In this note, we will prove the Minkowski's theorem on the boundedness for the norm of an ideal in the ideal class group. It is the fundamental tool to compute the Picard groups, which play a major roles in some kinds of Diophantine equations.
1. Minkowski's theorems. We recall that a lattice $L$ is a discrete subgroup of $\mathbb{R}^n$, it is of the form $\mathbb{Z}\omega_1\oplus...\oplus\mathbb{Z}\omega_d$, for some $d\le n$ and $\omega_i\in \mathbb{R}^n$, such that $\omega_i$ are linearly independent over $\mathbb{R}$. $L$ is called a lattice of rank $d$. We assume that $L$ is a full rank lattice, i.e. $d=n$. The fundamental domain of $L$ is defined as $\{a_1\omega_1+...+a_n\omega_n|a_i\in [0,1)\}$. The volume of $L$, denoted $vol(L)$ is defined as the volume of the fundamental domain. The Minkowski's convex body theory states that
Theorem 1.1. If $S\subset \mathbb{R}^n$ is a convex symmetric set with respect to $0$ satisfies one of the two conditions:
(1) $vol(S)>2^n vol(L)$ .
(2) $vol(S)\ge 2^n vol(L)$, and $S$ is compact.
Then $S$ always contains a lattice point of $L$ other than the origin.
This theorem play an important role in the geometry of numbers, and we will discuss one of its application in this section. First of all, we need to consider our ring of integers as a lattice. We fix some notations: $K$ is an algebraic number field of degree $n$, and $\sigma_1,...,\sigma_n$ are all embedding from $K$ to $\mathbb{C}$. If $\sigma_i(K)\subset\mathbb{R}$, we call $\sigma_i$ is a real embedding. Otherwise, $\sigma_i$ is called a complex embedding. Because the complex embedding always go in pair (they are conjugates of each other), the number of complex embeddings of $K$ is always an even integer, which is denoted $2r_2$, for some $r_2\in\mathbb{N}$. Also, the number of real embedding of $K$ is denoted $r_1$. We have $r_1+2r_2=n$. We can renumber $\sigma_1,...,\sigma_n$ such that $\sigma_1,...,\sigma_{r_1}$ are all real embeddings, and $\sigma_{r_1+i}=\overline{\sigma_{r_1+r_2+i}}$, for $1\le i\le r_2$ are all complex embeddings of $K$.
We are now ready to define a map $\sigma: K\to \mathbb{R}^{n}$ that send $x\in K$ to $(\sigma_1(x),...,\sigma_{r_1}(x), Re(\sigma_{r_1+1}(x)), Im(\sigma_{r_1+1}(x)),...,Re(\sigma_{r_1+r_2}(x)), Im(\sigma_{r_1+r_2}(x)))$. It can be seen by our previous argument that $\sigma$ is a well-defined map, which is called the canonical embedding of $K$. This map is actually an injective ring homomorphism.
Proposition 1.2. Let $M$ be a free $\mathbb{Z}$-module of rank $n$ in $K$, and $(x_1,...,x_n)$ is a $\mathbb{Z}$-base for $M$, then $\sigma(M)$ is a full-rank lattice in $\mathbb{R}^n$. Furthermore, $vol(\sigma(M))=2^{-r_2}|\det (\sigma_i(x_j))_{ij}|$.
Proof. One can see that
$$\sigma(x_j)=(\sigma_1(x_j),...,\sigma_{r_1}(x_j), Re(\sigma_{r_1+1}(x_j)), Im(\sigma_{r_1+1}(x_j)),...,Re(\sigma_{r_1+r_2}(x_j)), Im(\sigma_{r_1+r_2}(x_j)))$$
For all $1\le i\le r_2$, it can be seen that
$$Re(\sigma_{r_1+i}(x_j))=\frac{1}{2}(\sigma_{r_1+i}(x_j)+\overline{\sigma_{r_i+i}(x_j)})=\frac{1}{2}(\sigma_{r_1+i}(x_j)+\sigma_{r_1+r_2+i}(x_j))$$
Similarly,
$$Im(\sigma_{r_1+i}(x_j))=\frac{1}{2i}(\sigma_{r_1+i}(x_j)+\sigma_{r_1+r_2+i}(x_j))$$
By this, and the column transformations, we can see $vol(\sigma(M))=2^{-r_2}|\det(\sigma_i(x_j))_{ij}|$. Because $(x_i)_{i=1}^n$ is the $\mathbb{Q}$-base for $K$, due to our results from Part 1, $\det(\sigma_i(x_j))_{ij}\ne 0$. This yields $\sigma(x_i)_{i=1}^n$ is the basis for $\mathbb{R}^n$, i.e. $\sigma(M)$ is a lattice in $\mathbb{R}^n$. (Q.E.D)
Corollary 1.3. $vol(\sigma(\mathscr{O}_K))=2^{-r_2}|\Delta_K|^{1/2}$. If $I\subset \mathscr{O}_K$ is an ideal, $vol(\sigma(I))=2^{-r_2}N(I)|\Delta_K|^{1/2}$.
Proof. The first part follow easily from the previous proposition. For the second part, one can see that because $\sigma$ is an embedding, the index of $\sigma(I)$ in $\sigma(\mathscr{O}_K)$ is exacly $N(I)$. And hence, the fundamental domain of $\sigma(I)$ is obtained by tanking $N(I)$ copies of the fundamental domain of $\sigma(\mathscr{O}_K)$. (Q.E.D)
We are now ready for the following important theorem of Minkowski.
Theorem 1.4. Let $I\subset \mathscr{O}_K$ be an ideal of $K$, then there exists $x\in I$ such that $N(x)\le (\frac{4}{\pi})^{r_2}\frac{n!}{n^n}|\Delta_K|^{1/2}N(I)$.
Proof Sketch. We can see that, in fact, $\sigma$ is the embedding from $K$ to $\mathbb{R}^{r_1}\times \mathbb{C}^{r_2}$. Let $t\in\mathbb{R}_{>0}$, we consider the set $$S=\{x=(x_1,...,x_{r_1+r_2})\in\mathbb{R}^{r_1}\times \mathbb{C}^{r_2}|\sum_{i=1}^{r_1}|x_i|+2\sum_{j=r_1+1}^{r_1+r_2}|x_j|\le t\}$$
The set $S$ is convex, symmetric with respect to $0$, and compact. And $vol(S)=2^{r_1}(\frac{\pi}{2})^{r_2}\frac{t^n}{n!}$. We choose $t$ such that $vol(S)= 2^{n}vol(\sigma(I))$. By Corollary 1.3, $vol(\sigma(I))=2^{-r_2}|\Delta_K|^{1/2}N(I)$, i.e. $t^n=2^{n-r_1}\pi^{-r_2}n!|\Delta_K|^{1/2}N(I)=(\frac{4}{\pi})^{r_2}n!\Delta_K^{1/2}N(I)$. By Theorem 1.1, there exists $x\ne 0, x\in S\cap \sigma(I)$, and $$|N(x)|=\prod_{i=1}^{r_1}|\sigma_i(x)|\prod_{j=r_1+1}^{r_1+r_2}|\sigma_j(x)|^2$$
By using the Cauchy's inequality,
$$N(x)\le (\frac{1}{n}\sum_{i=1}^{r_1}|\sigma_i(x)|+\frac{2}{n}\sum_{j=r_1+1}^{r_1+r_2}|\sigma_j(x)|)^n\le \frac{t^n}{n^n}=(\frac{4}{\pi})^{r_2}\frac{n!}{n^n}\Delta_K^{1/2}N(I)$$
(Q.E.D)
This theorem has the most important
Corollary 1.5. Every ideal class of $K$ has an integral ideal $I\subset \mathscr{O}_K$, such that $N(I)\le (\frac{4}{\pi})^{r_2}\frac{n!}{n^n}\Delta_K^{1/2}$.
Proof. Recall that if $I, J$ are fractional $\mathscr{O}_K$-module, then $I\sim J$ in the ideal class group if $I=(x)J$ for some principal fractional $(x)$ as $\mathscr{O}_K$-module. Let $I$ be a representative of the given ideal class. Then one can see $I\sim I^{-1}$, because $II^{-1}=\mathscr{O}_K$, which is the ideal generated by $1$. Denote $J:=I^{-1}$, there exists $x\in \mathscr{O}_K\setminus \{0\}$ such that $(x)J$ is an ideal of $\mathscr{O}_K$, and $(x)J\sim J$. Hence, without loss of generality, we can assume that $J$ is an integral ideal.
Take $y\in J$, such that the condition of Theorem 1.4 holds, i.e. $N(y)\le (\frac{4}{\pi})^{r_2}\frac{n!}{n^n}|\Delta_K|^{1/2}N(J)$. Define $J':=(y)I$, then $J'\sim I$, and $J'$ is an integral ideal, since by definition $I^{-1}=J=\{y\in K|yx\in \mathscr{O}_K, \forall x\in I\}$. We then have
$$N(J')=N(y)N(I)\le (\frac{4}{\pi})^{r_2}\frac{n!}{n^n}|\Delta_K|^{1/2}N(J)=(\frac{4}{\pi})^{r_2}\frac{n!}{n^n}|\Delta_K|^{1/2}N(I)^{-1}$$
This yields $N(J')\le (\frac{4}{\pi})^{r_2}\frac{n!}{n^n}|\Delta_K|^{1/2}$. (Q.E.D)
From this, we can deduce the finiteness of the ideal class group.
Corollary 1.6. For any number field $K$, the ideal class group is finite.
Proof. It can be seen from Corollary 1.5 that for any ideal class of $K$, there always exists an integral ideal as its representative with the norm is bounded. And one can see that there exists only finitely many ideal of $\mathscr{O}_K$ with a given norm by the following argument. Assume that $q$ is a norm of an integral ideal $I$, then it can be seen that $q\in I$, and then exists only finitely many ideal lying over $(q)$ in $\mathscr{O}_K$, since $\mathscr{O}_K/(q)$ is a finite ring. (Q.E.D)
We now describe an algorithm to compute the ideal class group. The constant $M_K:=(\frac{4}{\pi})^{r_2}\frac{n!}{n^n}|\Delta_K|^{1/2}$ is called the Minkowski's constant. Because any ideal class have an integral ideal as its representative with $N(I)\le M_K$. Hence, we just need to determine the ideal class group from those ideals that have norm are equal or smaller than $M_K$. However, one can see $N(I)$ is an integer, and $I$ lies over $(N(I))$, and each $N(I)$ is product of prime numbers in $\mathbb{Z}$. Hence, it suffices to look at the factorization of primes smaller or equal to $M_K$. With each $p\le M_K$, we look at its factorization $p=\mathfrak{p_1}^{e_1}...\mathfrak{p_n}^{e_n}$ in $\mathscr{O}_K$, and determine with which smallest integer $n_p$, $\mathfrak{p}^{n_p}$ will be principal. And the similar method will continue for each $p\le M_K$. It is a very convenient way to compute the ideal class group of $K$. We will use it in the second section, with applications in solving some Diophantine equations.
2. Applications. In this section, we will first mention the Fermat's equation.
Problem 2.1. The equation $y^2=x^3-2$ has only two integral solutions, which are $(x,y)=(3,\pm 5)$.
Proof. Assume that $(x,y)$ is an integral solution of the equation above. If $x$ is even, then so is $y$, but then $y^2-x^3=2\equiv0\mod4$, a contradiction. Hence, both $x$ and $y$ are odd. From this, one can see that $x,y$ are coprime, i.e. there exists $a,b\in\mathbb{Z}$, such that $ax+by=1$. Hence, $(x),(y),(2)$ are co-prime in pair in any ring of integers of a number fields. It can be seen that $y^2+2=(y+\sqrt{-2})(y-\sqrt{-2})=x^3$, and this holds for ideals in $\mathscr{O}_{\mathbb{Q}(\sqrt{-2})}$. Assume that there exists a prime ideal of $\mathscr{O}_{\mathbb{Q}(\sqrt{-2})}$ such that $\mathfrak{p}|\gcd((y+\sqrt{-2}),(y-\sqrt{-2}))$, then $\mathfrak{p}|(2y)=(2)(y)$. Also, $\mathfrak{p}|(x)$, a contradiction, since $(x),(y),(2)$ are co-prime in pair.
This yields $(y+\sqrt{-2})=I^3, (y-\sqrt{-2})=J^3$, for some ideal $I,J\subset\mathscr{O}_{\mathbb{Q}(\sqrt{-2})}$. Now, if $I$ and $J$ are principal, the remaining part is very easy, we just need to compare the coordinates as $\mathbb{Q}$-vector space. Let us use the Minkowski's theorem at this point.
In this case, let $K:=\mathbb{Q}(\sqrt{-2})$, which is of degree 2 over $\mathbb{Q}$, and $K$ has no real embedding, and two complex embeddings, hence $r_2=1$. And because $-2\equiv2\mod 4$, $\mathscr{O}_K=\mathbb{Z}[\sqrt{-2}]$, and $|\Delta_K|=8$. Hence, the Minkowski's constant $M_K=(\frac{4}{\pi})\frac{2!}{2^2}8^{1/2}\approx 1.8$. From this, we can see that there exists only one element in the class group of $\mathscr{O}_K$, or equivalently, $\mathscr{O}_K$ is a P.I.D.
Now, assume $I=(a+b\sqrt{-2})$, then $I^3=(a+b\sqrt{-2})^3=(y+\sqrt{-2})$, i.e., there exists $u=c+d\sqrt{-2}\in\mathscr{O}_K^\times$, for some $c,d\in\mathbb{Z}$ (u is called unit on $\mathscr{O}_K$), such that $(a+b\sqrt{-2})^3=u(y+\sqrt{-2})$. For such $u$, it can be seen that $N(u)=\pm 1=c^2+2d^2$, and hence, the only units in $\mathscr{O}_K$ is just $\pm 1$.
If $u=1$, then $(a+b\sqrt{-2})^3=y+\sqrt{-2}$. In this case, $3a^2b-2b^3=1$, and it is easy to see that the only roots are $b=1,a=\pm 1$. Hence, $y=\pm 5$. Replace to the original equation, we get $x=3$.
If $u=-1$, then $3a^2b-2b^3=-1$ has solution $b=-1, a=\pm 1$. And we get the same value for $y=\pm 5$. It can be seen from our argument that the equation above only has integral solutions $(3,\pm5)$. (Q.E.D)
By a very similar method, one can prove that the equation $y^2=x^3-5$ has no integral solution.
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