Answer to Question #12701 in Abstract Algebra for Tsit Lam
Examine Z[i]/(2) on being a field. How many elements it has?
1
2012-08-10T09:02:33-0400
Notice that Z[i] consists of elements of the form a+bi, where a,b belong to
Z.
Therefore
Z[i]/(2)
consists of elments of the form
a+bi,
where a,b belong to Z_2 = {0,1}.
Hence
Z[i]/(2)
consists of the following 4 elements:
0, 1, i, 1+i
Let us
verify if it is a field.
Notice that
(1+i)(1+i) = 1 + i + i + i^2
=
= 1 + 2i - 1
= 2i = 0 mod (2),
Thus 1+i
is a zero divisor in Z[i]/(2),
and so Z[i]/(2) is not a field.
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