2

Describe each of following the rings:

a) $\mathbb{Z}[x] / (x^2 - 3, 2x + 4)$

b) $\mathbb{Z}[i]/ (2 + i)$

a) Well, $\mathbb{Z}[x]$ is the set of all polynomials with integer coefficients and $(x^2 - 3, 2x + 4)$ is the ideal generated by two polynomials. I know that $\mathbb Z[x] / (x^2 - 3, 2x + 4) = \{ r + (x^2 - 3, 2x + 4) : r \in \mathbb{Z}[x] \}$

b) I know that $\mathbb{Z}[i]$ is the set of all polynomials with the complex number $i$ as its variable. $\mathbb{Z}[i]/ (2 + i) = \{ r + (2 + i) : r \in \mathbb{Z}[i] \}$.

I already know that b) is $\mathbb Z$ modulo 5. What would a) be?

josh
  • 273

1 Answers1

5

b) The link didn't contain what I consider the cleanest approach to this problem, and this is a good warm-up for a), so I thought I'd answer it again. The trick is to think of $\mathbb{Z}[i]$ as $\mathbb{Z}[x]/(x^2+1)$. Then we can compute the quotient in two steps: $\mathbb{Z}[x]/(x^2+1,x+2) = \mathbb{Z}/((-2)^2+1) = \mathbb{Z}/(5)$.

a) We can write $(x^2-3,2x+4) = (x^2-3,2x+4,2x^2-6)=(x^2-3,2x+4,(-4)x-6)$ $=(x^2-3,2x+4,4x+6-2(2x+4))$ $=(x^2-3,2x+4,-2) = (x^2+1,2)$. You can rewrite this in terms of $\mathbb{Z}[i]$ if you like, but I think it's better to make the substitution $y=x+1$ to get the lovely expression $\mathbb{F}_2[y]/(y^2)$.

Hopefully this was useful, and hopefully you know that your grader probably also uses StackExchange and cares whether you properly cite your sources.

Andrew Dudzik
  • 30,074