2
$\begingroup$

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?

  • 2
    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
  • 0
    $f$ has to be irreducible, not just squarefree. Consider eg $(x^2+1)(x^2-3)$ over the integers.2011-06-28
  • 0
    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
  • 0
    Ah, now i get it, if f is not irreducible my stated isomorphy does not hold.2011-06-28
  • 0
    you should also have $deg(f)>0.$2011-06-28
  • 0
    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
  • 0
    @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
  • 0
    I can't see the reason for "we have $A[x]/(f) \cong A[b]$".2012-12-10

2 Answers 2