Question #134513
Let S = { a + ib / a , b ∈ Z ,b is even } Show that S is a
subring of Z[i] , but not an ideal of Z[i]
1
Expert's answer
2020-09-22T16:18:14-0400
SolutonSoluton

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)ca(bc)=(ab)c

and distributive laws

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


and

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

for every

a,b,cRa,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


SRS \subseteq R

that is a ring under the operations of R.

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


arAar \in A

and

raA,ra \in A,


we say that A is an ideal of R.

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

a+bi/a,bZ{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,bS,a,b \in S,


then,

ab,abSa-b, ab \in S

Let

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


Step 3

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

Since

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

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

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


Step 4

Similarly,

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


Step 5

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

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


Step 6

However,

ISI \in S

and

Ii=iS.Ii=i \notin S.

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


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