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I know that the ring of algebraic integers can be an answer because this ring has no irreducible elements.

So, is there an integral domain with nonempty set of irreducible elements and an element having no factorization into irreducibles?

user26857
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Springfield
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    See https://math.stackexchange.com/questions/2078464/algebraic-integers-form-a-ring-without-any-irreducible-elements-every-prime-ide – lhf Jan 25 '21 at 14:04
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    Related: https://math.stackexchange.com/questions/714719. This shows that such an example should be non-Noetherian. – user26857 Jan 25 '21 at 17:20

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Consider $R=\{P \in \mathbb{Q}[X],\, P(0) \in \mathbb{Z}\}$. $R$ has irreducible elements, such as $X^2+1$. However, $X$ has no factorisation into a product of irreducible elements. Indeed, any factorisation of $X$ is of the form $n \cdot \frac{X}{n}$, and $\frac{X}{n}$ is never irreducible (it can be written as $\frac{X}{(n+1)!} \frac{(n+1)!}{n}$ and none of the factors are units).

Aphelli
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