Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have a homomorphism $\phi : R \rightarrow S$ between integral domains, how can I show that if the kernel is non-zero then it is a maximal ideal in R?

share|improve this question
6  
It does not hold. Take for instance the map from $\mathbb{Z}[X]$ to $\mathbb{Z}$ sending $f(X)$ to $f(0)$. This is a homomorphism and the kernel is not 0, but the kernel is not a maximal ideal of $\mathbb{Z}[X]$ (it is $(X)$) –  Tobias Kildetoft Mar 22 '11 at 11:25
2  
@Tobias: Dear Tobias, You should make this an answer! Regards, –  Matt E Mar 22 '11 at 11:44
    
@jooiho9 - I think you wanted to say that it is also given that the kernel is a prime ideal. –  Pandora Mar 22 '11 at 12:40
2  
@Pandora: The kernel will automatically be a prime ideal since the image of the homomorphism is a domain. –  Tobias Kildetoft Mar 22 '11 at 13:30
    
@Tobias - Yes, my mistake. –  Pandora Mar 22 '11 at 14:28
add comment

3 Answers 3

It does not hold. Take for instance the map from $\mathbb{Z}[X]$ to $\mathbb{Z}$ sending $f(X)$ to $f(0)$. This is a homomorphism and the kernel is not $0$, but the kernel is not a maximal ideal of $\mathbb{Z}[X]$ (it is $(X)$). I realized you might mean $R$ to be a PID (since in that case it does hold). The reason it holds in this case is that a principal ideal is a prime ideal if and only if it is maximal among proper principal ideals, so in a PID, all non-zero prime ideals are maximal (and since the image of the homomorphism is a domain, the kernel must be a prime ideal).

share|improve this answer
add comment

The kernel of a ring homorphism $f\colon R\to S$ is maximal if and only if $f(R)$ is a simple ring (has no proper nontrivial two-sided ideals). In the case of a commutative ring with identity $1\neq 0$, a ring is simple if and only if it is a field.

So a homomorphism $\phi\colon R\to S$ between integral domains (which necessarily have $1\neq 0$) which sends $1$ to $1$ has kernel equal to a maximal ideal if and only if the image is a field. This is not true in general (as Tobias's example shows), but may be the case. E.g., $f\colon\mathbb{Q}[x]\to\mathbb{Q}[x]$ given by $f(p(x)) = p(0)$ has image equal to $\mathbb{Q}$, so the kernel, $(x)$, is maximal.

Since the image will always be an integral domain (being a subring of an integral domain that contains $1$), the kernel is always a prime ideal; if the morphism is not injective, then the kernel is always a proper nontrivial prime ideal.

In some cases, this may imply maximality. For example, if the ring $R$ has Krull dimension $1$ (this includes PIDs that are not fields), or if $R$ is an Artin ring (has the descending chain condition), then the conclusion you want will follow.

share|improve this answer
add comment

This is true for a class $\:\mathfrak R\:$ of rings iff every ring in $\:\mathfrak R\:$ has dimension at most one, i.e. prime ideals are minimal or maximal. Equivalently, by factoring out by the (necessarily) prime kernel, it is true iff every domain in $\:\mathfrak R\:$ is a field. Examples of such one-dimensional classes of rings are Dedekind domains (e.g. PIDs), and Artinian rings (see this prior question).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.