Prove that class of rings with all elements being nilpotent is not first order axiomatizable in signature $\{+, \cdot, 0, 1, = \}.$
I need some hints how to approach this problem. All similar problems I've solved before were about cardinality of the models (cyclic groups, finite groups, etc) and here it's seems like not an option.
Thanks!
To solve this problem, I think we have to assume the signature does not include the unit element "1", which is not nilpotent.
In that case, we can start with a commutative ring $R$ which exhibits "unbounded nilpotency", i.e., in which for every $n \in N$ there is $x \in R$ such that $x^n=0$ and $x^k \ne 0$ for $k=0, \dots ,n-1$. (Such a ring $R$ can easily be constructed as a direct sum of the rings $Z/p^k Z$ for $k \in N$.) Then, we define the subring $R_{0}=\{r \in R|(\exists n \in N) r^{n}=0\}$. (Some additional effort is required here to show that this is in fact a ring. The binomial theorem is useful.) $R_{0}$ exhibits the "unbounded nilpotency" property and also that every element in $R_{0}$ nilpotent.
Now assume there is a first order theory $T$ which axiomatizes rings all of whose elements are nilpotent. Then certainly $R_{0} \models T$. However, we may add a constant $a$ to the language of $T$ and the axioms $a^n \ne 0$ for $n \in N$. The resulting theory $T'$ is consistent by the compactness theorem and the fact that $R_{0} \models T$. But any model of $T'$ contains the non-nilpotent element assigned to the constant $a$.