I'm self-studying Model Theory via Hodges' book Model Theory and got stuck in the following problem (see below for notation):
Section 2.2, Problem 9: For each of the following classes, show that it can be defined by a single sentence of $L_{\omega_1\omega}$.
(a) Unique factorization domains.
(b) Principal ideal domains.
(c) Dedekind domains.
(d) Semisimple rings.
(e) Left coherent rings.
(f) Left artinian rings.
(g) Noetherian local commutative rings.
(h) Groups $G$ such that if $H,K$ are any two isomorphic finitely generated subgroups of $G$ then $H$ is congruent to $K$ in $G$.
The problem reduces to restate the properties in each item in a suitable manner, without dealing directly with ideals and subgroups but rather only with elements of the ring in question (finitely many of them at a time). See below for more details.
My problem is:
Problem: I don't know how to solve items (b), (c), (d), (f), and (g), because I don't know how to write conditions on ideals (and chains of these) in the given language.
Some notation: The signature $L$ for rings is $\langle 0,1,+,-,\cdot\rangle$. Given an ordinal $\kappa$, we define the language $L_{\kappa\omega}$ in the usual manner, but we allow disjunctions and conjunction on sets of cardinality $<\kappa$, i.e., if $I$ is a set of formulas of $L_{\kappa\omega}$ with $|I|<\kappa$, then $\bigvee_{\phi\in I}\phi$ and $\bigwedge_{\phi\in I}\phi$ are formulas of $L_{\kappa\omega}$ (but we can only put quantifiers over finitely many variables: that's what the $\omega$ in $L_{\kappa\omega}$ means).
For example, here's how we can do (h) (I believe the author meant "conjugate" instead of "congruent"): We use variables $x_i,y_i,z_i,\ldots$. Moreover, if the variables $x_1,\ldots,x_k$ are bound, we write $\overline{x}=x_1\ldots x_k$.
Given variables $x,y_1,\ldots,y_k$, let's write: $$x\in\langle \overline{y}\rangle:\bigvee_{n<\omega}\exists z_1\ldots z_n\left[\left[\bigwedge_{1\leq i\leq n}\bigvee_{1\leq j\leq k}(z_i=y_j)\lor (z_i=y_j^{-1})\right)\land x=z_1\cdots z_n\right]$$ which can be read, in the language of groups, as "$x$ belongs to the subgroup generated by the $y_i$".
We can also write a formula for two finitely generated subgroups to be isomorphic: Consider variables $p_1,p_2,\ldots$, and all terms $t(\overline{p})$ in the language of groups which have only these variables. There are only countably many of these terms. Say we have variables $x_1\ldots x_k$ and $z_1\ldots z_k$. We write $$\langle\overline{x}\rangle\simeq\langle\overline{z}\rangle:\bigwedge_{t(\overline{p})}(t(\overline{x})=1\leftrightarrow t(\overline{z})=1),$$ where $t(\overline{x})$ means that we change each occurence of the variable $p_i$ by $x_{i\bmod k}$, and similarly for $t(\overline{z})$. The formula above states that the maps $x_i\mapsto z_i$ extends to an isomorphism of the generated subgroups.
Finally, the condition in the exercise can be expressed by $$\bigwedge_{n<\omega}\forall x_1\ldots x_n,z_1\ldots,z_n\left[\langle\overline{x}\rangle\simeq\langle\overline{z}\rangle\rightarrow\exists g\left(\bigwedge_{1\leq i\leq k}gx_ig^{-1}\in\langle\overline{z}\rangle\land g^{-1}z_ig\in\langle\overline{x}\rangle\right)\right]$$ and the conjunction of this sentence with the usual group axioms is a formula in $L_{\omega_1\omega}$ which works.
EDIT: Here are a few solutions for some of the items.
Item (b): A ring is a PID iff every finitely generated ideal is principal, or equivalently iff every $2$-generated ideal is principal (Bézout domain), and it is a UFD (item (a)). All of these can be written in $L_{\omega_1\omega}$.
Item (d): first recall given a ring $R$, the ideal generated by some $x\in R$ is $(x)=\left\{ax+xb+cxd:a,b,c,d\in R\right\}$, and $R$ is simple iff it is the ideal generated by any nonzero element. Thus a semisimple ring is one for which there are $n$ and $x_1,\ldots,x_n$, with $(x_i)\cap (x_j)=0$ if $i\neq j$, and for which $(x_i)=(x)$ for any $x\in (x_i)$. These can all be written in $L_{\omega_1\omega}$.
Item (e): A ring $R$ is left coherent iff every finitely generated ideal is finitely presented. We can restate this as follows: For every $n$ and for every $a_1,\ldots,a_n$, there exist $m$ and terms $t_1(\overline{x}^1,\overline{p}),\ldots,t_m(\overline{x}^m,\overline{p})$, of the form $t_i(\overline{x}^i,\overline{p})=x^i_1p_1+\cdots+x^i_np_n$ (where $x^i_j$ and $p_j$ are variables) in the language of rings, and there exist $r^i_j\in R$, $1\leq i\leq m$, $1\leq j\leq n$, for which $t_i(\overline{r}^i,\overline{a})=0$ for all $i$, and such that for every $s_1,\ldots,s_n$, if $s_1a_1+\cdots+s_na_n=0$ then there are $q_1,\ldots,q_m$ such that for each $i$ $s_i=\sum_jq_jr_i^j$.
This means that the relations $t_i(\overline{r},\overline{p})=0$, which make sense for left $R$-modules, are satisfies by the generators $a_i$, the map from the free module generated by $p_1,\ldots,p_n$, mapping $p_i\mapsto a_i$, has precisely the module generated by $t_i(\overline{r},\overline{p})$ as its kernel.
