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One very convenient property of $\mathbb{R}$ as a ring is that there is an upper bound for the degree of irreducible polynomials in $\mathbb{R}[x]$, as

If $f\in\mathbb{R}[x]$ has degree larger than $2$, then $f$ is reducible.

However, the proof as I know depends highly on the fact that $\mathbb{C}$ is algebraically closed, and the very nice property: $z\in\mathbb{C}$ is in $\mathbb{R}$ if and only if $z=\bar{z}$.

This makes a generalization to other integral domains rather difficult. So the problem is

For what kind of integral domain $R$, we have a finite upper bound on the degree of irreducible elements in $R[x]$?

Some most familiar examples are ruled out: in $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{F}_{p}[x]$, there is not such a bound. Unfortunately these exhaust all integral domains about which I have a working knowledge.

Another result might help is Eisenstein's criterion and its generalized form, which says if we can find a prime ideal $\mathfrak{p}$ in $R$ such that $\mathfrak{p}^2\neq\mathfrak{p}$, then by picking $a\in\mathfrak{p}^2\backslash\mathfrak{p}$ we have an irreducible $a+x^d$, where $d$ can be arbitrary, and hence the upper bound is not possible.

So we only need to focus on domains where $\mathfrak{p}^2=\mathfrak{p}$ for all prime ideals. This seems to be a quite strong restriction but I am not sure what to make of it.

Can someone give a hint?

Thanks!

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If R is a UFD satisfying your property, and $K$ is its field of fractions, then every polynomial $f\in K[X]$ can be written as $f=ag$, with $a\in R\setminus0$ and $g\in R[X]$ primitive. If $f$ is irreducible in $K[X]$, then $g$ is irreducible in $R[X]$ by Gauss lemma, so $g$ has bounded degree. Therefore $f$ also has bounded degree, and now you are in the situation of Jacob Schlather's answer. –  Matemáticos Chibchas Feb 1 '13 at 18:12
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2 Answers 2

up vote 5 down vote accepted

Case 1: $R$ is a field

I will refer to these notes by Keith Conrad on the Artin Schreier theorem.

We will proceed by condition on whether or not $R$ is a perfect field. If $R$ is an imperfect field then there always exist irreducible polynomials of arbitrary large degree. This follows from Lemma 2.1 in Keith's notes which tell us that if $\alpha \in F \setminus F^{p^l}$, which is non-empty because $R$ is imperfect, then $p(t)=t^{p^l}-\alpha$ is irreducible.

If $R$ is perfect then we can deduce it has finite index in its algebraic closure. Let $A$ be the algebraic closure of $R$ then if $[A:R]=\infty$ we can construct arbitrarily large extensions by adjoining elements of a basis of $A$ to $R$. Then using the primitive element theorem we can find elements of arbitrarily large degree. So if every polynomial of degree larger than $n$ is reducible then the index of $R$ in $A$ is finite. Now if $R=A$ we are done otherwise the Artin-Schreier theorem tells us that if $1<[A:F] < \infty$ we have that $[A:F]=2$, $F$ is formally real closed and $A=F(i)$.

Case 2: R is not a field

Note that $R$ cannot be a UFD. A UFD is a field if and only if it has no prime elements. So if $R$ was a UFD that wasn't a field we could find some prime $p \in R$ and by the OP's observations we would have to have $(p)^2=(p)$. Any finitely generated idempotent ideal is generated by an idempotent, the only idempotents of an integral domain are $0$ and $1$. So $(p)=(0),R$ which is impossible, thereby $R$ cannot be a UFD if its not a field. This observation is probably not that important in the end.

Let $R$ be an integral domain such that every polynomial over $R[x]$ of degree greater than $n$ is reducible. If $K$ is the fraction field of $R$ then $K[x]$ necessarily satisfies the same condition, in particular $K$ is algebraically closed or satisfies the conclusion of Artin Shreier. One thing to notice is that if $K$ is algebraically closed then every polynomial splits over $R$ since any $a/b \in K$ satisfies $bx-a\in R[x]$. If $K$ is not algebraically closed then every polynomial splits over $R[i]$ for much the same reason. This MO question gives some details on how to construct these sorts of rings with valuation theory.

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Thanks for pointing me here. Unfortunately, I still can't see what precludes the closure from having infinite degree. I'm sure you can probably tell me here in a sentence or two, though(?) When I look, it seems like individual elements of the closure could stand at finite degree above $F$, and yet the entire closure might still be far above it. (But as I said, my field theory is rusted, atm.) –  rschwieb Feb 1 '13 at 16:29
    
@rschwieb Well in the perfect field case, for any finite extension you can always find a primitive element for any extension. So if the index is infinite in the algebraic closure, then you can find extensions of arbitrarily large degree, thereby primitive elements of arbitrarily large degree. I think you can cleverly construct irreducible polynomials of large degree when the field is imperfect. –  JSchlather Feb 1 '13 at 16:33
    
OK, that reiterates your meaning (which I already understood) but doesn't explain how. Fortunately in the meantime I think I do see how now... please tell me if I'm right! Choose an irreducible polynomial of maximal degree n and extend F to a degree n extension F'. If it weren't the entire algebraic closure, then you would be able to extend nontrivially again to F'', but then the primitive element theorem says that F'' is generated by a single root of degree more than n: contradiction. Is that right? –  rschwieb Feb 1 '13 at 16:41
    
@rschwieb Yes, precisely sorry that I misunderstood. –  JSchlather Feb 1 '13 at 16:42
    
I hope I don't come off as critical... the whole problem is just that a lot of things you know very well are atrophied for me, so it is mostly my fault :) Thanks for the lesson! –  rschwieb Feb 1 '13 at 16:43
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I'm not sure about integral domains in general, but the answer is pretty clear for a field $\Bbb F$:

There is a bound for the degrees of irreducible polynomials in $\Bbb F[x]$ if $\Bbb F$ has finite index in its algebraic closure.

This is the major feature that differentiates between the cases of $\Bbb R$ and $\Bbb F_p$. I'm not sure if the converse holds.


Update: You can find in the Jacob Schlather's answer (and the comments) the reasoning for why the convese must be true for perfect fields. That takes care of finite fields and rings of characteristic zero.

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I haven't learned algebraic closure yet. But a good motivation for me to move on. Thanks! –  Hui Yu Feb 1 '13 at 15:18
    
And I remember there is a strong charecterisation of such fields, right?IIRC it is due to Artin? –  awllower Feb 1 '13 at 15:18
    
At least it has ruled out all finite fields. –  Hui Yu Feb 1 '13 at 15:23
    
I'm wondering now if the converse holds: finite bound-> finite index in algebraic closure. It's tempting to believe but I have done field theory in ages... –  rschwieb Feb 1 '13 at 15:24
    
@rschwieb, See my answer. –  JSchlather Feb 1 '13 at 15:33
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