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Prove or disprove: The image of a ring homomorphism $\phi:R\to S$ is an ideal in $S$.

I only see examples where they use the image of an ideal, but I don't think this is the case for my question.

At first I tried using the ideal test, and I had a feeling that it wasn't working out the way it should, so now I am trying to find a counterexample. Any ideas for a simple counterexample?

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2 Answers 2

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It's false: Take $R=\mathbb{Z}$, $S=\mathbb{Q}$ and $\phi$ to be the inclusion.

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    $\begingroup$ Or more generally, $R$ any integral domain that is not a field, $S$ its field of fractions, and $\phi$ to be again the inclusion. $\endgroup$ Oct 29, 2018 at 4:48
  • $\begingroup$ Hello, may I ask what you mean by inclusion? $\endgroup$
    – user482939
    Oct 29, 2018 at 4:50
  • $\begingroup$ @numericalorange sure, take $n\in \mathbb{Z}$ to $n\in \mathbb{Q}$ $\endgroup$ Oct 29, 2018 at 5:00
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Another example I like : Take $R$ any ring and the polynomial ring $R[X]$. Now consider the morphism $f:R[X]\rightarrow R[X]$ which sends $X$ to $X^2$ (extend it).

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