Let I and J be ideals of a ring R
Prove that I + J is the smallest ideal of R containing both I and J
Can someone please help with this problem? Would appreciate some help.
Peter
typically, I + J = {x + y in R : x is in I, y is in J} is given as a definition.
this clearly contains I and J, as the sums x + 0, and 0 + y, for any x in I, y in J.
so, suppose K is an ideal containing I and J. we need to show that I + J is contained in K.
so suppose r is any element of I + J. then r = x + y, for some x in I, and y in J.
since x is in I, and K contains I, x is in K. since y is in J, and K contains J, y is also in K.
since K is an ideal, it is closed under addition, so x + y is likewise in K. thus K contains I + J.