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# Proving Local Rings

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i) Show that a non-zero ring R is local (i.e. it has a unique maximal ideal) if and only if x or 1 - x is a unit, where x is a member of R.

ii) Show why or why not are the following four rings local: {a/b belongs to Q s.t. b is not divisible by a prime p}, Z[x]/(X^3), F[[x]] (where F[[x]] is the ring formed by the set of all formal power series p(x) = a_0 + (a_1)(x) + (a_2)(x^2) + ... where a_i belongs to the field F), {p(x)/q(x) belongs to Q(x) s.t. q(0) = 0}.

iii) For the previous rings that you decided are not local, slightly change their definition (without proof) such that you get a local ring.

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#### Solution Preview

See the attachments.
(i) Show that a non-zero ring R is local if and only if x or 1-x is a unit, for all x in R.

Let's first prove the following proposition.
Proposition 1. A commutative ring R with 1≠0 is local if and only if the non-units of R form an ideal.
Proof: (⟹) Suppose that R is a local ring with a unique maximal ideal M. Since M is a proper ideal of R, it contains no units of R. Now let x be an element not in M. If x is not a unit, then x generates an ideal (x) which is contained in a maximal ideal N≠M. But this is impossible since M is a unique maximal ideal. Therefore, x is a unit of R. Therefore, M contains all the non-units of R.
(⟸) Suppose that R is a ring in which all the non-units form an ideal M. Then every proper ideal of R is contained in M since proper ideals cannot contain units. Thus M is a unique maximal ideal of R and therefore R is a local ring.

From Proposition 1, we immediately get this corollary.
Corollary 1. Let R be a local ring with 1≠0. Then its unique maximal ideal consists precisely of the non-units of R.

We now return to the proof.
Proposition 2. A ...

#### Solution Summary

This solution provides the detailed proofs and examples of local rings in attached .docx and .pdf formats.

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