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**JaneFairfax****Member**- Registered: 2007-02-23
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*Last edited by JaneFairfax (2008-08-03 14:28:47)*

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**JaneFairfax****Member**- Registered: 2007-02-23
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*Last edited by JaneFairfax (2008-08-03 10:53:19)*

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

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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**JaneFairfax****Member**- Registered: 2007-02-23
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Ricky wrote:

A ring R is a field if and only if

itsonly ideals are 1 and R.

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

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**JaneFairfax****Member**- Registered: 2007-02-23
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*Last edited by JaneFairfax (2008-08-03 14:58:57)*

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**JaneFairfax****Member**- Registered: 2007-02-23
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*Last edited by JaneFairfax (2008-08-03 15:00:20)*

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**Ricky****Moderator**- Registered: 2005-12-04
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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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**Ricky****Moderator**- Registered: 2005-12-04
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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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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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

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

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