In my textbook they say that if $f: R \rightarrow S$ is a ring homomorphism, then:

if $I \subset R$ is an left ideal of $R$, then $f(I)$ is a left ideal of $S$.

However, I think that this is a mistake and that $f$ should be surjective for this to hold.

I have the following argument

Let $I$ be the identity map from $(\mathbb{Z},+,\cdot)$ to $(\mathbb{R},+,\cdot)$. It is known that $n\mathbb{Z}$ is an ideal from $(\mathbb{Z},+,\cdot)$. Thus, $I(n\mathbb{Z}) = n\mathbb{Z}$ should be an ideal of $(\mathbb{R},+, \cdot)$. However, since $\mathbb{R},+,\cdot$ is a field, it only has trivial left ideals, which contradicts that $n\mathbb{Z}$ is a left ideal, this because it is nontrivial.

So, my question is whether it is a mistake in my book or whether there is a hole in my argument?

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    $\begingroup$ "I think this is a mistake and that $f$ should be surjective." You are not asserting that $f(I)$ is a left ideal of $S$, only of $f(R)$ (the image of $f$), and $f$ a function is always onto its image. $\endgroup$ – Arturo Magidin Dec 8 '11 at 4:24
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    $\begingroup$ $I(\mathbb{Z}) \ne \mathbb{R} \dots$ $\endgroup$ – Nate Eldredge Dec 8 '11 at 4:25
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    $\begingroup$ @ArturoMagidin Oh, I copied it wrong from the book. It had to be S instead of F(R). Thanks for correcting that. Now my argument does hold right? $\endgroup$ – sxd Dec 8 '11 at 4:27
  • $\begingroup$ If the book says $S$, then you are right: you need $f$ to be onto. Make sure your book actually says $S$, and that it does not say "onto" somewhere. $\endgroup$ – Arturo Magidin Dec 8 '11 at 4:32
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    $\begingroup$ Where exactly in what book? $\endgroup$ – Jonas Meyer Dec 8 '11 at 4:35

Indeed the statement "$f(I)$ is an ideal of $f(R)$" [the statement that was in the original version of the question] is true. First let's think what it means for $I$ to be to be a (left) ideal of $R$

1) $I \subset R$

2) For all $x \in I,r \in R, r \cdot x \in I$

So what about $f(I)$? Well

1) $f(I) \subset f(R)$ should be clear

For 2), let $x' \in f(I) = f(x)$ and $r' \in f(R) = f(r)$. Then

$$r' \cdot x' = f(r) \cdot f(x) = f(r \cdot x) \in f(I),$$

since $r \cdot x \in I$, and hence $f(I)$ is a left ideal of $f(R)$

Edit: I see you've updated the question. Indeed for it to be an ideal is $S$, it would have to be surjective

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    $\begingroup$ About your edit: No, it wouldn't have to be, this is just sufficient. Consider the zero map $R\to S$ for any two rings $R,S$. $\endgroup$ – Alex Youcis Dec 8 '11 at 4:47
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    $\begingroup$ This answer is now confused, probably due to edits. It seems that "Indeed the statement is true" refers to "... then $f(I)$ i a left ideal of $S$" (without surjectivity hypothesis), which is false. However the answer itself seems to be correct. $\endgroup$ – Marc van Leeuwen Dec 8 '11 at 7:09
  • $\begingroup$ @Marc: I've edited in a clarification $\endgroup$ – Arturo Magidin Dec 8 '11 at 16:12
  • $\begingroup$ @Juan: I've edited in a correction to take the change in the original post into account. Please double check that this is fine with you. $\endgroup$ – Arturo Magidin Dec 8 '11 at 16:12
  • $\begingroup$ @Arturo - That is fine - thanks for that $\endgroup$ – Juan S Dec 9 '11 at 4:23

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