By a standard argument* using ĊoĊ's ultraproduct theorem (either directly or through the compactness theorem), Noetherianness is not first order axiomatizable in the language of commutative unital rings. Is the same true of non-Noetherianness? (Of course, it must at the very least not be finitely axiomatizable.) I've tried the obvious thing but have gotten nowhere.
*For convenience write $n:=\left\{0,\dots,n-1\right\}$ and $\omega:=\left\{0,1,\dots\right\}$
Local rings whose maximal ideal squares to the zero ideal have the properties
that their maximal ideals are vector spaces over their residue fields in the obvious way.
(thus) that any generating set of their maximal ideals is a spanning set of said vector space and vice versa.
(thus) that they are non-Noetherian if(f) their maximal ideal admits arbitrarily large finite linearly independent sets under said obvious action of the residue field.
Now, for any natural $n\colon\omega$, the property "is a local ring whose maximal ideal squares to the zero ideal and whose maximal ideal contains an independent size of size $n$ under the obvious action by the residue field" is expressible by a sentence $\psi_{n}$ in the first order language of commutative unital rings. Clearly if $n_{0}\leq n_{1}$, then $\psi_{n_{1}}\implies \psi_{n_{0}}$. Each $\psi_{n}$ moreover admits a Noetherian model $A_{n}$ (e.g., $\mathbb{Q}\left[x_{i}\right]_{i\colon n}/\left(x_{i}\right)_{i\colon n}^{2}$).
The ultraproduct of these $A_{n}$s with respect to any non-principal ultrafilter on $\omega$ is then simultaneously a model of any sentence satisfied by all Noetherian commutative unital rings and a model of all the $\psi_{n}$s, whence non-Noetherian, thereby frustrating any first order axiomatization of the commutative unital rings.
