Understanding definable sets enter image description here
This is part of Marker's book on Model theory. I have a hard time understanding the example given there.
First of all, he says "Let $\mathcal M$ be a ring", which is confusing. In my understanding $\mathcal M$ is an $\mathcal L_r$-structure (i.e., a set together with interpretations of function, relation, and constant symbols from the language). Why does he refer to it as a ring? The standard definition of a ring is a set together with two operations, but it does NOT include interpretations, nor does the standard definition of a ring from algebra assume that there is some underlying language.
Second, I don't understand why $Y$ is $A$-definable for any $A$ containing $\{a_0,\dots,a_n\}$. (And to begin with, what is $A$ in the definition?) By definition, this means that there is a formula $\psi(\overline v,w_1,\dots,w_l)$ and $\overline b\in A^l$ such that $Y=\{\overline a\in M^n| \mathcal M ⊨ \psi(\overline a,\overline b)\}$ In our case $\psi$ is the $\phi$ defined in the text. The claim is that $\phi(v,a_0,\dots,a_n)$ defines $Y$. Is $\overline b$ supposed to be $(a_1,\dots,a_n)$? I still don't see why this formula is $A$-definable...
 A: When Marker says “let $\mathcal M$ be a ring” he is saying that it is an $\mathcal L_r$-structure that satisfies the ring axioms, not just that it is an $\mathcal L_r$-structure. The definition of satisfaction applied to the ring axioms is the exact same thing as the definition of a ring from abstract algebra, so this is the same thing as saying it is a ring.
The $a_i$ are the coefficients of the polynomial $p.$ As Marker shows, the only parameters you need to write down the formula that defines the set of zeros of the polynomial are the coefficients of the polynomial. $A$ is some subset of the ring that contains all of the coefficients. So it has all the parameters we need: the set of zeros is definable from it.
Yes the $\bar b$ in the formula corresponds to the parameters, in this case the coefficients. As for $\bar a,$ in this case $n=1$ so $\bar a$ has just one component, in other words we are defining as set, as opposed to a higher-arity relation. The $a$ that satisfy this formula are exactly the zeros of the polynomial.
