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I am looking for

an example of a ring where unique factorization does not hold. However the ring is supposed to have the property that every irreducible is already a prime.

So unique factorization should fail because one cannot decompose every element into a finite product of irreducibles. In particular the ring must not be Noetherian. What is an example of such a ring?

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2 Answers 2

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In the ring of all algebraic integers, there are no irreducibles (since, for example, if $a$ is an algebraic integer, so is $\sqrt a$, and $a=\sqrt a\sqrt a$), so every irreducible is (vacuously) prime. You cannot decompose any element into a finite product of irreducibles since, as noted, there aren't any irreducibles.

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  • $\begingroup$ For a link to a proof (and more) see this answer posted a week prior. $\endgroup$ Commented Feb 14 at 13:57
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How about $\mathbb C[X_1, X_2, \ldots]/(X_k-X_{k+1}^2\mid k\in\mathbb N)$? The only ireducible elements are the constants: Every polynomial in several variables $X_1,\ldots,X_n$ can be rewritten as one in $X_n$ only. Unless it is constant, we can even write it as a polynomial of degree $\ge 2$ in $X_{n+1}$ and this has a nontrivial linear factor $X_{n+1}-\alpha$. Since the product of constants is constant, not every element can be written as product of irreducibles.

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  • $\begingroup$ (nonzero)Constants are invertible, thus I can't see any irreducible in your ring. $\endgroup$
    – user26857
    Commented Dec 21, 2014 at 22:46

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