Answer to Question #133619 in Abstract Algebra for PRATHIBHA ROSE C S

Question

Let S={a+ib/a,b element of Z,b is even} show that S is a subring of Z[i],but not an ideal of Z[i]

Answer 1

$$Soluton$$

Step 1

A ring R is a set with two binary operations, addition and multiplication, satisfying several properties: R is an Abelian group under addition, and the multiplication operation satisfies the associative law

$a(bc)=(ab)c$

and distributive laws

$a(b+c)=ab+ac$

and

$(b+c)a=ba+ca$

for every

$a,b,c \in R$

The identity of the addition operation is denoted 0. If the multiplication operation has an identity, it is called a unity. If multiplication is commutative, we say that R is commutative.

A subring of a ring R is a subset

$S \subseteq R$

that is a ring under the operations of R.

Let A be a subring of a ring R. If, for every $a \in A$ and $r \in R$

$ar \in A$

and

$ra \in A,$

we say that A is an ideal of R.

The ring $Z[i]$ of Gaussian integers is defined as

${a+bi/ a,b \in Z}$

under ordinary addition and multiplication of complex numbers.

Step 2

The Subring Test states that a nonempty subset S of a ring R is a subring if and only if it is closed under subtraction and multiplication; that is, if

$a,b \in S,$

then,

$a-b, ab \in S$

Let

$a+bi, c+di \in S$

Step 3

$(a+bi)-(c+di)= (a-c)+(b-d)i$

Since

$a+bi, c+di \in S$

both b and d are even. Hence $b-d$ is even, and

$(a+bi)-(c+di) \in S$

Step 4

Similarly,

$(a+bi)( c+di) =ac+adi+bci-bd \implies (ac-bd)(ad+bc)i$

Step 5

Since b and d are even, $ad$ and $bc$, are even, and so is $ad+bc$. It follows that

$(a+bi)( c+di) \in S$

Step 6

However,

$I \in S$

and

$Ii=i \notin S.$

We conclude that S is not an ideal of $Z[i].$