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I was checking the 2nd answer of this How to deal with polynomial quotient rings. There is an isomorphism $F_2[X]/(X+1)^4 \cong F_2[Y]/Y^4$. How to explicitly construct this isomorphism? Could this be generalized to fields like rationals or reals?

user26857
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scsnm
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  • Have you tried verifying that $(X+1) \mapsto Y$ gives a homomorphism over $\mathbb{F}_{2}$? Does this help you see whether this will generalize to other fields? – xxxxxxxxx Dec 13 '20 at 15:35
  • $[X]\mapsto[Y-1]?$ – J. W. Tanner Dec 13 '20 at 15:35
  • @ Morgan Rodgers. Thanks. I seem to have seen that it could be generalized because any element of $F_2[x]$ can be expressed as a polynomial with variable $x+1$ by long division. And it does generalize, if I am correct? – scsnm Dec 13 '20 at 15:59

1 Answers1

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I seem to have seen that it could be generalized because any element of $F_2[x]$ can be expressed as a polynomial with variable $x+1$ by long division. And it does generalize, if I am correct?

The shift map $\,s(x)= x+c\,$ is a ring hom by the universal mapping property of polynomial rings (it implies there is a unique ring hom $\,s\,$ with $\,s(x) = x+c).\,$ $\,s\,$ is invertible $\,(s^{-1}(x) = x-c)\,$ thus it follows that $\,s\,$ is a ring automorphism. Ditto for $\,s(x) = u\:\! x + c\,$ if $\,u\,$ is a unit (invertible).

That the automorphism extends to the stated isomorphism of quotient rings follows by

Lemma $ $ If $\ s: R\to S$ is an injective ring homomorphism then $\,R/I\,\cong\, s(R)/s(I),$

Proof $\ $ See e.g. this answer.

Bill Dubuque
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