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Answer to Question #12702 in Abstract Algebra for Tsit Lam

Question #12702
Examine Z[i]/(3) on being a field. How many elements it has?
Notice that Z[i] consists of elements of the form a+bi, where a,b belong to
Z.
Therefore
Z[i]/(3)
consists of elments of the form

a+bi,
where a,b belong to Z_2 = {0,1,2}.

Hence
Z[i]/(3)

consists of the following 9 elements:

0,
1, 2,
i,
2i,
1+i, 1+2i,
2+i, 2+2i.

It is convenient to replace 2
with -1.
Then Z[i]/(3) will consists of the following elements:

0,
1, -1,
i, -i,
1+i, 1-i,
-1+i, -1-i.

Let us check
if it is a field.
Let us try to find inverses to each non-zero element of
Z[i]/(3).

1 * 1 = 1
(-1) * (-1) = 1
i * (-i) = 1
(1+i)(-1+i) =
-1 + i - i + i^2 = -1 - 1 = -2 = 1 (mod 3)
(-1-i)(1-i) = -1 + i - i + i^2 =
-1 - 1 = -2 = 1 (mod 3)

Thus all non-zero elements of Z[i]/(3) are
invertible, and so Z[i]/(3) is a field.

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