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I know that since $x^{101}+2$ and $x^{501}-2$ are both irreducible via Eisenstein, we have that $\mathbb{Q}[x]/(x^{101}+2)$ and $\mathbb{Q}[x]/(x^{501}-2)$ are actually fields. Since the kernel of any ring map is an ideal, any possible ring map will either be injective or trivial. So we are really looking for embeddings of $\mathbb{Q}(\alpha)$ into $\mathbb{Q}(\beta)$ where $\alpha,\beta$ are roots of $x^{101}+2$ and $x^{501}-2$ respectively. Any embedding will be a degree $101$ subextension of $\mathbb{Q}(\beta)$. By the Degree Formula, $$ 501=[\mathbb{Q}(\beta):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}(\alpha)]\cdot101, $$ which implies that $101$ divides $501$, a contradiction. Thus the only possible ring map is the trivial map. Is this solution correct?

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    $\begingroup$ Your reasoning is correct. As a loyal disciple of Jacobson's Basic Algebra I belong to the school whose creed includes that rings have 1, and that is preserved by ring homomorphisms. Therefore my answer would be that there are no ring homomorphisms between these two rings. $\endgroup$ Apr 11, 2020 at 20:50
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    $\begingroup$ That looks fine to me. (Aside to Matt: in the degree formula $[K: k]$ is the dimension of $K$ as $k$-vector space - no Galois groups are involved.) $\endgroup$
    – Rob Arthan
    Apr 11, 2020 at 20:53
  • $\begingroup$ @Sunyata Which part of the argument are you not sure about? $\endgroup$ Apr 12, 2020 at 16:16
  • $\begingroup$ The solution has been verified to be correct. $\endgroup$
    – Michael
    Apr 13, 2020 at 0:58

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