# Understanding definable sets

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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...

• Probably, the language under consideration is the one with 0-ary operations $0, 1$, unary operation $-$ (unary negation), binary operations $+, \cdot$. And then we also require the model to satisfy the ring axioms $\forall x,y,z: (x+y)\cdot z = x\cdot z + y\cdot z$, etc. – Daniel Schepler Oct 10 '18 at 17:29
• Similarly, we use the word "group" to refer to a structure in the language of groups satisfying the group axioms, etc. – Noah Schweber Oct 10 '18 at 19:58

## 1 Answer

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.

• I still don't understand why the set $\{x\in M | p(x)=0\}$ is equal to the set $\{x\in R| \mathcal M\models \phi(x,a_0,\dots,a_n)\}$. By definition, $\mathcal M\models \phi(x,a_0,\dots,a_n)$ means that the interpretation of the term $0$ is equal to the interpretation of the term $a_n x^n+\dots+a_1x+a_0$. The $a_i$s are constants, so their interpretation is $a_i$. The interpretation of $+$ is addition, and that of $\cdot$ is multiplication. But I still don't know how to show/see rigorously that the two sets are equal (if they are). – logic Oct 10 '18 at 21:56
• @logic “The interpretation of the term zero” in the ring is nothing more or less than the ring’s zero element. The interpretation of two terms being equal means that the ring elements that the terms represent are equal. (Also I’m not sure what distinction you’re making between $M$ and $R$ or if that is a typo.) – spaceisdarkgreen Oct 10 '18 at 22:25
• I understand this. I think what I don't understand is what is the interpretation of $a_nx^n+\dots+a_1x+a_0$ in $\mathcal M$. Is it supposed to be $p(x)$? In particular, how to interpret $x$? Is it considered as a constant and so $x^\mathcal M=x$? (Also $M=R$) – logic Oct 10 '18 at 22:30
• @logic $x$ is a variable. You interpret it by assigning some element of $M$ to it. The resulting sentence will be true under some assignments and not others. It will be true if and only if you assign a zero of the polynomial to $x.$ This is what we mean when we say the formula defines the set of zeros. – spaceisdarkgreen Oct 10 '18 at 22:36
• Thanks, that makes things more clear, though the interpretation of the term $a_nx^n+\dots+a_1x+x_0$ still seems somewhat vague for me. Would it be correct to say that the interpretation of this term is equal to (let me write $M$ instead of $\mathcal M$) $(a_nx\dots x)^M+\dots+(a_1x)^M+a_0^M=a_n^M x^M\dots x^M+a_1^Mx^M+a_0^M=a_n(x^M)^n+\dots+a_1x^M+a_0$, and the result depends on the interpretation of $x$? – logic Oct 10 '18 at 22:54