If $\phi : R \rightarrow R'$ is a homomorphism, is it true that $(\phi^{-1}(a))=\phi^{-1}((a))$? Thanks

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    $\begingroup$ $\phi^{-1}(a)$ is not a single element in general. $\endgroup$ – lhf Oct 2 '18 at 12:22
  • $\begingroup$ adding to @lhf and if it is not the claim is false. $\phi:R^2\rightarrow R$ with $\phi(x,y) = x$, choose any non-trivial $R$ you want. $\endgroup$ – Yanko Oct 2 '18 at 12:23

No, e.g. for $(x)\subset \mathbb{Z/2 Z}[x]$ and $\varphi:\mathbb{Z}[x]\to\mathbb{Z/2 Z}[x]$ you get $\varphi^{-1}((x))=(2,x)$.


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