I have to show J(I) is an ideal of R given that J(I) is the intersection of all maximal ideals of R. J is a subset of R containing I. I is an ideal of R.
So far I have, Let M,N be maximals ideals of R, then M intersected with N is in J(I). If m is an element of the intersection of M and N, then m is in N. Similarly, if n is an element of the intersection of M and N, then n is in M.
Consequently, with a and b elements of R, we have Ma+Nb an element of J(I) for all m in M and n in N. Hence, we can do similar process for all maximals ideals and by the above reasoning, J(I) is closed under addition.
I know I have to prove there's an identity and an inverse in J(I) which I am having trouble with.
Lastly, show that J(I) is an actual ideal of R.
I am still grasping the ideal of ideals, so I am not sure. Any suggestions and hints would be much appreciated.
You can do it by definition.
To see it's closed under addition, you only need to take $a,b\in J(I)$, then $a,b$ are both in every maximal ideal of $R$ containing $I$, and in each one you already have that $a+b$ is also in them. (1)
For the identity, let's take the ring $(R,+,\cdot)$, and let's call the identity of $R$, $0$ and the additive inverse of $x$, $-x$.
Then since a ring is just an abelian group with extra things, you need to prove that the intersection of these ideals is a subgroup of $(R,+)$ first.
So it's not enough to prove that there is an identity, you must prove that the identity of $J(I)$ is $0$. But this makes things easy too, so now you know the candidate of identity. Similarly, you must prove that for every $x\in J(I)$, the element $-x$ is also in $J(I)$. So it's a subgroup.
Then you must take $r\in R$, and $j\in J(I)$, and see that $rj,jr\in J(I)$. Since $j$ is in every maximal ideal containing $I$, you can repeat the argument in 1 to get the result.
You can see that you don't need to use the fact that $J(I)$ is an intersection of maximal ideals. So in this way you can also prove that any intersection of ideals is an ideal. But this is a special ideal, if $R$ is commutative it's called the Jacobson Radical of $I$.