Fields and gaussian integers

• Oct 8th 2010, 03:07 PM
ulysses123
Fields and gaussian integers
How many elements does the following field have? is u=1+i a primative element?

$\frac{Z[i]}{<2+3i>}$

I know that the norm of 2+3i =13, thus there will be 13 elements and this field will be isomorphic to GF(13), But i am not sure how to find the elements.
• Oct 8th 2010, 06:44 PM
tonio
Quote:

Originally Posted by ulysses123
How many elements does the following field have? is u=1+i a primative element?

$\frac{Z[i]}{<2+3i>}$

I know that the norm of 2+3i =13, thus there will be 13 elements and this field will be isomorphic to GF(13), But i am not sure how to find the elements.

Do as you do to find the elements of a ring of fractions of a polynomial ring divided by the principal

ideal generated by an irreducible pol.:

1) Divide any element $w=\alpha+\beta i\in\mathbb{Z}[i]$ by $2+3i$ with remainder (why is this possible?)

2) If $\alpha+\beta i=(a+bi)(2+3i)+r\,,\,\,r=c+di\in\mathbb{Z}[i]\,,\,\,r=0\,\,or\,\,c^2+d^2<2^2+3^2=13$ , then
$\alpha+\beta i\cong r\!\!\pmod{<2+3i>}$ ...

3) Deduce from the above that every element in the quotient ring can be represented by an element with norm < 13 or zero.

Tonio
• Oct 9th 2010, 06:44 PM
ulysses123
every element can be represented by one with norm <13, but we dont have to take this representation do we?
Using the element 2+3i and then rotating it by 90 degrees i formed a square with these two vectors then took all the elements in it as my represenatives, then using the element 1+i and generating its powers i found that it only has order 4.However for the second field i havnt been able to deduce if 1+i is a generator because one of the powers gives me an element thats not in the field or any of the representative cosets.So now i'm getting confused.