Prove that any element in a ring A can be written as a product of irreducible but A is not a UFD

Question

Let A = $Z$[$\sqrt{10}$] = {a+b$\sqrt{10}$$\mid$a,b $\in$ $Z$}.

Prove that any element in A an be written as a product of irreducible, but A is not a UFD.

Also a small question, I've proved that $4$+$\sqrt{10}$ and $4$-$\sqrt{10}$ are irreducible, but how to show that they cannot be associates?

Thank you so much!

Do you know about the norm on $A$? – William Stagner Nov 28 '17 at 19:06 — William Stagner, Nov 28 '17 at 19:06

Dietrich Burde · Accepted Answer · 2017-11-28T19:18:23.243

1

Since $\mathbb{Z}[\sqrt{10}]$ is Noetherian, every element can be written as a finite product of irreducible elements. For a proof, see the arguments here:

Every element of $Z (\sqrt{-5})$ is factorable into irreducible factors.

The other part, the main part, is proved here:

Proving $\mathbb{Z}[\sqrt {10}]$ is not a UFD

edited Nov 28 '17 at 19:18

answered Nov 28 '17 at 19:16

Dietrich Burde

130,978

Sorry for the duplication, and thanks for the source! – JacobsonRadical Nov 28 '17 at 19:18
You are welcome. – Dietrich Burde Nov 28 '17 at 19:18

Prove that any element in a ring A can be written as a product of irreducible but A is not a UFD

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