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## #1 2008-08-03 08:56:36

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Ideals

Last edited by JaneFairfax (2008-08-03 14:28:47)

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## #2 2008-08-03 10:50:34

JaneFairfax
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Registered: 2007-02-23
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### Re: Ideals

Last edited by JaneFairfax (2008-08-03 10:53:19)

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## #3 2008-08-03 13:44:27

Ricky
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### Re: Ideals

A ring R is a field if and only if it's only ideals are 0 and R.  In other words, fields aren't interesting in Ideal Theory.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #4 2008-08-03 14:05:25

JaneFairfax
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### Re: Ideals

Ricky wrote:

A ring R is a field if and only if its only ideals are 1 and R.

Last edited by JaneFairfax (2008-08-03 22:06:26)

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## #5 2008-08-03 14:31:38

JaneFairfax
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### Re: Ideals

Last edited by JaneFairfax (2008-08-03 14:58:57)

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## #6 2008-08-03 14:56:46

JaneFairfax
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### Re: Ideals

Last edited by JaneFairfax (2008-08-03 15:00:20)

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## #7 2008-08-03 15:04:07

Ricky
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### Re: Ideals

Let k be an algebraically closed field.  M is a maximal ideal in the polynomial ring k[x_1, x_2, ..., x_n] if and only if M = (x_1 - a_1, ..., x_n - a_n) for some a_1, ..., a_n in k.

More specifically, when solving systems of n-variable polynomial equations over the complex field, there exists a solution if the polynomials form a proper ideal in C[x_1, ..., x_n].

Last edited by Ricky (2008-08-03 15:14:01)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #8 2008-08-04 17:16:37

Ricky
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Registered: 2005-12-04
Posts: 3,791

### Re: Ideals

The proof of the theorem in post #4 is a lot more straightforward if you use a bit more of the machinery.

Just as in group theory, the lattice isomorphism theory holds for rings modulo ideals.  Specifically, there is an inclusion preserving bijection between subrings of R that contain I and the set of subrings of R/I.  Furthermore, A is an ideal containing I in R if and only if A/I is an ideal in R/I.  In other words, the lattice composed of subrings that contain I and the lattice of R/I "look" exactly the same.  The proof of this is rather tedious, but I'm told not difficult.

Now, armed with this fact and my theorem in post #3, the proof follows right away.  M is maximal if and only if there aren't any nontrivial ideals in R containing M, excluding M itself.  There aren't any nontrivial ideals in R containing M excluding M itself if and only if there aren't any nontrivial ideals in R/M.  Finally, there aren't any nontrivial ideals in R/M if and only if R/M is a field.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #9 2008-08-05 02:39:15

JaneFairfax
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## #10 2008-08-05 03:20:59

JaneFairfax
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## #11 2008-08-05 03:42:02

JaneFairfax
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