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In field theory, an important concept is being able to extend a field $k$ by the root of an irreducible polynomial $f$, for instance by considering the field extension $k \subseteq k[x] / (f)$. This extension is unique up to isomorphism, in the sense that if $\alpha$ is a root of $f$ in a larger field $F$, then $k(\alpha) \cong k[x] / (f)$.

If I'm not mistaken, we can do a similar thing for domains. If $R$ is a domain and $f$ is a nonconstant irreducible polynomial of $R$, then $S = R[x] / (f)$ is a domain such that 1) $R$ is a subring of $S$ and 2) $f \in S[x]$ has a root in $S$. For example, if $R = \mathbb{Z}$ and $f(x) = 2x - 1$, then $S$ is the smallest subring of $\mathbb{Q}$ that contains both $\mathbb{Z}$ and $1 / 2$.

Does the uniqueness still hold in this case? Namely, if $R \subseteq T$ and $f$ has a root $\alpha \in T$, is it the case that $S \cong R[\alpha]$ (where $R[\alpha]$ is the image of $R[x]$ under the evaluation map)? If not, how should I interpret notation such as $\mathbb{Z}[i]$, $\mathbb{Z}[1 / 2]$, and $\mathbb{C}[x, x^{-1}]$? Is it perhaps true that uniqueness holds in these special cases?

Frank
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  • It holds if $R$ is a UFD, as in your examples. Because in that case, the kernel of the morphism $R[x]\to T,;P(x)\mapsto P(\alpha)$ will be equal to $(f)$ and not only contain $(f).$ – Anne Bauval Nov 20 '22 at 09:00
  • @Frank A word of warning: the mere assumption that $f$ is nonconstant irreducible in $R[X]$ does not ensure that the quotient $R[X]/(f)$ is necessarily a domain. The claim that the quotient be a (non-zero) domain is equivalent to claiming that the ideal w. rsp. to which you take the quotient be prime which is furthermore equivalent - in this particular case of a principal ideal - to claiming that $f$ itself be a prime element of $R[X]$. In general there is no guaranty that irreducible elements are prime (while the quasi-converse is always true, non-zero primes are necessarily irreducible). – ΑΘΩ Nov 20 '22 at 09:03
  • @ΑΘΩ You are completely correct; I missed this detail. – Frank Nov 20 '22 at 20:39

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All you need is that Kernel of the evaluation map be equal to $<f>$. i.e., $g(\alpha) = 0$ must imply $f(x) \ | \ g(x)$. For this to work we need $R[x]$ be a Unique Factorization Domain (UFD).

Please see: Does $A$ a UFD imply that $A[T]$ is also a UFD?

So i think it works if $R$ is a UFD.

Balaji sb
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  • I agree: this was my first comment above. But I didn't type it as an answer because the less obvious part of the question was to find a counterexample in the general case. – Anne Bauval Nov 20 '22 at 11:53
  • Is the proof of this fact given by working in the field of fractions? That is, $f$ would still be irreducible in the $K[x]$ (where $K$ is the field of fractions of $R$). Then $f$ would then be the minimal polynomial of $\alpha$, so $f \mid g$ in $K[x]$, and by Gauss's lemma $f \mid g$ in $R[x]$. Is there a more elementary proof that you had in mind? – Frank Nov 20 '22 at 20:48
  • @Frank What u said is the fact i had in mind. – Balaji sb Nov 21 '22 at 02:31