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I am trying to solve the following question.

Let $R$ be a principal ideal domain (PID). Suppose that $2$ is a unit in $R$. Let $c_1, ..., c_t$ be irreducible elements of $R$ and let $c=c_1 ... c_t$. Show that the ring $R[x]/(x^2-c)$ is a Dedekind domain.

A Dedekind domain is a Noetherian ring of dimension $1$, which is integrally closed.

So far I have that $R$ is a PID $\Rightarrow$ R is a Dedekind domain, so in particular $\dim(R)=1$.

Also $R$ Noetherian $\Rightarrow$ $\dim(R[x]) = 2$ and $R[x]$ Noetherian.

$R[x]$ Noetherian $\Rightarrow$ $R[x]/(x^2-c)$ Noetherian.

I think I'm supposed to somehow use the fact that we are taking the quotient by an irreducible polynomial to show that the dimension is reduced by $1$, but I'm not quite sure how. Also I'm not sure how to use the assumption that $2$ is a unit, nor how to go about showing that the resulting ring is integrally closed.

Robin
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Orlly
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    You meant $c$ is squarefree. Use Gauss lemma to say that if $a+bx\in Frac(R)[x]/(x^2-c),b\ne 0$ is integral over $R$ then its minimal polynomial $(T-a-bx)(T-a+bx)$ is in $R[x]$. You'll get that $2a\in R$, $a\in R,b^2c\in R$. Writing $b=u/v$ with $u,v\in R,(u,v)=(1)$ then $c/v^2\in R$ so that $v\in R^\times$ and $b\in R$. – reuns Apr 24 '21 at 16:04

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