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?
kernel maximal ideal
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ring-theory
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6It 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 $(X)$) – 2011-03-22
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2@Tobias: Dear Tobias, You should make this an answer! Regards, – 2011-03-22
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0@jooiho9 - I think you wanted to say that it is also given that the kernel is a prime ideal. – 2011-03-22
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2@Pandora: The kernel will automatically be a prime ideal since the image of the homomorphism is a domain. – 2011-03-22
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0@Tobias - Yes, my mistake. – 2011-03-22