Determining Prime Numbers in the Ring of Algebraic Integers $mathbb{Z}[frac{1 sqrt{13}}{2}]$

Rational integers are integral numbers, and the ring of algebraic integers, specifically the ring $mathbb{Z}[frac{1 sqrt{13}}{2}]$, is a fascinating area of study in number theory. This article will explore how to determine whether a number is prime within this ring, using key theorems and methods such as Dedekind's Theorem and quadratic reciprocity.

Introduction to $mathbb{Z}[frac{1 sqrt{13}}{2}]$

Algebraic integers are numbers that are roots of monic polynomials with integer coefficients. In this case, we are dealing with the ring $mathbb{Z}[frac{1 sqrt{13}}{2}]$, where the element $alpha frac{1 sqrt{13}}{2}$ is a root of the irreducible polynomial $x^2 - x - 3 0$. This is a unique factorization domain (UFD), which means prime elements here are unique up to multiplication by units in the ring.

Prime Ideals in $mathbb{Z}[frac{1 sqrt{13}}{2}]$

We will analyze prime ideals in this ring. Prime ideals are principal, which means they can be represented by a single element. The prime elements in this ring are unique up to multiplication by units. We employ Dedekind's Theorem for factoring ideals of the form $langle p rangle$, where $p$ is a positive prime in the integers $mathbb{Z}$. This theorem is crucial for determining the behavior of prime numbers in algebraic extensions.

Case 1: Prime $p 2$

For $p 2$, the minimal polynomial $x^2 - x - 3$ modulo 2 is $x^2 - x - 1 equiv x^2 - x 1 pmod{2}$. This polynomial is irreducible over $mathbb{F}_2$ because it does not have roots in $mathbb{F}_2$. Therefore, $langle 2 rangle$ remains prime in $mathbb{Z}[frac{1 sqrt{13}}{2}]$.

Case 2: Prime $p 13$

For $p 13$, we can directly see that $langle 13 rangle langle sqrt{13} rangle^2$. This means that $13$ is not a prime in $mathbb{Z}[frac{1 sqrt{13}}{2}]$ because it is the square of a non-unit element.

General Case: Prime $p eq 2, 13$

For any odd rational prime $p eq 13$, the minimal polynomial of $alpha$ can be rewritten as $frac{1}{4} 2x - 1^2 - 13$. We then consider the reduction of this polynomial modulo $p$. This works because $frac{1}{4}$ exists modulo $p$.
If $13$ is a quadratic nonresidue modulo $p$, the polynomial remains irreducible, and thus $langle p rangle$ is prime in $mathbb{Z}[frac{1 sqrt{13}}{2}]$.
If $13$ is a quadratic residue modulo $p$, the polynomial factors, and $langle p rangle$ is not a prime ideal in $mathbb{Z}[frac{1 sqrt{13}}{2}]$.

Quadratic Residue and Nonresidue Analysis

To determine if $13$ is a quadratic residue modulo $p$, we use properties of the Legendre symbol and quadratic reciprocity. Specifically, we have:

[ left(frac{13}{p}right) left(frac{p}{13}right) ]

The value of $left(frac{p}{13}right)$ can be computed directly:

[ left(frac{p}{13}right) begin{cases} 1 text{if } p equiv pm 1, pm 3, pm 4 pmod{13} -1 text{if } p equiv pm 2, pm 5, pm 6 pmod{13} end{cases} ]

Using these results, we can conclude the primality of $langle p rangle$ in $mathbb{Z}[frac{1 sqrt{13}}{2}]$ more concretely. Specifically,

- If $p 2$ or $p equiv pm 1, pm 3, pm 4 pmod{13}$, then $langle p rangle$ remains a prime ideal in $mathbb{Z}[frac{1 sqrt{13}}{2}]$. - If none of the above conditions are met, then $langle p rangle$ is not a prime ideal in this ring.

Conclusion

This article has explored the methodology for determining if a number is prime in the ring of algebraic integers $mathbb{Z}[frac{1 sqrt{13}}{2}]$ using key theorems and properties of quadratic residues. By analyzing the reductions of the minimal polynomial and applying properties of the Legendre symbol and quadratic reciprocity, we can effectively determine the primality of prime numbers in this specific algebraic extension. This insight is valuable for further study in number theory and provides a clear framework for understanding prime ideals in more complex algebraic structures.