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Let $f: R_1 \rightarrow R_2$ be a ring homomorphism. Does $r \in R_1$ being non-invertible mean $f(r)$ is non-invertible?

This seems incorrect, but I find it difficult finding counter-example...

Thanks for any assistance!

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Hint: Consider the inclusion $\Bbb Z \to \Bbb Q$.

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    $\begingroup$ Both familiar and dramatic at the same time :) $\endgroup$
    – rschwieb
    Jun 5 '14 at 1:49
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Let's take a direct product and use a forgetful homomorphism.

Consider $f : \mathbb{R}\times (\mathbb{Z}/6\mathbb{Z}) \longrightarrow \mathbb{R}$, defined as $f: (a,b) \mapsto a$ (which is why we call it forgetful - it just forgets part of the structure).

Clearly, $\mathbb{R}$ is a field, so every element in $\mathbb{R}$ is invertible. But $(*,2)$ is not invertible in $\mathbb{R}\times \mathbb{Z}/6\mathbb{Z}$. In fact, every element in $\mathbb{R}$ is the image of a noninvertible element.

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Hint: Start with the ring $\mathbb{Q}[x]$. of polynomials with rational coefficients. The polynomial $x+1$ is not invertible. You can find a nice homomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}$ that sends $x+1$ to $1$.

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