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Let $A$ be a domain, $f \in A[x]$ irreducible and $b$ a zero of $f$, so we have $A[x]/(f) \cong A[b]$. I want to show, under certain circumstances, that $b$ integral over $A$ implies that $f$ has an invertible leading coefficient.

I consider the following: $A = K[Y]$, the coordinate ring of an affine variety $Y$. If $A$ would be integrally closed, then we would know that $b$ had a monic minimal polynomial $p$ over $A$ and would be done ($f = u \cdot p$, $u$ unit in $A[x]$, i.e. constant).

But here, as a coordinate ring, $A$ in general will not be integrally closed. How do we proceed here?

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    The notation $A[b]$ doesn't seem to me to be well-defined if $f$ isn't squarefree. For example, if $f(x) = x^2$ then what is $A[b]$? Is it $A$ or is it $A[x]/x^2$?2011-06-28
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    $f$ has to be irreducible, not just squarefree. Consider eg $(x^2+1)(x^2-3)$ over the integers.2011-06-28
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    I do not fully understand your question. $A[b]$ means the ring adjunction of $b$ to $A$ which is given by $A[b] = \{g(b), g \in A[x]\}$ and the stated consequence is given by the homomorphism theorem, so my answer would be $A[x]/(x^2)$.2011-06-28
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    Ah, now i get it, if f is not irreducible my stated isomorphy does not hold.2011-06-28
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    you should also have $deg(f)>0.$2011-06-28
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    Isn't it by definition that an element is integral over a ring $A$ if it is the root of a *monic* polynomial with coefficients in $A$?2011-06-28
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    @user7475 Glad you "got it" that's great. I hope the responses to this question help you to understand some of the maths which has most inspired me!2011-06-28
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    I can't see the reason for "we have $A[x]/(f) \cong A[b]$".2012-12-10

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So, if $A$ is a domain with $b$ integral over $A$, and if $f(x)\in A[x]$ is irreducible with $b$ as a root, then is the leading coefficient of $f$ invertible in $A$?

In general, no. Let $A=\mathbb{Z}[\sqrt{-3}]=\{a+b\sqrt{-3}\,\vert\,a,b\in \mathbb{Z}\}$, and let $\omega=\frac{-1+\sqrt{-3}}{2}$ be a primitive third root of unity. Note that $\omega\notin A$ and $\omega$ is a root of the monic polynomial $g(x)=x^2+x+1\in A[x]$, thus $\omega$ is integral over $A$. Also, $g(x)$ is irreducible in $A[x]$, since the only roots of $g$ are $\omega$ and $\overline{\omega}:=\frac{-1-\sqrt{-3}}{2}$.

However, $f(x)=2x-2\omega = 2x+1-\sqrt{-3}\in A[x]$ is irreducible (and has $\omega$ as a root), but $2\notin U(A)=\{\pm 1\}$.

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Isn't it by definition that a element is integral over a ring $A$ if it is the root of a monic polynomial with coefficients in $A$?

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    If the domain $A$ isn't integrally closed then minimal polynomials of elements integral over $A$ need not be monic. In fact $A$ is integrally closed iff every $A$-integral element has a monic minimum polynomial $\in A[x]$.2011-06-28
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    Yes, lets call this monic polynomial $p$. How do i then follow that $f$ equals $u \cdot p$ for $u$ constant?2011-06-28
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    @lhf A question shouldn't be an answer. This should have been a comment.2011-06-28
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    @Bill, you're right. Shall we move it and its comments to comments? I'll wait until the comments are moved and then I'll delete the answer.2011-06-28
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    Ah i think it could be quite easy: let $p \in S[x]$ be the nomic polynomial with $p(\alpha) = 0$ with minimal degree. Now we can use euclidean division to derive that $p$ divides $f$ and so f = $p \cdot u$, $u$ unit, because $f$ is irreducible.2011-06-28
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    Would it also be possible like this: Choose $p$ as above (with minimal degree), then $S[\alpha] \cong S[x]/(f) \cong S[x]/(p) \Rightarrow (f) = (p)$ (is this valid?) and now we can follow again $f = p \cdot u$, $u$ unit, because we are inside a domain.2011-06-28
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    @use Did you see my comment above? When $A$ is not integrally closed there exist $A$-integral elements with no monic minimal polynomial.2011-06-28
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    Ah sorry, then i guess i am lost. Anyone has an idea?2011-06-28
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    I am completely confused: If $b$ (above i exchanged $b$ with $\alpha$ and $A$ with $S$ sorry) is integral over $A$ then there exists one monic polynomial $p$ with $p(b) = 0$. But then we can choose the monic polynomial that is minimal (in degree) under these monic polynomials $g$ with $g(b) = 0$, this would be the argument i tried above. Isnt this valid?2011-06-28
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    Ah, but then i would have to argue, that the remainder r(x) (in the euclidean division) is monic, which i dont see how... too bad2011-06-28