This definition seems backwards, relative to the definition of a prime element, but it's not. An ideal in the integers is prime iff it consists of the multiples of a prime number. Using numbers, p divides ab means p divides a or p divides b. But containment reverses things. The multiples of ab are contained in the multiples of p, and that means either the multiples of a are contained in the multiples of p, or the multiples of b are contained in the multiples of p.
If P contains a finite product of ideals it contains one of them. Use induction on the number of ideals in the product.
There is another equivalent definition of prime ideal, which is often easier to verify or refute. We will show that "xRy in P implies x is in P or y is in P" is equivalent to "P is prime".
Assume P is prime and A and B are principal, generated by x and y respectively. If xwy is in P for every w in R, i.e. xRy is in P, then P contains RxRyR, and all finite sums thereof, P contains the product ideal A*B, P contains either A or B, and P contains either x or y. This holds for every x and y in the ring R. Thus a prime ideal satisfies the xRy criterion.
Conversely, assume xRy in P implies x is in P or y is in P, for every x and y in R. Let A and B be two left ideals with A*B in P. If A is in P we are done, at least for left ideals. Assume A is not entirely in P, and let x be an element in A-P. Select y in B. For every w in R, xwy lies in P. This because A*B lies in P. Now apply the xRy criterion, and either x or y lies in P. Yet x does not lie in P, so y must. This holds for all y in B, hence B lies in P.
A similar argument can be made for right ideals.
Since ideals are also left ideals, we are done. The xRy test is necessary and sufficient for P to be prime.
If R is commutative the test simplifies to xy in P implies x is in P or y is in P. Use the xy test to show that the principal ideal generated by p in a commutative ring is prime iff p is a prime element in the ring.
Nothing in the above requires R to contain 1. (In the case of the previous paragraph, an element is not prime if it generates all of R, thus {p} must be proper.) So any ring, with or without 1, noncommutative or commutative, may contain prime ideals, and the xRy (or xy) test applies.
Let's consider another ring, the n by n matrices over any domain, such as the integers. Let x and y be any two nonzero matrices. Think of x as a collection of vectors, each column a vector. Multiplication by a matrix w produces a new set of column vectors, where each column in the product xw is a prescribed linear combination of the vectors in x. All linear combinations are possible; it depends on the entries in w. Multiplication by y does the same thing. Let y use the jth column of xw to build one of its linear combinations. It may scale this column by the element t. We will make sure the jth column of xw, and only the jth column of xw, is nonzero. Construct w so that the jth linear combination copies a nonzero column from x, and ignores the other columns in x. Thus w contains one 1, and the rest of the matrix is 0. Now one of the columns in xwy is a scale multiple of one of the columns of x, where t is the scaling factor. Since this is a domain, the scaled column is nonzero. Therefore xwy is nonzero.
For every x and y, there is some matrix w with xwy nonzero. We don't have to worry about xRy = 0; it never happens. Thus the zero matrix is a prime ideal.
Now place 1 in the upper left of the zero matrix and let that generate the left ideal A. Place 1 in the lower right of the zero matrix and let that generate the right ideal B. Now A*B always = 0, and is contained in the zero prime ideal, even though A and B are not.
This reasoning holds for left ideals or right ideals.
If a maximal ideal, or maximal left ideal, does not include all of Ri and Rj, increase it so that Hi is all of Ri. Since Hj is still a proper subset of Rj, the resulting ideal remains proper. Therefore a maximal ideal has Hi = Ri in every component except 1, and in this component Hi is maximal.
Consider the product of two ideals G and H. Within the ith component, the resulting ideal is GiHi. All linear combinations must be present, hence gh becomes the direct sum over GiHi. In other words, product and direct sum commute.
Let P be a prime ideal in R, where Pi and Pj are proper ideals of Ri and Rj respectively. Now PiRj times RiPj lies in P, while neither factor lies in P. This is a contradiction, hence only one component, Pi, can be a proper ideal in its component ring. Clearly this component must be prime, hence we have characterized the prime ideals of R.
An infinite direct sum may leave you feeling a bit uneasy, since R does not contain 1, and rings (on this website) usually contain 1. Still - it's a perfectly good ring - and we've got a handle on its prime and maximal ideals, assuming each Ri contains 1.
If R is the infinite direct product of component rings, it contains 1, but there are ideals, even maximal ideals, that we haven't characterized. The direct sum of rings is now a proper ideal in R, and it ratchets up to a maximal ideal in R whose projection onto any Ri is all of Ri. It's hard to know what such a maximal ideal might look like; we only know it exists, thanks to the axiom of choice.