Integral Basis In Quadratic Fields: A Discriminant Approach

Alright, guys, let's dive into the fascinating world of algebraic number theory! Today, we're tackling a fundamental problem: finding an integral basis for the ring of integers in quadratic fields. We'll leverage the discriminant of a basis to make our lives easier. Buckle up; it's going to be a fun ride!

Understanding Quadratic Fields and Rings of Integers

Before we jump into the nitty-gritty, let's make sure we're all on the same page regarding quadratic fields and rings of integers. Think of quadratic fields as extensions of the rational numbers, $\mathbb{Q}$, obtained by adjoining the square root of some integer m. We denote this field as $\mathbb{Q}(\sqrt{m})$. So, any element in $\mathbb{Q}(\sqrt{m})$ can be written in the form $a + b\sqrt{m}$, where a and b are rational numbers.

Now, what about the ring of integers? Well, given a field extension, the ring of integers consists of all elements in that field that are roots of monic polynomials with integer coefficients. In simpler terms, these are the "integer-like" elements within our field. For example, in $\mathbb{Q}(\sqrt{m})$, we want to find all elements $\alpha$ such that $\alpha$ is a root of a polynomial like $x^2 + cx + d = 0$, where c and d are integers. This ring of integers is usually denoted as $\mathcal{O}_K$, where K is the field (in our case, $\mathbb{Q}(\sqrt{m})$).

It's known that for $\mathbb{Q}(\sqrt{m})$, the ring of integers takes one of two forms depending on m modulo 4:

• If $m \equiv 2, 3 \pmod{4}$, then $\mathcal{O}_K = \{a + b\sqrt{m} : a, b \in \mathbb{Z}\}$.
• If $m \equiv 1 \pmod{4}$, then $\mathcal{O}_K = \{\frac{a}{2} + \frac{b}{2}\sqrt{m} : a, b \in \mathbb{Z}, a \equiv b \pmod{2}\}$. This means a and b are either both even or both odd.

Why do we care about the ring of integers? Well, it's the natural setting to study arithmetic in these fields. It's where we can generalize many concepts from the integers, like prime factorization (though it's not always unique!).

The Discriminant of a Basis: Our Secret Weapon

Okay, now let's introduce our secret weapon: the discriminant of a basis. This is a powerful tool that helps us determine whether a given set of elements forms an integral basis for our ring of integers. Given a basis $\{\alpha_1, \alpha_2, ..., \alpha_n\}$ of a field extension K of degree n over $\mathbb{Q}$, the discriminant of this basis is defined as:

$D(\alpha_1, \alpha_2, ..., \alpha_n) = (det(\sigma_i(\alpha_j)))^2$

Where $\sigma_i$ are the n distinct embeddings of K into the complex numbers $\mathbb{C}$. In simpler terms, we apply all possible ways of mapping our field K into the complex numbers while keeping the rational numbers fixed, and then we compute a determinant. Squaring the determinant ensures that the discriminant is always a rational number.

For our specific case of quadratic fields $\mathbb{Q}(\sqrt{m})$, we have a basis of the form $\{1, \sqrt{m}\}$. The embeddings are:

• $\sigma_1(a + b\sqrt{m}) = a + b\sqrt{m}$ (the identity embedding)
• $\sigma_2(a + b\sqrt{m}) = a - b\sqrt{m}$ (the conjugate embedding)

Thus, the discriminant of the basis $\{1, \sqrt{m}\}$ is:

$D(1, \sqrt{m}) = \begin{vmatrix} 1 & \sqrt{m} \\ 1 & -\sqrt{m} \end{vmatrix}^2 = (-2\sqrt{m})^2 = 4m$

The discriminant of the basis $\{1, \sqrt{m}\}$ is $4m$. This value is incredibly useful because it's related to the discriminant of the field itself, often denoted as $d_K$. The discriminant of the field is an intrinsic property of the field, independent of the choice of basis. If $\{\alpha_1, \alpha_2, ..., \alpha_n\}$ is an integral basis, then $D(\alpha_1, \alpha_2, ..., \alpha_n) = d_K$.

The main idea here is that if we have any basis $\{\alpha_1, \alpha_2\}$ for $\mathbb{Q}(\sqrt{m})$ (as a vector space over $\mathbb{Q}$), then $D(\alpha_1, \alpha_2) = d_K \cdot i^2$ for some integer i. Thus, we can determine whether $\{\alpha_1, \alpha_2\}$ is an integral basis if we know the discriminant of the field $d_K$.

Calculating the Integral Basis

Now, let's put everything together to calculate the integral basis for $\mathbb{Q}(\sqrt{m})$. We'll consider the two cases:

Case 1: $m \equiv 2, 3 \pmod{4}$

In this case, we suspect that the integral basis is $\{1, \sqrt{m}\}$. We already know that $D(1, \sqrt{m}) = 4m$. We also know that in this case, $d_K = 4m$. Since the discriminant of our basis matches the discriminant of the field, we can confidently say that $\{1, \sqrt{m}\}$ is indeed an integral basis.

Thus, if $m \equiv 2, 3 \pmod{4}$, the ring of integers is $\mathcal{O}_K = \{a + b\sqrt{m} : a, b \in \mathbb{Z}\}$.

Case 2: $m \equiv 1 \pmod{4}$

In this case, we suspect that the integral basis is $\{1, \frac{1 + \sqrt{m}}{2}\}$. Let's calculate the discriminant of this basis:

$D(1, \frac{1 + \sqrt{m}}{2}) = \begin{vmatrix} 1 & \frac{1 + \sqrt{m}}{2} \\ 1 & \frac{1 - \sqrt{m}}{2} \end{vmatrix}^2 = \left(\frac{1 - \sqrt{m}}{2} - \frac{1 + \sqrt{m}}{2}\right)^2 = (-\sqrt{m})^2 = m$

In this case, $d_K = m$. Since the discriminant of our basis matches the discriminant of the field, we can confirm that $\{1, \frac{1 + \sqrt{m}}{2}\}$ is an integral basis.

An alternative basis is $\{\frac{1 - \sqrt{m}}{2}, \frac{1 + \sqrt{m}}{2}\}$. Let's calculate the discriminant of this basis:

$D(\frac{1 - \sqrt{m}}{2}, \frac{1 + \sqrt{m}}{2}) = \begin{vmatrix} \frac{1 - \sqrt{m}}{2} & \frac{1 + \sqrt{m}}{2} \\ \frac{1 + \sqrt{m}}{2} & \frac{1 - \sqrt{m}}{2} \end{vmatrix}^2 = \left( (\frac{1 - \sqrt{m}}{2})^2 - (\frac{1 + \sqrt{m}}{2})^2 \right)^2 = \left( \frac{1 -2\sqrt{m} + m}{4} - \frac{1 + 2\sqrt{m} + m}{4} \right)^2 = \left( -\sqrt{m} \right)^2 = m$

We get the same discriminant of $m$.

Thus, if $m \equiv 1 \pmod{4}$, the ring of integers is $\mathcal{O}_K = \{\frac{a}{2} + \frac{b}{2}\sqrt{m} : a, b \in \mathbb{Z}, a \equiv b \pmod{2}\}$. Notice that we can rewrite any element in $\mathcal{O}_K$ as $a + b(\frac{1 + \sqrt{m}}{2})$ where $a, b \in \mathbb{Z}$.

Examples

Let's solidify our understanding with a couple of examples:

Example 1: $\mathbb{Q}(\sqrt{2})$

Here, $m = 2$, so $m \equiv 2 \pmod{4}$. Thus, the integral basis is $\{1, \sqrt{2}\}$ and the ring of integers is $\mathcal{O}_K = \{a + b\sqrt{2} : a, b \in \mathbb{Z}\}$.

Example 2: $\mathbb{Q}(\sqrt{5})$

Here, $m = 5$, so $m \equiv 1 \pmod{4}$. Thus, the integral basis is $\{1, \frac{1 + \sqrt{5}}{2}\}$ and the ring of integers is $\mathcal{O}_K = \{\frac{a}{2} + \frac{b}{2}\sqrt{5} : a, b \in \mathbb{Z}, a \equiv b \pmod{2}\}$.

Conclusion

Alright, that's a wrap! We've seen how to calculate an integral basis for the ring of integers in quadratic fields using the discriminant of a basis. By understanding the properties of quadratic fields and the discriminant, we can efficiently determine the integral basis and describe the ring of integers explicitly.

Remember, the key steps are:

1. Determine m modulo 4.
2. Identify the potential integral basis based on the congruence of m.
3. Calculate the discriminant of the proposed basis.
4. Compare the discriminant of the basis with the discriminant of the field. If they match, you've found your integral basis!

This technique provides a powerful approach to understanding the arithmetic structure of quadratic fields. Keep practicing, and you'll become a pro in no time!