I just turned in this homework and I want to know if I got it right. The proof is pretty simple, but I think I might be using a theorem in the wrong way.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

[tex]\{P_{i} : i \in \Lambda \}[/tex] is a family of prime ideals in a ring, R. Prove that [tex]R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex] is a multiplicative system.

2. Relevant theorems

- Let R be a ring. An element [tex]u \in R[/tex] is called a
unitif it has an inverse in R.- Let R be a ring. The union of all maximal ideals of R is the set of non-units in R.
- Any Maximal ideal is also a prime ideal .

3. The attempt at a solution

Any maximal ideal is also a prime ideal. Hence,

[tex]\cup_{M max} M \subseteq \cup_{i \in \Lambda} P_{i}[/tex].

Then every element in

[tex]R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex]

is a unit. (We have removed all non-units and, possibly, some units since there may be prime ideals that are not maximal.) The set of units is closed under multiplication because, if a and b are units, then c=ab has an inverse [tex]c^{-1}=b^{-1}a^{-1}[/tex].

[tex]1 \in R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex].

1 is not in any of the prime ideals, hence,

[tex]R \setminus \{ \cup_{i \in \Lambda} P_{i} \}[/tex]

is a multiplicative system.

So would this work? It bothers me that when we remove [tex] \cup_{i \in \Lambda} P_{i} [/tex] from R we are not just taking out the maximal ideals, but also the prime ideals that may not be maximal. Could this lead to some non-units remaining in the set?

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# Homework Help: Rings, ideals, prime and maximal

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