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Recall that an integral domain $R$ is atomic if every nonzero nonunit admits at least one factorization into irreducible elements. (Indeed, hard-core factorization theorists have replaced the word "irreducible" by "atom".)

From prior reading, I happen to know that there exist atomic integral domains $R$ such that the univariate polynomial ring $R[t]$ is not atomic. This is a somewhat surprising pathology, because the implication is true if both instances of "atomic" are replaced by "UFD", "Noetherian" or "Ascending Chain Condition on Principal Ideals".

But I don't know a precise example or a reference, and I would like one for an expository article I'm writing. Of course, the chronologically earlier and logically simpler the example, the better.

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+1, but I removed the LaTeX from the title, as it served no purpose, as far as I can tell. –  Harry Gindi May 19 '10 at 9:06

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According to the book "Non-Noetherian commutative ring theory" by S.T. Chapman and S. Glaz the question was first asked in "Factorization of integral domains" by D.D. Anderson, D.F. Anderson, M. Zafrullah, Journal of Pure and Applied Algebra 69 (1990) 1-19 (question 1). An answer was given here by M. Roitman.

There it was conjectured that $R[X]$ atomic $\implies$ $R[X,Y]$ is also atomic.

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Thanks, Gjergji. I will acknowledge you in my article. Since it is expository, there is no guarantee that it will see the light of day -- or rather, traditional publication -- but it will be internet available, at least. –  Pete L. Clark May 19 '10 at 9:49
    
Thanks! Your articles are really nice, may I ask what this next one will be about? –  Gjergji Zaimi May 19 '10 at 10:15
    
I'm working on revising / slightly expanding math.uga.edu/~pete/factorization.pdf, which was first written about a year ago. –  Pete L. Clark May 19 '10 at 11:21

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