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Let $R$ be an integral domain with field of fractions $K$, and let $L/K$ be a field extension. Suppose that $\alpha\in L$ is an element, algebraic over $K$, whose minimal polynomial $f\in K[T]$ has coefficients in $R$. Show that there is an isomorphism of $R$-algebras $$R[T]/(f)\to R[\alpha]$$ sending $T$ to $\alpha$

If I consider $\phi:R[T]\to R[\alpha]$ with $\phi(g(T))=g(\alpha)$, I get $(f)\subseteq\ker\phi$, but $\ker\phi\neq R[T]$ because any constant polynomial $g(T)=c, c\neq0$ is not mapped to zero. For the other direction, it seems, I have to make use of the fact that $f$ has coefficients in $R$ and $R$ is a domain, or am I wrong?

inequal
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  • We had this a few days ago: http://math.stackexchange.com/questions/1163874/minimal-polynom-in-a-integral-domain/1163941#1163941

    Division with remainder works in an integral domain, if the leading coefficient is a unit. Of course, one should point out that the monic minimal polynomial over $K$ has coefficients in $R$. (You can always clear denominators, but the polynomial is not monic after that)

     – MooS Feb 28 '15 at 11:37

1 Answers1

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For the sake of completion, let me show (details will be omitted) that division with remainder works in an integral domain if the leading coefficient is a unit (Assume it to be $1$ for the sake of notation).

Let $f=x^n+f_{n-1}x^{n-1} + \dotsb + f_0$ and $g \in R[x]$. Division with remainder works in the quotient field $K$ of $R$, hence we get $g = hf+r$ with $h,r \in K[x]$ and $r$ has smaller degree than $f$. Let us show $h,r \in R[x]$.

Let $h=h_mx^m + \dotsb h_0$. The leading coefficient of $hf+r$ is $h_m$, hence $h_m \in R$. The next coefficient is $h_mf_{n-1}+h_{m-1}$, so we get $h_{m-1} \in R$, since $h_mf_{m-1} \in R$.

We proceed by downward induction, at some step we reach the coefficient of $x^n$ (The coefficients of $r$ have not played a role until here, since the degree of $r$ is at most $n-1$) and deduce that $h_0 \in R$, hence $h \in R[x]$. Then we are done due to $r = g-hf \in R[x]$

MooS
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