关于Artin《代数》中ℚ[√d]代数整数证明的两处疑问

Let's start by recapping the core setup here, since it's foundational to Artin's proof:

An element $\alpha \in \mathbb{Q}[\sqrt{d}]$ is an algebraic integer if and only if it's a root of a monic polynomial with integer coefficients. Every element of $\mathbb{Q}[\sqrt{d}]$ can be written as $\alpha = a + b\sqrt{d}$ where $a,b \in \mathbb{Q}$. The minimal polynomial of $\alpha$ is:

x² - 2a x + (a² - b²d) = 0

For $\alpha$ to be an algebraic integer, the coefficients $2a$ and $a² - b²d$ must both be integers. That's the starting point for Artin's case split.

Question 1: Where does $d \equiv 2$ or $3 \pmod{4}$ get used in Case 1? #

First, let's walk through Case 1's logic with the condition in mind:

1. From $2a \in \mathbb{Z}$, let $2a = m$ where $m$ is an integer, so $a = m/2$.
2. Substitute into the second coefficient condition: $a² - b²d \in \mathbb{Z}$ → $\frac{m²}{4} - b²d \in \mathbb{Z}$. Rearranged, this gives $b²d = \frac{m²}{4} - k$ for some integer $k$, so $4b²d = m² - 4k$ (which is an integer).

Now here's where $d \equiv 2$ or $3 \pmod{4}$ comes into play:

• We need to rule out the possibility that $b$ is a half-integer (i.e., $b = n/2$ where $n$ is an odd integer). Suppose $b$ were a half-integer: then $b²d = \frac{n²d}{4}$. Since $n$ is odd, $n² \equiv 1 \pmod{4}$.
  • If $d \equiv 2 \pmod{4}$, then $n²d \equiv 1*2 = 2 \pmod{4}$, so $\frac{n²d}{4} = \text{integer} + \frac{1}{2}$.
  • If $d \equiv 3 \pmod{4}$, then $n²d \equiv 1*3 = 3 \pmod{4}$, so $\frac{n²d}{4} = \text{integer} + \frac{3}{4}$.
• Now plug back into $a² - b²d$: if $a = m/2$, $m$ is integer, $a²$ is either integer (if $m$ even) or $\text{integer} + \frac{1}{4}$ (if $m$ odd). Neither case will subtract with $b²d$ (which is integer + 1/2 or 3/4) to give an integer. That's a contradiction.

So the condition $d \equiv 2$ or $3 \pmod{4}$ forces $b$ to be an integer (not a half-integer). Then, since $2a$ is integer and $a² - b²d$ is integer, $a$ must also be an integer (if $a$ were a half-integer, $a²$ would be integer + 1/4, and subtracting integer $b²d$ would give non-integer). Hence in Case 1, algebraic integers are exactly $a + b\sqrt{d}$ with $a,b \in \mathbb{Z}$.

Question 2: Why is $b²d \in \mathbb{Z} + \frac{1}{4}$ if and only if $d \equiv 1 \pmod{4}$? #

Let's break this equivalence down in the context of Case 2:

Forward direction ($\implies$): If $b²d \in \mathbb{Z} + \frac{1}{4}$, then $d \equiv 1 \pmod{4}$ #

• If $b²d = t + \frac{1}{4}$ for some integer $t$, multiply both sides by 4: $4b²d = 4t + 1$. The right-hand side is odd, so $4b²d$ must be odd.
• Since $d$ is square-free (a standard assumption for $\mathbb{Q}[\sqrt{d}]$), $b$ can't be an integer (otherwise $4b²d$ would be divisible by 4, hence even). So $b$ must be a half-integer: $b = n/2$ where $n$ is odd.
• Substitute back: $4*(n²/4)d = n²d = 4t + 1$. Since $n$ is odd, $n² \equiv 1 \pmod{4}$, so $1d \equiv 1 \pmod{4}$ → $d \equiv 1 \pmod{4}$.

Reverse direction ($\impliedby$): If $d \equiv 1 \pmod{4}$, then there exists $b$ such that $b²d \in \mathbb{Z} + \frac{1}{4}$ #

• Take $b$ as a half-integer: $b = n/2$, $n$ odd. Then $b²d = \frac{n²d}{4}$. Since $n² \equiv 1 \pmod{4}$ and $d \equiv 1 \pmod{4}$, $n²d \equiv 1*1 = 1 \pmod{4}$. So $n²d = 4k + 1$ for some integer $k$, hence $b²d = \frac{4k+1}{4} = k + \frac{1}{4}$, which is exactly $\mathbb{Z} + \frac{1}{4}$.

This is exactly the key that lets Case 2 include algebraic integers of the form $\frac{m}{2} + \frac{n}{2}\sqrt{d}$ where $m,n$ are integers (since when $d \equiv1 \pmod{4}$, $a = m/2$ and $b =n/2$ will make $a² -b²d = \frac{m² -n²d}{4}$, which is integer because $m² -n²d \equiv 1 -1*1=0 \pmod{4}$ if $m,n$ are odd, or even minus even if $m,n$ are even).

内容的提问来源于stack exchange，提问作者user5826