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Let $R$ be an integral domain which is not a field.

Does $R$ necessarily have an irreducible element?

I suspect the answer is no, but I couldn't find an example showing that...

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    $\begingroup$ On the other hand, the answer is yes if $R$ is noetherian. $\endgroup$
    – user18119
    Feb 7, 2013 at 21:51
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    $\begingroup$ Because if $R$ is Noetherian then the family of principal ideals has a maximum $(a)$, and thus $(a)$ is irreducible? $\endgroup$
    – Lior B-S
    Feb 8, 2013 at 7:30
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    $\begingroup$ Yes, this is correct. One can even show that any non-zero element is a finite product of irreducible elements in a similar way. $\endgroup$
    – user18119
    Feb 8, 2013 at 14:47

3 Answers 3

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Hint $\ $ There are rings, not fields, closed under square roots (e.g. the ring of all algebraic integers). In such a ring there are no irreducibles since every element factors: $\rm\ a = \sqrt{a} \sqrt{a}.$

A concrete example: $ $ let $\rm\, D = \Bbb Q[x,x^{1/2},x^{1/4},x^{1/8},\ldots].\,$ Every element $\ne0$ is writable in the form $\rm\ d_i x^{i}\! + d_j x^j\! +\cdots\! + d_k x^k\! = x^i\, f\ \,$ for $\rm\,\ f\in D,\,\ d_i\in\Bbb Q,\,\ d_i\ne 0,\:$ and $\rm\:i < j < \ldots < k,\:$ e.g.

$$\rm a\, x^{1/8}\! + b\, x^{1/4}\! + c\, x^{1/2} =\, x^{1/8} (a + b\, x^{1/8}\! + c\, x^{3/8})\, =\, x^{1/8}\, f,\quad a\ne 0$$

All exponents of $\rm\,x\,$ in terms of $\rm\,f\,$ remain of form $\rm\,n/2^m\!,\,$ being differences of such. We can force all the cofactors $\rm\,f\,$ to be units by adjoining inverses of all such elements, as follows. Since $\rm\,f\,$ has "constant" term $\rm\:a\ne 0,\,\ f\,$ is not in the maximal ideal $\rm\, M = (x,x^{1/2},x^{1/4},x^{1/8},\ldots).$ Therefore, localizing at $\rm\,M,\,$ i.e. adjoining inverses of all elements $\rm\not\in M,\,$ turns all the $\rm\,f\,$ cofactors into units, yielding a domain $\rm\,\bar D,\,$ not a field $\rm(x^{-1}\!\not\in \bar D),\,$ where all elements have form $\rm\,x^k\, u,\,$ for some unit $\rm\,u.\,$ Since $\rm\ a = x^k = (x^{k/2})^2 = \sqrt{a}\sqrt{a},\ $ this domain has no irreducibles.

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  • $\begingroup$ Of course YACP example, the ring of algebraic integers, is closed under square roots... $\endgroup$
    – Lior B-S
    Feb 8, 2013 at 7:31
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    $\begingroup$ This is also remarked in the first sentence of this answer. $\endgroup$ Feb 8, 2013 at 10:19
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It is proven here that the ring of all algebraic integers has no irreducible elements.

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Another class of examples is given by valuation rings whose value group is divisible.

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  • $\begingroup$ Sorry, I don't quite understand your first paragraph. How do you get, for example, a holomorphic square root of $f(z)=z$ on an open disk around zero? $\endgroup$ Feb 7, 2013 at 17:33
  • $\begingroup$ You are right, this doesn't work. The answer by Math Gems contains what I wanted to say. $\endgroup$ Feb 7, 2013 at 20:32

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