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Given a ring $A$ and a polynomial $p\in A[x]$, write $Z(p,A)$ for the simple roots in $A$ of $p\in A[x]$. On the other hand, consider the set $\mathrm{Idemp}(A[x]/(p))$ of idempotents of the quotient.

What's the relation between $Z(p,A),\mathrm{Idemp}(A[x]/(p))$? (Perhaps assuming $p$ is monic?) Are the simple roots perhaps in bijection with the connected components of $(A[x]/(p))$?

Given a simple root $\alpha\in A$ of $p$ I thought to look at $(x-\alpha)\in A[x]$ and use the Chinese remainder theorem...

If there isn't a "pointwise" relation, perhaps the following properties of the quotient $A\twoheadrightarrow A/I$ are equivalent?

  1. For each (monic?) $p\in A[x]$ the set-function $Z(p,A){\longrightarrow} Z(\overline p,\frac AI)$ is surjective.
  2. For each (monic?) $p\in A[x]$ the boolean algebra morphism $\mathrm{idemp}(A[x]/(p))\to \mathrm{idemp}(\frac AI[x]/(\overline p))$ is surjective.
  3. For each (monic?) $p\in A[x]$ the analogous set function between connected components is surjective.

Update. As answered by Mohan, the simple roots are certainly not in bijection with connected components. The question remains about the equivalence of (1) and (2) above.

Arrow
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  • If $p=x-a$ I can't see any connection. Do you? – user26857 Dec 27 '18 at 16:37
  • Dear @user26857, I certainly don't see anything you don't. I ask because here there is a characterization of Henselian local rings via induced bijection on idempotents, while in example 6.1 here it's claimed that being Henselian is also equivalent to inducing a bijection between simple roots. – Arrow Dec 27 '18 at 16:54
  • Example 6.2.... – user26857 Dec 27 '18 at 19:23
  • @user26857 ah yes, sorry! – Arrow Dec 27 '18 at 19:29
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    Dear @user26857, it seems Lemma 10.1.9 in the book Separable Algebras by Ford proves (rather indirectly) that condition (2) in my question implies condition (1). (I think lifts of simple roots in the residue field are simple in the local ring.) I have no idea about the converse. – Arrow Dec 27 '18 at 20:17

1 Answers1

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If $f(x),g(x)\in\mathbb{Q}[x]$ are irreducible and coprime of degree at least 2 and $p=fg$, then $p$ has no roots in $\mathbb{Q}$, but the quotient $\mathbb{Q}[x]/p(x)$ is not connected and has two connected components.

Mohan
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  • Dear Mohan, thank you for your answer. I think I have read conditions that 1,2 are equivalent (also to $(A, I)$ being a Henselian pair). What do you think? – Arrow Dec 27 '18 at 15:33
  • Dear Mohan, I linked to the assertions which I'd like to understand in this comment. – Arrow Dec 27 '18 at 17:01