Is this what casus irreducibilis actually is?

Question

Suppose we are working over the complex numbers $\mathbb{C}$. I know that every quadratic equation $x^2 + ax + b = 0$ with complex coefficients $a$ and $b$ has a solution expressible in terms of the constants $0$ and $1$, the field operations $+$, $-$, $\times$, $\div$, and the multiple-valued operation $\sqrt{}$, which takes one complex number and returns its two square roots. Specifically, the formula is $(-a + \sqrt{a^2-4b})/2$. I know that there is usually a plus-or-minus sign attached to the square root, but since the square root operation over the complex numbers is already multiple-valued, we don't need it. This means that, to put it in a novel way, the multiple-valued function that takes an ordered pair of complex numbers $(a,b)$ and returns the root(s) of the quadratic equation $x^2 + ax + b = 0$ belongs to the clone generated by $\{0,1,+,-,\times,\div,\sqrt{}\}$. Yes, I know that clones of functions are usually reserved for single-valued functions, but I am generalizing the definition to include partial and multiple-valued functions. Basically, under my generalized definition, a clone of partial/multivalued functions is a class of such functions closed under projections and under composition of functions, similar to the standard single-valued function case.

For cubic equations over the complex numbers, there is also an analogous formula, involving also the multi-valued cube root. But now, here is my question. Suppose we are working in the real numbers $\mathbb{R}$, and I am only interested in the real roots of the cubic equation $x^3 + ax^2 + bx + c = 0$, with $a,b,c$ all real. Does the function that takes the ordered triple $(a,b,c)$ and returns the real root(s) belong to the clone generated by $\{0,1,+,-,\times,\div,\sqrt{}, \sqrt[3]{}\}$, where, in this case, the square root refers to the real number square root function which gives the two roots for positive numbers and is undefined for negative numbers, and the cube root is the single valued cube root function defined all over the real line? My understanding is that it is not, and that is what casus irreducibilis means. Is that true?

Also, bonus question, has anyone generalized the definition of clone of functions to partial/multi-valued functions, like I do here, and developed theory based on it?

Answer 1

Does the function that takes the ordered triple $(a,b,c)$ and returns the real root(s) belong to the clone generated by $\{0,1,+,−,×,÷,\sqrt{\phantom{}},\sqrt[3]{\phantom{}}\}$, where, in this case, the square root refers to the real number square root function which gives the two roots for positive numbers and is undefined for negative numbers, and the cube root is the single valued cube root function defined all over the real line?

The casus irreducibilis is the case of a cubic with a real polynomial and three distinct real roots (and no rational root), so no matter what formula you have for the roots, the roots themselves are all real numbers. And here's the key point: when the roots are distinct and real, the two square roots appearing in the cubic formula are square roots of negative real numbers, so those square roots don't exist in $\mathbf R$. The sum of the terms in the root formulas is a sum of two complex conjugates, so in the sum those imaginary parts cancel out, resulting in real numbers even thought parts of the sum are not real numbers.

In the language of field theory, when $f(x)$ is a cubic irreducible with rational coefficients and all real roots, there is no radical tower over $\mathbf Q$ containing the three roots that lies entirely inside $\mathbf R$ even though the roots themselves are in $\mathbf R$. The dream of trying to express those three roots using only ordinary algebraic operations plus root extractions can't be realized by working only within $\mathbf R$.

Answer 2

Let me add essentially a footnote to KCd's answer to clarify the (non-)role of clones here.

A clone, classically, is a set $\mathscr{F}$ of finite-arity functions on a given "base set" $A$ such that all projections are in $\mathscr{F}$ and $\mathscr{F}$ is closed under multi-ary composition. Generalizing to allow clones to consist of partial multivalued functions is not too common but it is straightforward, so I'll talk about "pmv (functions/)clones" going forward without any fanfare.

Given a set of pmv functions $F$ on a base set $A$, let $\langle F\rangle_A$ be the pmv clone on $A$ generated by $F$. Clones are natural objects, but also quite complicated even when the base set is small and we require totality and single-valuedness (e.g. consider Post's lattice); because of this, I think we should generally make an effort to distinguish between when a problem really is about clones and when they are not actually relevant. In my opinion, the current question is of the latter type, but let's press on.

Suppose we have algebras (in the universal algebra sense) $\mathfrak{A}\subseteq\mathfrak{B}$ and a distinguished set of pmv functions $E$ on $\mathfrak{B}$. Writing $g_\mathfrak{X}$ for the bi-restriction of $g$ to $\mathfrak{X}$ (that is, restrict both domain and codomain), there are two natural ways to "$\mathfrak{A}$-ify" the clone $\langle E\rangle_\mathfrak{B}$:

• The "bottom-up" approach is $$\langle E\rangle_{\mathfrak{A}\subseteq\mathfrak{B}}^-:=\langle\{f_\mathfrak{A}: f\in E\}\rangle_\mathfrak{A}.$$

• On the other hand, we also have the "top-down" approach $$\langle E\rangle_{\mathfrak{A}\subseteq\mathfrak{B}}^+:=\{f_\mathfrak{A}: f\in\langle E\rangle_\mathfrak{B}\}.$$

It's easy to check that these are each pmv clones on $\mathfrak{A}$ and that we're guaranteed to have $\langle E\rangle_{\mathfrak{A}\subseteq\mathfrak{B}}^-\subseteq \langle E\rangle_{\mathfrak{A}\subseteq\mathfrak{B}}^+$ (hence the choice of superscripts); on the other hand, it is meaningful and nontrivial to ask how big a gap - if any - exists between them.

In the present situation, the gap is indeed huge. Specifically, say that a triple $(\mathfrak{A},\mathfrak{B},E)$ as above is super-gappy iff there is some subalgebra $\mathfrak{C}\subseteq\mathfrak{A}$ such that $\mathfrak{C}$ is $\langle E\rangle^-_{\mathfrak{A}\subseteq\mathfrak{B}}$-closed but $\mathfrak{C}$ is not $\langle E\rangle^+_{\mathfrak{A}\subseteq\mathfrak{B}}$-closed. In a super-gappy situation, the gap between the two $\mathfrak{A}$ifications of $\langle E\rangle_\mathfrak{B}$ is so great that it's even visible at the level of individual elements; essentially, all the "higher-type" reasoning that makes clones complicated is superfluous. And that's exactly the situation we have here, with $\mathfrak{B}=\mathbb{C},\mathfrak{A}=\mathbb{R}$, $E=\{0,1,+,\cdot,-,\div,\sqrt{-},\sqrt[3]{-}\}$, and $\mathfrak{C}$ the $E_\mathfrak{A}$-closure of $\mathbb{Q}$. So yes there's a "clone gap," but it's so coarse that it's really better understood by talking about individual elements (and "nice" field extensions).

I would argue as a sort of "closing moral" that clones provide us with a super-heavy-duty language to discuss issues of function closure/definedness/etc. ... which is usually massive overkill for concrete situations. So yes, technically the casus irreducibilis indicates that there is a huge gap between two natural pmv clones, but this is just making the situation more mysterious than it needs to be.