Abstract Algebra Proofs help

if I is an ideal in R and J is an ideal in the ring S, prove that I x J is an ideal in the ring R x S

Since both I and J are ideals, 1. they are closed under subtraction and 2. they absorb multiplication.
We need to check for these 2 properties in IxJ

Let's start.
RxS= {(x,y): x in R, y in S}

So we need to check two things: properties 1&2 mentioned above.

Let's check for property 1: closed under subtraction?
Pick any two elements in IxJ
They look like this: (x,y) and (i,j) ; where x,i are in I & y, j are in J
Need to check that (x,y)-(i,j) is in IxJ

(x,y)-(i,j)=(x-i, y-j).
Since I and J are both Ideals, they are closed under subtraction, so x-i is in I, and y-j is in J,
and so (x-i, y-j) is in IxJ
So yes, IxJ is closed under subtraction.

Now to check for property 2: does IxJ absorb multiplication?
pick any element in RxS, let's call it (a,b) (where of course a is in R, b is in S)
and any element in IxJ, let's call it (i,j)
Need to check that (a,b)*(i,j) is still in IxJ
(a,b)*(i,j)=(ai, bj);
ai we know is in I since I was an ideal in R, and bj we know is in S since J was an ideal in S
Therefore, (a,b)*(i,j) is in IxJ
IxJ absorbs multiplication!

Putting these two together, we get that IxJ is an ideal in the ring RxS