Problem: Let $ R,S $ be integral domains and $ f: R \to S $ a unit-preserving homomorphism. Assume that $ x \in S $ is irreducible. Then does the pre-image $ {f^{-1}}[\{ x \}] $ contain only irreducible elements?

My research effort

Assume that $ f(x) $ is irreducible and $ x = y \cdot z $. Then $ f(x) = f(y \cdot z) = f(y) \cdot f(z) $. Hence, either $ f(y) $ is invertible or $ f(z) $ is invertible. However, this does not imply that either $ y $ or $ z $ is invertible. For example, $ 5 $ is irreducible (hence not a unit) in $ \mathbb{Z} $, but its image $ \bar{5} $ in $ \mathbb{Z} / 7 \mathbb{Z} $ is a unit.

Could anyone tell me if the statement is true/false and give a hint for a proof/counterexample?

  • $\begingroup$ Thank you for your answers. @Pavel, I have been thinking about this, but I don't understand your answer yet. I'm not familiar with localisations. I have read something about on wikipedia. Does it have something to do with fields of fractions? I hope one of you can clarify this. $\endgroup$ – Koenraad van Duin Nov 7 '13 at 21:55
  • $\begingroup$ Well, yes. Basically, a localization by some set S is (in the integral domain case) the smallest subring T of the field of fractions such that all the elements of S are invertible in T. But never mind that, I adjusted the hint to a concrete example (in particular, to a localization of $\mathbb{Z}$ by the set $\{2, 2^2, 2^3, \dots\}$). $\endgroup$ – Pavel Čoupek Nov 8 '13 at 9:57

The statement is not true, i.e. a non-irreducible element can be mapped to an irreducible element.

Hint: Think of embedding $f$ of $R$ into some localization of $R$ (by some multiplicative set) and the fact that some element $y$ of the localization usually is a unit despite the fact that $f^{-1}(\{y\})$ is not. Such $y$ can be then used to find counterexample.

More detailed hint:

Ok, so I would still use the localization example, but in a concrete situation to make it clear. Let $f$ be an embedding of the ring $\mathbb{Z}$ to the following subring of $\mathbb{Q}$: $$ S=\{ \frac{a}{2^k} \; | \; a\in \mathbb{Z}, k \in \mathbb{N}_0 \}.$$

Try to find the example in this situation. That is, try to find an integer which is a product of at least two primes, but is irreducible as an element of the ring $S$.

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    $\begingroup$ If $x$ in the localization is a unit, then it is not irreducible and hence does not satisfy the hypothesis of the question. $\endgroup$ – RghtHndSd Nov 6 '13 at 22:29
  • $\begingroup$ I should pribably have written $y$ instead of $x$ to avoid confusion. Of course, such $x$ is not the counterexample, however, such $x$ can be used to find one. In other words it's only a hint. I changed the answer to avoid further confusion. $\endgroup$ – Pavel Čoupek Nov 6 '13 at 22:47
  • $\begingroup$ This won't work. If $x\in R$ is irreducible in $S^{-1}R$ then it's irreducible in $R$. $\endgroup$ – anon Nov 6 '13 at 22:53
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    $\begingroup$ @anon: Take $xy$ in $R$ such that $y$ becomes a unit in the localization but $x$ does not (essentially). $\endgroup$ – RghtHndSd Nov 7 '13 at 0:43
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    $\begingroup$ Right, I am being very silly. $\endgroup$ – anon Nov 7 '13 at 0:56

The statement is correct if both $R$ and $S$ are local rings and if $f$ is a homomorphism of local rings, i.e. $f(\mathfrak{m}_{R}) \subset \mathfrak{m}_{S}$, where $\mathfrak{m}_{R}, \mathfrak{m}_{S}$ denotes the maximal ideal of $R$, $S$, respectively.

Proof: The set of non-units in $R$ is the maximal ideal $\mathfrak{m}_{R}$. Since $f(\mathfrak{m}_{R}) \subset \mathfrak{m}_{S}$, it hence follows that if $a$ is a non-unit then $f(a)$ is a non-unit. Let $b\in S$ be irreducible. Suppose that there exists an $x\in f^{-1}(b)$ which is reducible. Then there exist $y, z \in \mathfrak{m}_{R}$ so that $x = y \cdot z$, which yields a factorization of $b= f(x) = f(y) \cdot f(z)$, and since neither $f(y)$ nore $f(z)$ is a unit, this gives contradiction to the irreducibility of $b$.


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