Ideal preliminaries
by math_explorer, Apr 12, 2015, 10:48 AM
Wait, I finally figured out the misunderstanding that confused me for a long time; I had my criteria for ideals and quotient rings mixed up. Darn. Also, it's so nice to be able to assume rings are commutative...
(again this is easy stuff you should know this already, self)
As you recall, an ideal
of a commutative ring 
is a subset of that is closed under addition (with another element of ) and under multiplication by any element of . Given an ideal and a ring, you can get a new ring — a quotient ring (also called a factor ring among other names) — by taking the quotient 
, which contains as elements sets of the form 
for 
. It is easy/boring to check that ring operations are well-defined, so I won't do that.
is defined to be prime if 
implies 
or 
. Some authors additionally require that be proper, i.e. not equal to itself.
→ is prime iff is an integral domain. This is a straightforward translation of the definition. The additional requirement depends on if you accept the single-element 
, in which 
serves as both the additive and the multiplicative identity, to be a ring.
is defined to be maximal if is proper and no ideal 
exists such that ![\[ I \subsetneq J \subsetneq R. \]](//latex.artofproblemsolving.com/a/d/5/ad5cbb0eea3fdadfe4cf354730038e0bb0cd0787.png)
→ is maximal iff is a field. This is more or less what we proved in the last post, once we note that any ideal of satisfying the above equation gives rise to an ideal 
in and vice-versa, which is one of those intuitive homomorphism theorems.
(again this is easy stuff you should know this already, self)
As you recall, an ideal
of a commutative ring 
is a subset of that is closed under addition (with another element of ) and under multiplication by any element of . Given an ideal and a ring, you can get a new ring — a quotient ring (also called a factor ring among other names) — by taking the quotient 
, which contains as elements sets of the form 
for 
. It is easy/boring to check that ring operations are well-defined, so I won't do that. is defined to be prime if 
implies 
or 
. Some authors additionally require that be proper, i.e. not equal to itself.→
is prime iff is an integral domain. This is a straightforward translation of the definition. The additional requirement depends on if you accept the single-element 
, in which 
serves as both the additive and the multiplicative identity, to be a ring. is defined to be maximal if is proper and no ideal 
exists such that ![\[ I \subsetneq J \subsetneq R. \]](http://latex.artofproblemsolving.com/a/d/5/ad5cbb0eea3fdadfe4cf354730038e0bb0cd0787.png)
→
is maximal iff is a field. This is more or less what we proved in the last post, once we note that any ideal of satisfying the above equation gives rise to an ideal 
in and vice-versa, which is one of those intuitive homomorphism theorems.
May 2020
January 2020