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Question 26947: PLEASE NEED HELP
show that the set of all integers of form 36a-90b+72c forms an ideal in Z(intergers) and find the generator for the ideal.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! LET US DENOTE THE SET OF ALL INTEGERS OF FOR
36K-90L+72M BY S.
K,L,M ARE INTEGERS
TO SHOW THAT S IS AN IDEAL OF Z.
LET S1 AND S2 BE 2 ELEMENTS OF S SO THAT
S1=36K1-90L1+72M1 AND S2=36K2-90L2+72M2
THEN S1-S2=36(K1-K2)-90(L1-L2)+72(M1-M2)
SINCE K1-K2,L1-L2,M1-M2 ARE INTEGERS ,S1-S2 IS IN THE FORM
S=36K-90L+72M
SO S IS A SUB GROUP OF Z UNDER ADDITION.
NOW LET Z1 BE ANY ELEMENT OF Z ; AND S1 AS ABOVE AN ELEMENT OF S.THEN
WE HAVE Z1S1=Z1*36K1-Z1*90L1+Z1*72M1…
SINCE MULTIPLICATION OF INTEGERS IS ASSOCIATIVE AND COMMUTATIVE IN Z,WE GET
Z1S1=S1Z1=36(K1Z1)-90(L1Z1)+72(M1Z1),WHICH BEING OF THE FORM S
ARE ELEMENTS OF S.
HENCE S IS LEFT AND RIGHT IDEAL OF Z ,THAT IS AN IDEAL OF Z..
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S=36K-90L+72M
THIS MEANS S IS THE GCD OF 36,90,72 OR MULTIPLE OF THEIR GCD.
SO GCD OF 36,72,90 IS A GENERATOR OF THIS IDEAL.THAT IS 18 IS A GENERATOR.
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