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Let $R$ be a unique factorization domain with quotient field $K$, let $K \subset L$ be a field extension, and let $\alpha$ be an element of $L$ that is algebraic over $K$. Consider the subring $R[\alpha]$ of $L$. Find an ideal $I$ of the polynomial ring $R[X]$ so that $R[\alpha] \cong R[X]/ I.$

My proof is given below. I never knowingly used that $R$ is a UFD. Have I done something that uses that, or is something wrong somewhere?


Let $f$ be the minimal polynomial of $\alpha$ over $K$, and consider $I = (f)$. Let $$\varphi\colon R[X]\to R[\alpha]$$ via $f(x)\mapsto f(\alpha).$ Then $\varphi$ is a homomorphism of rings: that it is an additive group homomorphism follows from the way addition of polynomials is defined. Moreover, given monomials $aX^n$, $bX^m$, we have $$\varphi(aX^nbX^m) = \varphi(abX^{n+m}) = ab\alpha^{n+m} = a\alpha^n b\alpha^m = \varphi(aX^n)\varphi(bX^m),$$ and clearly $1\mapsto 1$. It is also clear that $\varphi$ is surjective. Now, $$\ker\varphi = \{g\in R[X]\mid g(\alpha) = 0\} = \{g\in R[X]\mid f\,|\,g\}.$$ Hence $\ker\varphi = (f) = I.$ Now by the first isomorphism theorem, we have $R[X]/I\cong R[\alpha].$

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  • $\begingroup$ @user26857 Good point. So if $f$ is irreducible in $K[X]$ with $f\notin R[x],$ then we can multiply $f$ by the $a$ = lcm of the denominators of the coefficients to get $af\in R[X]$. After factoring out the content of $af$, we find $f_1\in R[X]$ which is primitive in $R[X]$ and irreducible in $K[X].$ Since $R$ is a UFD, we can say that $f_1$ is irreducible in $R[X].$ Thus w.l.o.g., we may assume $f = f_1$. Yeah? $\endgroup$ – user346096 Aug 4 '16 at 20:01
  • $\begingroup$ Actually $f\in R[X]$ :) $\endgroup$ – user26857 Aug 4 '16 at 20:09
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The assumption that $R$ is a UFD is not needed: if $R \subseteq L$ is any inclusion of rings and $\alpha \in $L, there is a (unique) homomorphism $h$ from the polynomial ring $R[X]$ onto the subring $R[\alpha]$ of $L$ generated over $R$ by $\alpha$, such that $h(r_0 + r_1 X + \ldots r_n X^n) = r_0 + r_1 \alpha + \ldots r_n \alpha^n$, for any $r_0, \ldots, r_n \in R$. If you put $I = \ker h$, then $R[\alpha] \cong R[X]/I$ by the first isomorphism theorem.

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