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$R$ is a ring which is a PID [i.e., $R$ is an integral domain in which every ideal is generated by a single element] and we are given with a map $f:R\to S$ which is a homomorphism, i.e.

$f(a + b) = f(a) + f(b)$ for all a and b in $R$,

$f(ab) = f(a) f(b)$ for all a and b in R,

$f(1_R) = 1_S. $

Is it necessary that image of ring $R$, i.e. $f(R)$, also PID?

rschwieb
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Pitt HarmanN - FreshmaN
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    My intuitions say that it is an PID. If i am right then we need to prove that $f(D)$ is an integral domain and its every ideal is generated by a single element. – Pitt HarmanN - FreshmaN Nov 20 '13 at 06:04
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    If it is so then we need to show that the product of any two non zero elements in $f(R)$ is nonzero. [Since integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero.] – Pitt HarmanN - FreshmaN Nov 20 '13 at 06:11
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    Lets suppose that we have two non zero elements, $f(r)$ and $f(s)$ lets take $f(r).f(s)= 0$ then according to the property 2. $f(rs)=0$. Where $ r$ and $s \in R$. ! how will we proceed now? – Pitt HarmanN - FreshmaN Nov 20 '13 at 06:18
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    I think one to one is needed. Then it would mean it's isomorphic to it's image which would do it. – TheNumber23 Nov 20 '13 at 06:22

1 Answers1

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This is not the case in general. Note for example that $\Bbb Z$ is a PID. Consider quotient rings of $\Bbb Z$ to find homomorphic images of $\Bbb Z$ that are not PIDs (in fact, not even integral domains).

However, if we know that a ring homomorphism $f:R\to S$ is one-to-one, then of course it is true, since $R$ is then isomorphic to $f(R)$.

Cameron Buie
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  • Thank you Cameron. Do you mean that factor rings, like $Z/2Z$ is holomorphic image of $Z$ but $Z/2Z$ is not PID! – Pitt HarmanN - FreshmaN Nov 20 '13 at 06:45
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    Not quite, but you have the idea. It turns out that $\Bbb Z/2\Bbb Z$ is in fact a field, so a Euclidean domain, so a PID. However, there are many $n\in\Bbb Z$ for which $\Bbb Z/n\Bbb Z$ is not even an integral domain (for all others, it is either a field, or $\Bbb Z$ itself, or the zero ring). Do you know which $n$ are which? – Cameron Buie Nov 20 '13 at 06:49
  • Nope, which n are which Cameron? I think those n are prime no.'s right. It is just a guess. – Pitt HarmanN - FreshmaN Nov 20 '13 at 06:57
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    $\Bbb Z/n\Bbb Z$ is a field if and only if $n$ is prime (unless you consider the zero ring to be a field, in which case $n=\pm 1$ also make $\Bbb Z/n\Bbb Z$ a field). When $n=0,$ we have $\Bbb Z/n\Bbb Z\cong\Bbb Z,$ and if $n\in\Bbb Z$ is non-$0$, non-unit, and non-prime, then $\Bbb Z/n\Bbb Z$ is not even an integral domain. It's a good exercise to prove all these claims. – Cameron Buie Nov 20 '13 at 07:00
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    Now I gotcha. So in our case when n is non zero, non-unit and non prime then this image of $Z$ i.e. $Z/nZ$ is not even ID. Poor factor ring :) thank you so much dear Cameron. good night – Pitt HarmanN - FreshmaN Nov 20 '13 at 07:09
  • You are very welcome. – Cameron Buie Nov 20 '13 at 07:10