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This problem has Hartshorne Height 2; problem I.1.6 can help solve this one. No big commutative algebra results to apply, just work through the topology.
a) Use exercise I.1.6 and show if are distinct and closed in an arbitrary subset then are distinct in X. In fact take a point of that is not in and show this point is still not in the closure.
b) The hard part is to show there is an i such that . The idea is to take a chain or closed irreducible subsets in X, and pick a intersecting . Now show the chain still has the same length as a chain in . Basically if this wasn't the case you can find two disjoint open subsets in an irreducible closed subset.
c) It can be done by constructing a simple two point topological space.
d) If then for every chain in Y, there is a strictly longer chain in X by taking X to be the last element of the chain.
e) Think: countably many finite dimensional noetherian topological spaces.