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RECAP - WHAT I KNOW AND HOW I STUDIED IT

I started my logic/algebra course from studying propositional logic, then first order logic, then set theory, and then we started “building” natural, integer, rational and finally real numbers (as approximations by default, not by excess and default as Dedekind sections, see notes$^1$). Then, we started talking about monomials, polynomials and, finally, rational functions.

We study that a rational (or algebraic) function is defined as the equivalence class of a “rational term”, where a rational term is a couple $(p,q)$ of polynomials in a single variable with coefficients in a generic field $k$, named “polynomial fraction”, that satisfies certain rules. The equivalence relation is the one defined by $\frac{p}{q} \sim \frac{p’}{q’} \iff pq’=p’q$ and the quotient set is the field of rational functions.

For rational terms we can then define an “algebraic equation in one variable” as being a formula of type $s(x)=t(x)$ where $s$ and $t$ are rational terms with coefficients in a field $k$ and $x$ is the only variable. Then, we build the whole equation theory on this ”constraint” that everything that appears in this formula must be algebraic terms, and all the results are based on the concepts related to a rational term.

QUESTION

I have always took for granted that the ideas of “equivalent equations” and the “two principles of equations” translate to non algebraic equations, so equations in which appear anything that is not a rational term (such as n-th roots of the variable, trascendental functions with the variable as an argument etc.). But now I want to know: how can we study non algebraic equations in a rigorous way? Can you give me a preview? Is there a textbook or a source that you recommend?

  • Note that I’m only interested in a rigorous way to define what a non-algebraic equation is and in seeing that you can still define the “equivalence” and “principles” of equationa. I’m not interested in general solving-methods, which, from what I know, don’t even exist

  • As “two principle of equations” I refer to the result that operating on both sides of an equations, if you satisfy the correct hypothesises, yields an equivalent equation, as written here. I’m intersted particularly in this because it’s what gives us the ability to manipulate equations in the most basic way.

NOTES

$^1$ This method of constructing the rational numbers is called "ideals" by my textbook, so an ideal is defined as "a subset $\mathfrak a \subset \mathbb{Q}$ such that, if $x \in \mathfrak a$ and $y \leq x$ then $y \in \mathfrak a$". Then we define intervals, majorant and minorants, "closed ideals" etc. For example the "principal ideal of a $q \in \mathbb{Q}$ is $<q> = (-\infty, q]$. I see that it is a different definition of ideal than the wikipedia one, which seems more general.

selenio34
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    You might be interested in equality in first-order logic, the axiom of extensionality, and extension by definitions. Together these establish the logical foundations for manipulating equations involving set-theoretically defined objects. – Karl Aug 03 '23 at 17:29
  • @Karl ah thanks, so the idea becomes that there isn’t anything even “similar” to algebra anymore, and we just treat those “non rational terms” as functions (but we could insert any set-theoretically defined object)? So the definition of a non algebraic real equations would just be a formula of type $f=g$ where $f, g$ are real functions in one real variable? Then, two equations are equivalent iff they have the same solutions (same set made by $x\in \mathbb{R}$ that satisfy the equation) and then we demonstrate also the two principles of equation using this? – selenio34 Aug 03 '23 at 17:42
  • Yeah, you can think of an equation as a relation, i.e. a subset of the cartesian product of the domains of its variables. – Karl Aug 03 '23 at 17:57
  • @Karl mm but how does that help me, instead of considering it a formula of the first order logic theory of set theory? – selenio34 Aug 03 '23 at 18:05
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    OK, your book is generalizing the term "ideal" as used in the theory of Boolean algebras to ordered sets. This approach (for constructing $\mathbb{R}$) is essentially the same as using Dedekind cuts, although superficially different. If you have an ideal $\mathfrak{a}$, then let $\mathfrak{b}$ be the complement of $\mathfrak{a}$; then $(\mathfrak{a}|\mathfrak{b})$ is a Dedekind cut. Conversely, if $(\mathfrak{a}|\mathfrak{b})$ is a Dedekind cut, then $\mathfrak{a}$ is an ideal. – Michael Weiss Aug 07 '23 at 17:57
  • see https://math.stackexchange.com/questions/349709/proving-a-problem-has-a-closed-form-solution – IV_ Oct 02 '23 at 21:26
  • see https://math.stackexchange.com/questions/467667/is-there-an-algorithm-to-determine-if-a-closed-form-solution-exists – IV_ Oct 02 '23 at 21:33

1 Answers1

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There are two ways to define something like (say) $\sqrt{1-x^3}$ rigorously. The first approach you will find in most (abstract) algebra textbooks. It can be presented in superficially different but essentially equivalent ways. I can only sketch the ideas very briefly. We start with a field $K$ and an irreducible monic polynomial $f(z)\in K[z]$; we want to construct a field $E\supseteq K$ containing an element $\alpha$ such that $f(\alpha)=0$ in $E$. This is called adjoining a root.

The more concrete construction you will find, for example, in Emil Artin's Galois Theory, starting on p.26. We let $E$ be the set of all polynomials $g(z)$ of degree $<n$, where $n=\deg f$. $K\subseteq E$ by regarding elements of $K$ as polynomials of degree 0. Addition in $E$ is defined as usual, but we define multiplication (call it $\times$) in $E$ as follows: $$g\times h=gh\bmod f$$ where $gh$ is the usual polynomial multiplication, and $gh\bmod f$ is the remainder upon dividing by $f$. With this definition it can be shown that $E$ is a field, and that $f(z)=0$, where "$f(z)$" means the result of evaluating $f$ at $z$ using $\times$ for multiplication.

This follows the same pattern as used to define the field of $p$ elements where $p$ is a prime. This is $\mathbb{F}_p=\{0,\ldots,p-1\}$, with multiplication defined by $a\times b=ab\bmod p$.

In both cases, we can reframe this in a more abstract manner. Let's take the case of $\mathbb{F}_p$ first. Let $(p)$ be the set of all multiples of $p$, the so-called principal ideal generated by $p$. We say $a\equiv b\pmod{p}$ ($a,b\in\mathbb{Z}$) if $p|(a-b)$, i.e., if $a-b\in(p)$. $\mathbb{F}_p=\mathbb{Z}/(p)$ is defined as the set of equivalence classes. Addition and multiplication are defined by $[a]+[b]=[a+b]$ and $[a][b]=[ab]$. Each equivalence class $[a]$ has a unique representative in the set $\{0,\ldots,p-1\}$. The more concrete construction just deals always with these representatives.

It works the same way for adjoining a root: we say $g\equiv h\pmod{f}$ if $f|(g-h)$; $E=K[z]/(f)$, the set of equivalence classes; each equivalence class has a unique representative among the polynomials of degree $<\deg f$. It can be shown that if $\alpha$ is the equivalence class of the polynomial $z$, then $f(\alpha)=0$ in $E$.

This works fine for any field $K$. If $K$ happens to be $k(x)$, the field of rational functions in $x$, and (say) $f(z)=z^2-(1-x^3)$, then we have a field $E$ with an element $\alpha$ that we could denote $\sqrt{1-x^3}$, since $\alpha^2=1-x^3$.

You can iterate this procedure as many times as you like. Using either transfinite induction or Zorn's lemma, you can even define rigorously the algebraic closure of $k(x)$.

Besides Emil Artin's Galois Theory, most abstract algebra textbooks will cover this one way or another. This MSE question lists a bunch.

The other approach, not as commonly seen, is using so-called Puiseaux series. These are formal series where the exponents are allowed to be both negative and fractional. This works only in characteristic zero, but apparently there is a generalization (Hahn series) that works in all characteristics. I leave the details to the linked Wikipedia page.

You ask about transcendental functions, like say $\log x$. Here there are lots of approaches, not equivalent but related, depending on what you want to achieve. To mention just one: it's not hard to define the field of formal power series $K[[z]]$ over a field $K$.

  • Finally, I wanted to ask you: your comment defines something like the root of a polynomial as the zero of another polynomial (correct?). So now, because this is still algebra, we should be able at least in theory to make an "extended" theory of equations that is able to demonstrate the two equation principles including roots of polynomials. Is this correct? – selenio34 Aug 07 '23 at 14:04
  • Thanks for your answer, it took me a little bit to understand it since I have a lot of knowledge missing. I updated the "ideal" notion that I studied (see the notes) and I saw the "correct" ideal (the one of wikipedia). If I understood correctly, the polynomial function $f(z)$ has a principal ideal $(f(z))$ defined as ${ f(z)g(z):g(z) \in K[z] }$ where $K=k(x)$ is the field of rational functions - so elements $f(z) \in K[z]$ are polynomials, and the variable z is "something" with, as coefficients, other polynomials (rational functions actually) in variable x? – selenio34 Aug 07 '23 at 14:34
  • If yes, I have two questions: 1) $z$ is the element of our set that includes rational functions and their n-ths roots, right? 2) I don't understand the role of the equivalence relations in your comment - what is the equivalence class $\alpha$ of $z$? What equivalence relation does $(f(z))$ induce on $K[z]$? – selenio34 Aug 07 '23 at 14:41
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    I've rewritten my answer; hopefully the new version will answer these questions. – Michael Weiss Aug 07 '23 at 19:01
  • Thanks, the similarities with $\mathbb{Z}/p$ - which I was more familiar with - made it clearer. I still struggle to have an intuition on how to get those results, but I probably have to study more first, and at least I have an idea of how to rigorously define the things I asked. – selenio34 Aug 07 '23 at 20:54
  • Only one last question: by proving that nth-roots of polynomials and trascendental functions are elements each of their "own kind" of fields, do we already prove that the two principles of equations are valid? If yes, it must be because if we can prove them for a general field - can we? - they must be true also for our specific fields, right? – selenio34 Aug 07 '23 at 20:58
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    Yes, that's correct. – Michael Weiss Aug 07 '23 at 23:14