Show that the set of all multiples of 3 is an ideal in the ring of integers. Ext

Question

Show that the set of all multiples of 3 is an ideal in the ring of integers. Extend your proof to show that the set of all multiples of any given integer m is an ideal.

Explanation / Answer

We directly prove that is the ideal for all m. The proof works for 3 by just substituting m as 3. I = = {km | k in Z}. To show that I is an ideal, we have to show that, I is an additive subgroup of Z. And that for any r in Z, u in I, ru and ur are in Z. Proof: I is additive subgroup. Associativity is guaranteed by the fact that integers are associative. The fact that 0 = 0m => 0 is in Z. The additive identity Closure: am + bm = (a+b)m hence the set is closed. Inverse: if element is am, its inverse is (-a)m in I. Their sum is am + (-a)m = am -am = 0. hence done. to show that ru and ur in Z. u in I and r in Z, then u = mv for v in Z. Hence, ru = rmv = m(rv) in I - as its a multiple of m. for the other half - ur = mvr = m(vr) in I as its a multiple of m. Hence proved. Message me if you have any doubt.

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