# some question related to $\mathbb{Z}[\sqrt{-5}]$ and UFD

could any one give me examples/proofs/counter examples against or for of the followings?

1.Homomorphic image of a UFD is again a UFD

2.The element $2\in\mathbb{Z}[\sqrt{-5}]$ is irreducible

3.Units of the ring $\mathbb{Z}[\sqrt{-5}]$ are units of $\mathbb{Z}$

4.$2$ is a prime element in $\mathbb{Z}[\sqrt{-5}]$

for 1, I know that $\mathbb{Z}$ is a UFD, to show the statement is false from $\mathbb{Z}$ to where I should construct a homomorphism?

for 2, In abstract algebra, a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units.

3 is true.

4 is false

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While one easily finds trivial non-domain counterexamples to $(1)$, I suspect that the point of this set of exercises is to find a nontrivial domain counterexample, using the other exercises in the set. Namely, examining norms shows $(2)$ and $(3),$ and then $(4)$ is refuted by $\rm\:2\:|\:ww',\ 2\nmid w,w',$ for $\rm\, w = 1+\sqrt{5}.\:$ Therefore $\rm\, \Bbb Z[\sqrt{5}]\,$ is not a UFD, since $2$ is irreducible, but not prime.

Finally, to refute $(1)$ it suffices to present $\rm\,\Bbb Z [\sqrt{5}]\,$ as the image of a UFD. Here is a big hint: any simple adjunction $\rm\,R[\alpha]\,$ of an element $\rm\,\alpha\,$ to a ring $\rm\,R\,$ is a homomorphic ring image of the universal simple adjunction of an indeterminate, the polynomial ring $\rm\,R [x],$ which is a UFD if $\rm\,R\,$ is a UFD. You probably already know that $\rm\,\Bbb Z[x]\,$ is a UFD. Though you do not need it here, it deserves to be better known that the proof of the general case has a beautiful conceptual proof by pulling back the UFD property from $\rm\,Q[x],\:$ using localization (Nagata's Lemma).

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The question was about $\mathbb{Z}[\sqrt{-5}]$, not $\mathbb{Z}[\sqrt{5}]$, right? – Martin Brandenburg Jan 31 '14 at 9:46

HINT for (1): Every UFD is an integral domain; can you think of a quotient of $\Bbb Z$ that has zero-divisors?

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$\mathbb{Z}/n\mathbb{Z}$ when $n$ is not prime – Un Chien Andalou Oct 26 '12 at 13:53
so $f:\mathbb{Z}\rightarrow \mathbb{Z}/n\mathbb{Z}$ when $n$ is not prime, define a homo $f(n)=[n]$, that will work? – Un Chien Andalou Oct 26 '12 at 13:55
@Flute: Yes, it will. – Brian M. Scott Oct 26 '12 at 13:59
@Briab, Thank you – Un Chien Andalou Oct 26 '12 at 14:00
@Flute: You’re welcome. – Brian M. Scott Oct 26 '12 at 14:00