25,5+5\sqrt{-34},-33+2\sqrt{-34}) \\&=(25,71+\sqrt{-34},-33+2\sqrt{-34})&17+2\sqrt{-34}&=(5+5\sqrt{-34})-2ร(-33+2\sqrt{-34}) \\&=(25,-4+\sqrt{-34},-33+2\sqrt{-34})&-4+\sqrt{-34}&=(71+\sqrt{-34})-3ร25 \\&=(25,-4+\sqrt{-34},-25)&-25&=(-33+2\sqrt{-34})-2ร(-4+\sqrt{-34}) \\&=(25,-4+\sqrt{-34}) \end{align*} then \begin{align*} (4+\sqrt{-34})๐ญ_5^2&=(100+25\sqrt{-34},50) \\&=(50,25\sqrt{-34})&25\sqrt{-34}=100+25\sqrt{-34}-2ร50 \\&=(25)๐ญ_2 \end{align*} so $[๐ญ_2]=[๐ญ_5]^2$.)
but we proved $(2)=๐ญ_2^2$, so $[๐ญ_2]^2=e$, so $[๐ญ_5]^4=[๐ญ_2]^2=e$.
So $\Cl(K)โ C_4$ is generated by $[๐ญ_5]$. - Find $\Cl(K)$, where $K=๐(\sqrt{65})$.
Since $dโก1\pmod4$, $๐ช_K=๐[ฮฑ],ฮฑ=\frac{1+\sqrt{65}}2$ and $โ_K=65$. Therefore $M_K=\frac12\sqrt{65}<5$. Thus generators of $\Cl(K)$ may be found amongst the ideal prime factors of (2) and (3).
The minimal polynomial $m(X)$ for $ฮฑ$ is $X^2-X-16$.
Over $๐ _2$ this factors as $X(X+1)$, so by Dedekind $(2) = ๐ญ_2๐ฎ_2$ where $๐ญ_2=(2,ฮฑ),๐ฎ_2=(2,1+ฮฑ)$ have norm 2 so not principal [$N(a+bฮฑ)=2โ(2a+b)^2-65b^2=ยฑ8$ which has no integer solutions $(a,b)$ since ยฑ8 is not quadratic residue mod 5].
Over $๐ _3$ this has no roots, so is irreducible, so by Dedekind (3) is inert.
Since $[๐ญ_2]=[๐ฎ_2]^{-1}$, the class group is generated by $[๐ญ_2]$. \begin{align*} ๐ญ_2^2&=(4,2ฮฑ,16+ฮฑ)&ฮฑ^2&=16+ฮฑ \\&=(4,2ฮฑ,ฮฑ)&ฮฑ&=(16+ฮฑ)-4ร4 \\&=(4,ฮฑ) \\&=(4,ฮฑ-4) \\&=(ฮฑ-4)&4&=(ฮฑ+3)(ฮฑ-4) \end{align*} So $[๐ญ_2]^2=e$. So $\Cl(K)โ C_2$ is generated by $[๐ญ_2]$. - Find all integer solutions to the equation $y^2+74=x^3$.
Let $K=๐(\sqrt{-74})$. It turns out that $h_K=10$. In particular, $๐ช_K$ does not have unique factorisation.
The equation factors in $๐ช_K$ as $(y+\sqrt{-74})(y-\sqrt{-74})=x^3$. We do not have unique factorisation into elements of $๐ช_K$, only into ideals, so we think of this as an equation \[\tag{1} (y+\sqrt{-74})(y-\sqrt{-74})=(x)^3 \] of ideals.
We are going to prove that the two ideals on the left are coprime. Suppose some prime ideal $๐ญ$ divides both terms on the LHS. Then $y+\sqrt{-74}, y-\sqrt{-74} โ ๐ญ$, and so, taking the difference, $2 \sqrt{-74} โ ๐ญ$. Therefore $๐ญ โฃ(2 \sqrt{-74})$. (Here, we are using the fact that containment and division of ideals are the same thing, Theorem 5.2.)
Taking norms, we have \[\tag{2} N(๐ญ) โฃ N(2 \sqrt{-74})=2^3 โ 37 . \] Also, since $๐ญ โฃ(y+\sqrt{-74})$, we have $๐ญ โฃ(x)^3$ and so \[\tag{3} N(๐ญ) โฃ N\left((x)^3\right)=x^6 . \] We claim that neither 2 nor 37 divides $x$.
If $2 โฃ x$ then $8 โฃ x^3$, so $y^2=x^3-74 โก 2\pmod 4$, a contradiction.
If $37 โฃ x$ then $37 โฃ y$, and so $37^2 โฃ y^2-x^3=74$. This is also a contradiction.
From these facts and $(2),(3)$ we have $N(๐ญ)=1$, which is impossible; therefore we are forced to conclude that $๐ญ$ does not exist, so the ideals $(y+\sqrt{-74}),(y-\sqrt{-74})$ are indeed coprime.
Now we return to (1). By unique factorisation of ideals, both $(y+\sqrt{-74})$ and $(y-\sqrt{-74})$ are cubes of ideals. Suppose that $(y+\sqrt{-74})=๐^3$. In particular, $[๐]^3=e$. Since the order of the class group $h_K=10$ is coprime to 3, $[๐]$ is trivial, or in other words $๐$ is a principal ideal. Thus we have an equation \[ (y+\sqrt{-74})=(a+b \sqrt{-74})^3 \] for some $a, b โ ๐$. This means that \[ y+\sqrt{-74}=u(a+b \sqrt{-74})^3 \] in $๐ช_K$, where $u$ is a unit. The only units are $ยฑ 1$; by replacing $a, b$ with $-a,-b$ if necessary, we may assume that $u=1$.
Since $\sqrt{-74}$ is irrational, comparing coefficients we obtain \[ y=a(a^2-3ร74b^2),โb(3 a^2-74b^2)=1. \] The second of these implies that $b= ยฑ 1$ and hence that $3 a^2-74= ยฑ 1$, so $a=ยฑ5,y=ยฑ197ร5=ยฑ985,x=99$. - Show that the ring of integers in $๐(2^{1/3})$ is a principal ideal domain (any results about this field established on previous sheets may be used without proof).
proven on Sheet 1 $๐ช_K=๐[2^{1/3}]$. To prove the class group is trivial by using the Minkowski bound: $$ \left|ฮ_K\right|= \left|\text{disc}(1, 2^{1/3}, 2^{2/3})\right| = 108 = 2^23^3, $$ so that the Minkowski bound for $K$ is given by $$ \frac{3!}{3^3}โ \frac{4}{ฯ}{\textstyle\sqrt{\left|ฮ_K\right|}}=\frac{16}{\sqrt{3}ฯ}< 3 $$ The minimal polynomial $m(X)$ for $2^{1/3}$ is $X^3-2$.
Over $๐ _2$ this factors as $X^3$, so by Dedekind $(2) =(2^{1/3})^3$, so the class group of $K$ is trivial.
NEXTTikz Test