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Let me detail the title of the question. I'm trying to give students an intuition of what the class number is.

Let $K=\mathbb{Q}(\sqrt{-d})$, with $d>0$ a square-free integer, be a quadratic imaginary field. Let $\mathcal{O}_K$ be its ring of integers. It is of the form $\mathbb{Z}[\tau]$ with $\tau=\sqrt{-d}$ or $\tau=\frac{1+\sqrt{-d}}{2}$ depending on the value of $d$ mod $4$.

So let us think of $\mathcal{O}_K$ as the lattice of $\mathbb{C}$ generated by $1$ and $\tau$. Then ideals should correspond to sublattices of $\mathcal{O}_K$ and two of them should define the same class in the class group if one can pass from one to another by multiplying by an element of $\alpha$, isn't it?

Could anybody help me to precisemake this analogy precise? For instance, how can one see that $\mathbb{Q}(i)$ has class number 1 but $\mathbb{Q}(\sqrt{-5})$ doesn't just by looking at the corresponding lattices? The (non-equivalent) decompositions

$2.3=(1+\sqrt{5}i)(1-\sqrt{5}i)$

suggest to consider the lattices $\mathbb{Z}\cdot 2+\mathbb{Z}\cdot(1+\sqrt{5}i)$ and $\mathbb{Z}\cdot 3+\mathbb{Z}\cdot(1-\sqrt{5}i).$ Is that what I have to do?

Thanks!

Let me detail the title of the question. I'm trying to give students an intuition of what the class number is.

Let $K=\mathbb{Q}(\sqrt{-d})$, with $d>0$ a square-free integer, be a quadratic imaginary field. Let $\mathcal{O}_K$ be its ring of integers. It is of the form $\mathbb{Z}[\tau]$ with $\tau=\sqrt{-d}$ or $\tau=\frac{1+\sqrt{-d}}{2}$ depending on the value of $d$ mod $4$.

So let us think of $\mathcal{O}_K$ as the lattice of $\mathbb{C}$ generated by $1$ and $\tau$. Then ideals should correspond to sublattices of $\mathcal{O}_K$ and two of them should define the same class in the class group if one can pass from one to another by multiplying by an element of $\alpha$, isn't it?

Could anybody help me to precise this analogy? For instance, how can one see that $\mathbb{Q}(i)$ has class number 1 but $\mathbb{Q}(\sqrt{-5})$ doesn't just by looking at the corresponding lattices? The (non-equivalent) decompositions

$2.3=(1+\sqrt{5}i)(1-\sqrt{5}i)$

suggest to consider the lattices $\mathbb{Z}\cdot 2+\mathbb{Z}\cdot(1+\sqrt{5}i)$ and $\mathbb{Z}\cdot 3+\mathbb{Z}\cdot(1-\sqrt{5}i).$ Is that what I have to do?

Thanks!

Let me detail the title of the question. I'm trying to give students an intuition of what the class number is.

Let $K=\mathbb{Q}(\sqrt{-d})$, with $d>0$ a square-free integer, be a quadratic imaginary field. Let $\mathcal{O}_K$ be its ring of integers. It is of the form $\mathbb{Z}[\tau]$ with $\tau=\sqrt{-d}$ or $\tau=\frac{1+\sqrt{-d}}{2}$ depending on the value of $d$ mod $4$.

So let us think of $\mathcal{O}_K$ as the lattice of $\mathbb{C}$ generated by $1$ and $\tau$. Then ideals should correspond to sublattices of $\mathcal{O}_K$ and two of them should define the same class in the class group if one can pass from one to another by multiplying by an element of $\alpha$, isn't it?

Could anybody help me to make this analogy precise? For instance, how can one see that $\mathbb{Q}(i)$ has class number 1 but $\mathbb{Q}(\sqrt{-5})$ doesn't just by looking at the corresponding lattices? The (non-equivalent) decompositions

$2.3=(1+\sqrt{5}i)(1-\sqrt{5}i)$

suggest to consider the lattices $\mathbb{Z}\cdot 2+\mathbb{Z}\cdot(1+\sqrt{5}i)$ and $\mathbb{Z}\cdot 3+\mathbb{Z}\cdot(1-\sqrt{5}i).$ Is that what I have to do?

Thanks!

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how to visualize the class number of an imaginary quadratic field?

Let me detail the title of the question. I'm trying to give students an intuition of what the class number is.

Let $K=\mathbb{Q}(\sqrt{-d})$, with $d>0$ a square-free integer, be a quadratic imaginary field. Let $\mathcal{O}_K$ be its ring of integers. It is of the form $\mathbb{Z}[\tau]$ with $\tau=\sqrt{-d}$ or $\tau=\frac{1+\sqrt{-d}}{2}$ depending on the value of $d$ mod $4$.

So let us think of $\mathcal{O}_K$ as the lattice of $\mathbb{C}$ generated by $1$ and $\tau$. Then ideals should correspond to sublattices of $\mathcal{O}_K$ and two of them should define the same class in the class group if one can pass from one to another by multiplying by an element of $\alpha$, isn't it?

Could anybody help me to precise this analogy? For instance, how can one see that $\mathbb{Q}(i)$ has class number 1 but $\mathbb{Q}(\sqrt{-5})$ doesn't just by looking at the corresponding lattices? The (non-equivalent) decompositions

$2.3=(1+\sqrt{5}i)(1-\sqrt{5}i)$

suggest to consider the lattices $\mathbb{Z}\cdot 2+\mathbb{Z}\cdot(1+\sqrt{5}i)$ and $\mathbb{Z}\cdot 3+\mathbb{Z}\cdot(1-\sqrt{5}i).$ Is that what I have to do?

Thanks!