# Why is the noetherian ring property not definable in first-order logic?

I am reading this paper on the connection between model theory and algebraic geometry.

https://math.uchicago.edu/~may/REU2015/REUPapers/Zhang,Victor.pdf

On page 9, I have trouble understanding Example 19: it says:

This shows the property of being a noetherian ring is not expressible as a set of sentences in first-order logic, otherwise $$\text{Th}(\mathbb Z)$$ would model such a set and our example model would be a noetherian ring.

I do not know why "otherwise $$\text{Th}(\mathbb Z)$$ would model such a set and our example model would be a noetherian ring". May I please ask for a more expanded explanation? Thank you for any help!

• The answer below is spot on! I think there might be a small typo causing you confusion - with my understanding of the terminology, as well as what's stated in Definition 9, it should say ".. otherwise $\Bbb Z$ would model such a set (ie $\mathrm{Th}(\Bbb Z)$ would contain such a set) ..", and the idea is $M$ models the same sentences as $\Bbb Z$ (they're "elementarily equivalent"). This is a standard way to show that some property is not axiomatisable. As an aside this isn't the only way to construct such an $M$.. I discuss this a bit here. Dec 3, 2023 at 12:49
• Thank you for the reference!
– Y.X.
Dec 4, 2023 at 3:16

If being a noetherian ring can be expressed as a set of sentences in first-order logic of ring theory, that means, there is a set of first-order sentences $$\mathcal S$$ (in the language of ring theory), such that a ring $$R$$ is noetherian iff $$\mathcal S\subset \text{Th}(R)$$.
If so, then since $$\mathbb Z$$ is noetherian, $$\mathcal S\subset \text{Th}(\mathbb Z)$$. As $$\text{Th}(\mathbb Z)\subset \text{Th}(M)$$ (where $$M$$ is the special example given in Example 19, which is shown to be non-noetherian), we have $$\mathcal S\subset\text{Th}(M)$$, so $$M$$ is noetherian, a contradiction.