a) ( I,+) is an additive subgroup (show closure, identity, and inverses) b) Given s S, a I then as I and sa I. Solution A left ideal of a ring S is a subset I of S, such that I is an additive subgroup, and is closed under left multiplication by elements of the ring S, i.e, for any s S, a I, sa I. SImilarly a right ideal of a ring S is a subset I of S, such that I is an additive subgroup, and is closed under right multiplication by elements of the ring S, i.e, for any s S, a I, as I. Thus, the definition given by your teacher means that a subset of a ring is called an ideal if it is both a left ideal and a right ideal. Nevertheless, we need to consider intersection of two left ideals, and that of two right ideals. Firstly, let I and J be two left ideals of the ring S. Let their intersection be denoted by P.Note that both I and J contain the additive identitiy 0, so that 0 is in P, i.e., P is non-empty. Now, if there are two elements a and b in P, then by definiton, a and b are in both I and J. In particular, since I and J are closed under addition, (a+b) is also in both I and J, i.e, it is in P. Hence P is closed under addition. Also, as mentioned above, P contains the additive identity 0. Also, if a is in P, then a is in both I and J, which are additive subgroups, so that I and J both contain the additive inverse (-a) also, so that (-a) is also in P, whenever a is in P. Thus P also contains additive inverse of all of its elements. Thus, we have shown that P is closed under addition, contains the additive identity, and also contains the additive inverse of all of its elements. Hence, P is an additive subgroup. Also, for any element a of P, a must be in both I and J, which are left ideals, so that the product (sa) is in both I and J for any element s of S. Hence, (sa) is in P for any element s of S(as it is in both I and J). THus we have shown that P is an additive subgroup of S, which is closed under left multiplication by elements of S. Hence, by definition, it is a left ideal, as desired. The proof for intersection of right ideals is essentially the same as above, with left multiplication replaced by right multiplication at each place. So, I have omitted proof for that. Let me know if ou have any doubts, though. .

1. a) ( I,+) is an additive subgroup (show closure, identity, and inverses)
b) Given s S, a I then as I and sa I.
Solution
A left ideal of a ring S is a subset I of S, such that I is an additive subgroup, and is closed under
left multiplication by elements of the ring S, i.e, for any s S, a I, sa I.
SImilarly a right ideal of a ring S is a subset I of S, such that I is an additive subgroup, and is
closed under right multiplication by elements of the ring S, i.e, for any s S, a I, as I.
Thus, the definition given by your teacher means that a subset of a ring is called an ideal if it is
both a left ideal and a right ideal. Nevertheless, we need to consider intersection of two left
ideals, and that of two right ideals.
Firstly, let I and J be two left ideals of the ring S. Let their intersection be denoted by P.Note that
both I and J contain the additive identitiy 0, so that 0 is in P, i.e., P is non-empty. Now, if there
are two elements a and b in P, then by definiton, a and b are in both I and J. In particular, since I
and J are closed under addition, (a+b) is also in both I and J, i.e, it is in P. Hence P is closed
under addition. Also, as mentioned above, P contains the additive identity 0. Also, if a is in P,
then a is in both I and J, which are additive subgroups, so that I and J both contain the additive
inverse (-a) also, so that (-a) is also in P, whenever a is in P. Thus P also contains additive
inverse of all of its elements. Thus, we have shown that P is closed under addition, contains the
additive identity, and also contains the additive inverse of all of its elements. Hence, P is an
additive subgroup.
Also, for any element a of P, a must be in both I and J, which are left ideals, so that the product
(sa) is in both I and J for any element s of S. Hence, (sa) is in P for any element s of S(as it is in
both I and J).
THus we have shown that P is an additive subgroup of S, which is closed under left
multiplication by elements of S. Hence, by definition, it is a left ideal, as desired.

2. The proof for intersection of right ideals is essentially the same as above, with left multiplication
replaced by right multiplication at each place. So, I have omitted proof for that. Let me know if
ou have any doubts, though.