Pandora Charms proper principal ideals

18 Jan 15 - 21:35

Mathematics stack exchange

Mathematics stack exchange

It hold.For examples the map from $mathbbz[Times]$ to $mathbbz$ providing $f(Times)$ if you want to actually Cheap Pandora Bracelets Sale $f(0)$'.It's really a homomorphism and the kernel is not $0$, but the kernel is not a maximum ideal of $mathbbz[By]$ buck(It can $(By)$bucks).

I realized may well mean $r$ to be a pid(Since so it does See Results About Pandora Charms hold).

The reason it holds in such cases is that a principal ideal is a prime ideal if and only if it is maximal among Pandora Charms proper principal ideals, so inside pid, all non zero prime ideals are maximum(And since the of the homomorphism is a domain, the kernel ought to be a prime ideal).

The keRnel of a Ring homoRphism $fcolon Rto s$ is maximal if in suppoRt of if $f(R)$ is an easy ring(Has no recommended nontrivial two sided Hanging Charms with Gems ideals).In the event of a commutative ring with identity $1neq 0$, a ring is not hard if and only if it is a field.

Since the image have been an integral domain(Being a subring of an integral domain that has $1$), The kernel is definitely a prime ideal;If the morphism certainly not injective, then the kernel is consistently a proper nontrivial prime ideal.

Occasionally, it could imply maximality.As an example, if the ring $r$ has krull dimensions $1$(This includes pids aren't fields), Or if $R$ may possibly be Artin ring(Has the climbing down chain condition), Then the final outcome you want will follow.

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