Understanding Irreducibility in Z[√-3]: The Cases of 7 and 11

In this article, we will explore the concepts of irreducibility in the ring of Eisenstein integers, denoted as Z[√-3]. Specifically, we will investigate whether the integers 7 and 11 can be factored further within this ring.

Introduction to Eisenstein Integers

Eisenstein integers are a special type of complex numbers. They are elements of the form u vω, where u and v are integers and ω is a complex cube root of unity, satisfying ω^3 1 but ω ≠ 1. One common representation is ω -1 √-3/2, and its complex conjugate is ω^2 -1 - √-3/2.

Norm in Z[√-3]

To determine irreducibility, we use the concept of the norm. The norm of an Eisenstein integer u vω is defined as the product of the integer with its complex conjugate, N[u vω] u^2 - u v ω v^2 ω^2. Simplifying this, we obtain N[u vω] u^2 - u v - v^2.

Irreducibility of 7 in Z[√-3]

Consider the Eisenstein integer 3 ω. We calculate its norm as follows:

Norm of 3 ω: N[3 ω] 3^2 - 3 * 1 - 1^2 9 - 3 - 1 5.

However, we need to show that 7 is not irreducible. We start by deconstructing 7:

7 (2ω)(2ω^2) 1 7, confirming that 7 is indeed reducible in Z[√-3].

Irreducibility of 11 in Z[√-3]

To prove that 11 is irreducible in Z[√-3], we assume the contrary. Suppose 11 (a bω)(c dω), where neither factor is a unit.

Conjugating this equation, we get 11 (a - bω)(c - dω). Multiplying the two equations, we have 11^2 (a^2 3b^2)(c^2 3d^2).

For non-unit factors, it must be true that a^2 3b^2 c^2 3d^2 11.

Taking modulo 3, we find that a^2 ≡ 2 (mod 3). Since 2 is not a perfect square modulo 3, this leads to a contradiction.

Therefore, either a bω or c dω must be a unit, and the claim follows. Irreducibility of 11 in Z[√-3] is confirmed.

Conclusion

In conclusion, the integer 7 is reducible in Z[√-3] because it can be expressed as a product of other Eisenstein integers. On the other hand, 11 is irreducible in Z[√-3] based on our proof utilizing the properties of the norm and modular arithmetic.