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This is a homework problem so it would be much better to receive a hint than a solution. I've tried everything I can think of and I'm completely stumped. I need to show $\mathbb{Z}[\sqrt {10}]$ satisfies the ascending chain condition on principle ideals.

I'm pretty confident this needs to be done by contradiction, so I assume there exists a chain $$(a_1) \subsetneqq (a_2) \subsetneqq \cdots $$ I've deduced that the norm $$N:x+y\sqrt {10}\mapsto x^2-10y^2$$ assigns associates the same values and if $a\vert b$ in the ring, $N(a)\vert N (b) $ in $\mathbb {Z} $.

I know this leads to the chain $$(N (a_1)) \subseteq (N (a_2)) \subseteq \cdots $$ in $\mathbb {Z}$, which must stabilize. But the stability of this chain doesn't seem to imply much about the original chain. I've deduced that if $(N (a_i))=(N (a_j))$, then $x_i^2\equiv x_j^2 \pmod {10} $ where $x_i$ and $x_j $ are the rational parts of $a_i $ and $a_j $, but this fact doesn't seem particularly useful. I know there are some elements which are not associates that are assigned the same norm, so the stability of this chain doesn't guarantee association of the $a_i $.

I also know $\mathbb {Z} $ is not a UFD, so satisfaction of ACCP is not guaranteed.

I've also considered $$\bigcup_{i =1}^\infty (a_i)$$, which I know is an ideal containing every element of the chain. But this hasn't led to anything useful, either.

I'm completely out of ideas. A push in the right direction would be greatly appreciated.

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    You need a lemma: if $a|b$ and $(N(a))= (N(b))$, then $(a)=(b)$. – 2017-01-31
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    Crostul is right. Since you assumed that the chain that you started with was strictly ascending, the chain of principal ideals in Z generated by the norms must also be strictly ascending. But this cannot happen since Z is a PID and hence a UFD. – 2018-01-23

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The ring is Noetherian. In particular, $k[x]$ is Noetherian (by the basis theorem.) Applying the homomorphism $x \mapsto \sqrt{10}$ (substitution really), we get a surjective homomorphism onto $k[\sqrt{10}]$. Hence, we can consider its kernel and by taking quotients, see that $$\mathbb Z[\sqrt{10}] \cong \mathbb Z[x]/\ker \phi$$ which is again Noetherian, implying accp