Find all prime ideals and maximal ideals of $\displaystyle \mathbb{Z}_2 \ \times \ \mathbb{Z}_4$.
start looking at the (additive) subgroups, first. the first thing to look for is cyclic subgroups of order 4: these are generated by elements of order 4. we have:
<(0,1)> = <(0,3)> = {(0,0),(0,1),(0,2),(0,3)} = {0} x Z_{4}
<(1,1)> = <(1,3)> = {(0,0),(1,1),(0,2),(1,3)}
next, we have non-cyclic subgroups of order 4, generated by any two elements of order 2:
<(1,0),(0,2)> = {(0,0),(1,0),(0,2),(1,2)} since we have only 3 elements of order 2, this is the only non-cyclic subgroup of order 4.
finally, we have 3 subgroups of order 2:
<(1,0)> = {(0,0),(1,0)}, <(0,2)> = {(0,0),(0,2)} and <(1,2)> = {(0,0),(1,2)}, and the trivial subgroup {(0,0)}.
if our multiplication is (a,b)*(a',b') = (aa',bb'), it's clear we don't have an integral domain, so {(0,0)} is neither a prime nor a maximal ideal (it is an ideal however).
it's also clear that the entire ring is an ideal, but by definition is not maximal nor prime.
so that leaves us with 6 additive subgroups to check for "multiplicative absorption".
now it's clear that (a,b)(0,k) = (0,bk), so {0} x Z_{4} is an ideal.
since there's no additive subgroups between this and the entire ring, there can be no ideals, either. so this is a maximal ideal.
however, consider <(1,1)>. note that (0,1)*(1,1) = (0,1), which is NOT in <(1,1)>, in fact, (1,1) is the identity of our ring. so any ideal containing (1,1) is the entire ring, so <(1,1)> (the additive subgroup) is NOT an ideal.
<(1,0),(0,2)> is a bit more interesting:
suppose (c,d) is an element of (additive) order 2 in our ring. then (a,b)*(c,d) + (a,b)*(c,d) = (ac,bd) + (ac,bd) = (ac+ac,bd+bd) = (a(c+c),b(d+d)) = (a(0),b(0)) = (0,0), so that (a,b)*(c,d) is again, either an element of additive order 2, or (0,0). since this subgroup contains ALL elements of order 2, we conclude that this, too, is an ideal, and since it has 4 elements, is maximal.
the proof that <(1,0)> = Z_{2} x {0} is an ideal is similar to the one that {0} x Z_{4} is an ideal, and so is the one that <(0,2)> is an ideal. for <(1,2)>, note that (0,1)*(1,2) = (0,2), so this subgroup is again NOT an ideal.
so we have 4 non-trivial proper ideals, 2 of which are maximal. which of these are prime ideals?
well, for {0} x Z_{4}, if (a,b)*(c,d) = (0,k), we have ac = 0 (mod 2), which can only happen if either a or c = 0 (since 1*1 = 1 (mod 2)). so this is a prime ideal.
i leave it to you to decide if J = {(0,0),(1,0),(0,2),(1,2)} is prime. you might want to decide if (Z_{2} x Z_{4})/J is an integral domain, rather than do a case-by-case analysis.
so let's look at <(1,0)>: we have (0,2)*(0,2) = (0,0), which is in <(1,0)>, but neither factor is, so this ideal is not prime.
similarly, for <(0,2)> we have (1,2)*(0,1) = (0,2), which is in <(0,2)>, but neither factor is, so this ideal is not prime, either.