# Answer on Abstract Algebra Question for Tsit Lam

Question #12702

Examine Z[i]/(3) on being a field. How many elements it has?

Expert's answer

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.

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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