# Answer to Question #12701 in Abstract Algebra for Tsit Lam

Question #12701

Examine Z[i]/(2) 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]/(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.

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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