# Rational functions in one indeterminate - useful concept?

## Summary:

Is the concept of rational functions in one indeterminate useful in algebra?

## Main Question or Discussion Point

The examples of "formal" power series and polynomials in one indeterminate are familiar and useful in algebra. However, I don't recall the example of rational functions (ratios of polynomials) in one indeterminate being used for anything. Is that concept useful? - or trivial? -or equivalent to something that's called by a different name?

Related Linear and Abstract Algebra News on Phys.org
fresh_42
Mentor
2018 Award
What do you mean by "useful"?

What are quotient fields of integral domains good for?
What are transcendental field extensions good for?
Where are rational transformations $t \longmapsto \dfrac{ax+b}{cx+d}$ used?

What do you mean by "useful"?
By useful, I mean of sufficient instructive value to be used in texts on algebra as an interesting example of some well known abstract structure - such as a ring or module or field.

What are quotient fields of integral domains good for?
Are you accusing me of being interested in applied mathematics? I am, but my question isn't about practical applications.

fresh_42
Mentor
2018 Award
I do not accuse anybody. I just don't understand the question. What is Lüroth's theorem good for? It exists. And quotient fields also exist, so does $\mathbb{Q}(x)$. Same with transcendental extensions in contrast to algebraic extensions. They are algebraically useful for their insertion homomorphisms $x \longmapsto \vartheta$. The restriction on one variable as well as on $\mathbb{Q}$ is what narrows down the usefulness. Otherwise we would enter algebraic geometry and general quotient rings.

I just don't understand the question.
I don't understand the replies formulated as questions. I can't tell whether your point is ask about the utility of algebraic structures in general or whether you mean each of your questions to hint at a case where rational functions in one indeterminate are a good example. If you can confirm that each of your questions is meant to hint at a use of rational functions in one indeterminate as an example , I can probably look up the details.

fresh_42
Mentor
2018 Award
There isn't much to it. You said rational functions, i.e. quotients of polynomials. Now we have
$$\underbrace{\mathbb{Q}[x]}_{integral\;domain} \subseteq \underbrace{\mathbb{Q}(x)}_{quotient \;field}=\underbrace{\mathbb{Q}[x,x^{-1}]}_{transcendental \;extension}$$
It is one specific example of a field. One can generalize it by changing the field of rationals to arbitrary rings, by more than one variable, by algebraic extensions instead of transcendental, by series instead of polynomials. But as is, it is just a certain field. To ask about its uses is similar to ask about the use of any other field.

Math_QED
Homework Helper
What do you mean with rational functions? Quotients of polynomials? If yes, here are some applications:

- Möbius transformations in the complex plane.

- Rational functions are widely used in algebraic geometry in the theory of local rings to describe the local behaviour of curves (think about intersection numbers).

- Quotient fields give a way to canonically embed any domain in a field. This is often useful! If you want to prove something about a domain, embed it in its quotient field and use that you have more structure there to get information on the original ring. Gauss' lemma in ring theory comes to mind.

Last week I was wondering if we can embed every ring in a semisimple ring. For domains, this is easy because you can embed a domain in their fields of fractions. This made me realise I needed to find a non-domain if the answer is negative. That way, I quickly came up with the counterexample $\mathbb{Z}/4 \mathbb{Z}$ so yes, these fraction fields have good advantages and give good examples of rings.

Last edited:
What do you mean with rational functions? Quotients of polynomials?
As mentioned in the OP, I don't mean quotients of polynomials that are functions. I mean quotients of polynomials that are regarded only as "expressions" - the so-called polynomials "in one indeterminate". For example, the function $f(x) = \frac{(x-3)(x-2)}{(x-6)}$ does not have a multiplicative inverse function that is defined on the same domain. However, the formal expression $f([x]) = \frac{([x] -3)([x]-2)}{([x]-6)}$ has the multiplicative inverse expression $f[x] = \frac{([x]-6)}{([x] -3)([x]-2)}$. There is no worry about dividing by zero, because the operations are defined as algorithms applied to symbolic expressions - the way a naive algebra student would manipulate them.

fresh_42
Mentor
2018 Award
As mentioned in the OP, I don't mean quotients of polynomials that are functions. I mean quotients of polynomials that are regarded only as "expressions" - the so-called polynomials "in one indeterminate". For example, the function $f(x) = \frac{(x-3)(x-2)}{(x-6)}$ does not have a multiplicative inverse function that is defined on the same domain. However, the formal expression $f([x]) = \frac{([x] -3)([x]-2)}{([x]-6)}$ has the multiplicative inverse expression $f[x] = \frac{([x]-6)}{([x] -3)([x]-2)}$. There is no worry about dividing by zero, because the operations are defined as algorithms applied to symbolic expressions - the way a naive algebra student would manipulate them.
You don't need absolute values. It works for $f(x)=\dfrac{(x-3)(x-2)}{x-6}$ as well, and it is what I described in post #6. It's a transcendental extension of degree one, the quotient field of the integral domain, the polynomial ring in one variable. @Math_QED didn't mean functions, except perhaps the Möbius transformation. As it is a quotient field, it is a standard example how they are created. The way from integers to rationals is the same. $x$ isn't a variable here. it is an element of a ring, resp. a field.

You don't need absolute values.
Agreed. I typed square brackets to hint that "$[x]$" is not a numerical argument.

As I understand the notation "$\mathbb{Q}[x]$" and "$\mathbb{Q}[\sqrt{2}]$" can denote distinct concepts. With "$\mathbb{Q}[\sqrt{2}]$" there is the fact that $(\sqrt{2})(\sqrt{2}) = 2 \in \mathbb{Q}$. However, with "$\mathbb{Q}[x]$" there is no implication that some polynomial in $x$ can be identified with an element of $\mathbb{Q}$. Is that correct?

fresh_42
Mentor
2018 Award
No, both are rings, so the concept is the same. $\mathbb{Q}[\alpha]$ denotes the ring of rational expressions $a_0 \alpha^n+\ldots +a_{n-1}\alpha+a_{n}$ regardless whether $\alpha = \sqrt{2}$ or $\alpha=x$. In the first case we have an algebraic extension of the rationals where $\alpha^2-2=0$ holds, in the second a transcendental where no such algebraic identification exists. We can as well write $\mathbb{Q}[x]\cong\mathbb{Q}[\pi]\cong\mathbb{Q}[e]$.

In the first case we have an algebraic extension of the rationals where $\alpha^2-2=0$ holds,
That's what I was asking about. The notation "$Q[\sqrt{2}]$" conveys to the reader that the special condition $\alpha^2 -2 = 0$ is understood to hold.

Although $\mathbb{Q}[x]$ is distinct from a set of functions, it is still possible to think of composing members of $\mathbb{Q}[x]$ using the manpulations from elementary algebra that implement substituting one expression in for another. What is the proper term for $\mathbb{Q}[x]$ with this added structure? Or am I being too optimistic that that composition of members of $\mathbb{Q}[x]$ can be well defined?

fresh_42
Mentor
2018 Award
I do not understand what you mean. There is an "insertion" ring homomorphism
$$\varphi_r\, : \,\mathbb{Q}[x] \longrightarrow \mathbb{Q}[r]$$
where the polynomials are evaluated at $x=r\in \mathbb{Q}$.

If in the previous example with $\alpha^2-2=0$, then we can consider the ideal $\mathcal{I}\subseteq \mathbb{Q}[x]$ generated by the polynomial $p(x)=x^2-2$. In this case we get an isomorphism
$$\mathbb{Q}[x]/\mathcal{I}=\mathbb{Q}[x]/\langle x^2-2\rangle \cong \mathbb{Q}[\sqrt{2}]=\mathbb{Q}[\alpha]$$
The (principle) ideal $\mathcal{I}$ consists of all polynomials of $\mathbb{Q}[x]$ which are divisible by $x^2-2\,.$ So in a way $\mathbb{Q}[x]$ is the special case where $p(x)=0$. For $p(x)=1$ we get $\{\,0\,\}$, and for $p(x)=x$ we get $\mathbb{Q}$. This all is basic algebra, or commutative algebra in this case.

Math_QED
Homework Helper
Agreed. I typed square brackets to hint that "$[x]$" is not a numerical argument.

As I understand the notation "$\mathbb{Q}[x]$" and "$\mathbb{Q}[\sqrt{2}]$" can denote distinct concepts. With "$\mathbb{Q}[\sqrt{2}]$" there is the fact that $(\sqrt{2})(\sqrt{2}) = 2 \in \mathbb{Q}$. However, with "$\mathbb{Q}[x]$" there is no implication that some polynomial in $x$ can be identified with an element of $\mathbb{Q}$. Is that correct?
I think it is clearer to introduce this notation in a more formal setting. Let $K,F$be fields with $F\subseteq K$. Let $\alpha\in K$.

Then $F[\alpha]$ is by definition the smallest subring of $K$ containing $\alpha$ and $F$.

So, it really can be thought of as: adjoin $`\alpha\in K$ to $F$ and see what ring it generates.

In our case,we just add $\sqrt{2}$ to $\mathbb{Q}$ and see what the smallest ring is containing both $\mathbb{Q}$ and $\sqrt{2}$.

One can prove that:

$$F[\alpha]=\{f(\alpha)\mid f\in F[X]\}$$

and you can then visualise this ring as
a polynomial ring $F[X]$ with $X$ replaced by $\alpha$, but with possibly more relations! Hence the notation we give it makes sense. At least that'the way I view it.

I do not understand what you mean. There is an "insertion" ring homomorphism
$$\varphi_r\, : \,\mathbb{Q}[x] \longrightarrow \mathbb{Q}[r]$$
where the polynomials are evaluated at $x=r\in \mathbb{Q}$.
I was thinking about a substitution that amounts to composing functions. For example, $f(x) = x^3 + 2x + 1, g(x) = x^2 - 6$ , $f(g(x)) = g(x)^3 + 2(g(x)) + 1$ = some 6th degree polynomial.

I think it is clearer to introduce this notation in a more formal setting. Let $K,F$be fields with $F\subseteq K$. Let $\alpha\in K$.

Then $F[\alpha]$ is by definition the smallest subring of $K$ containing $\alpha$ and $F$.
I agree that giving all the background information is clearer. As I interpret tradition, it's traditional to leave it up to the reader to supply the identity of $K$ based on inferring what familiar field contains $\alpha$.

Math_QED
Homework Helper
I agree that giving all the background information is clearer. As I interpret tradition, it's traditional to leave it up to the reader to supply the identity of $K$ based on inferring what familiar field contains $\alpha$.
Yes, because it doesn't matter, in the following sense:

Let $F \subseteq K \subseteq L$ and $\alpha \in K$. Then you can form $F_K[\alpha], F_L[\alpha]$ as I explained earlier (in the fields $K,L$). You can then show that $F_K[\alpha] = F_L[\alpha]$, so the notation $F[\alpha]$ is justified.

So, whether you consider $\sqrt{2}\in \mathbb{R}$ or $\sqrt{2}\in \mathbb{C}$ or $\sqrt{2} \in \mathbb{C}(X)$, it won't affect what $\mathbb{Q}[\sqrt{2}]$ is.

Let F⊆K⊆LF \subseteq K \subseteq L and α∈K\alpha \in K.
I think of "$\mathbb{Q}[x]$" as denoting a ring of polynomials and "$\mathbb{Q}[\sqrt{2}]$" as denoting a field. Is there a feature of the notation that I'm missing which distinguishes between ring and field?

If I consider "$[3]$" to denote a member of $\mathbb{Z}_7$ then (to me) the notation "$\mathbb{Q}[[3]]$" conveys the idea of forming all possible polynomials in powers of $[3]$ and then simplifying the powers according to the arithmetic of $\mathbb{Z}_7$. So I'm trying to extend the field $\mathbb{Q}$ by an element that is not in a field containing $\mathbb{Q}$. Is there special notation or terminology for this type of extension?

fresh_42
Mentor
2018 Award
I think of "$\mathbb{Q}[x]$" as denoting a ring of polynomials and "$\mathbb{Q}[\sqrt{2}]$" as
$\mathbb{Q}[\sqrt{2}]$ is constructed as ring, yes, which is why the brackets are squarish. But
$$\dfrac{1}{\sqrt{2}} = \dfrac{1}{2} \cdot \sqrt{2}$$
implies, that it is at the same time a field, i.e. $\mathbb{Q}[\sqrt{2}]=\mathbb{Q}(\sqrt{2})$ which is not the case for transcendental extensions.

IIRC the notation $\mathbb{Q}[[\cdot]]$ is reserved for power series. And you must not switch between different characteristics. How should arithmetic be possible in such a set?

Math_QED
Homework Helper
I think of "$\mathbb{Q}[x]$" as denoting a ring of polynomials and "$\mathbb{Q}[\sqrt{2}]$" as denoting a field. Is there a feature of the notation that I'm missing which distinguishes between ring and field?
Let's go back to our formal setting.

$F[\alpha]$ is by definition the smallest subring of $K$ containing $F$ and $\alpha$
$F(\alpha)$ is by definition the smallest subfield of $K$ containing $F$ and $\alpha$.

It is well known that $F[\alpha] = F( \alpha)$ if and only if $\alpha$ is algebraic over $F$, i.e. $\alpha$ is a root of a non-zero polynomial with coefficients in $F$.

So, you are right $\mathbb{Q}[\sqrt{2}]$ is a field, because $\sqrt{2}$ is algebraic over $\mathbb{Q}$ (it is a zero of $X^2-2$). Thus $\mathbb{Q}[\sqrt{2}]= \mathbb{Q}(\sqrt{2})$.

As @fresh_42 correctly pointed out, it is in general not true that $F[\alpha]$ is a field. Indeed, $\mathbb{Q}[\pi]$ is not a field because $\pi$ is not algebraic over $\mathbb{Q}$.

As @fresh_42 correctly pointed out, it is in general not true that $F[\alpha]$ is a field. Indeed, $\mathbb{Q}[\pi]$ is not a field because $\pi$ is not algebraic over $\mathbb{Q}$.
A conceptual distintion that I see between extending $\mathbb{Q}$ by an "indeterminate" and extending it by a transcendental number is that extending by an indeterminate opens the possibility of defining an operation on two elements of $\mathbb{Q}[x]$ that corresponds to the composition of two functions, as mentioned in post #15. In $\mathbb{Q}[\pi]$ it would be awkward to think of substituting an arbitrary element of $\mathbb{Q}[\pi]$ for $\pi$.

The Wikipedia article on "composition ring" https://en.wikipedia.org/wiki/Composition_ring includes a polynomial ring $R[x]$ as an example. However, I think the article does not define composition in a technically correct way. The article treats the elements of $R[x]$ as if they are necessarily functions. Given how tedious it is to define operations in $R[x]$ without assuming the elements are functions, I forgive the authors.

mathwonk
Homework Helper
A few remarks: if f(x,y) is an irreducible polynomial with complex coefficients, then the locus f=0 defines a plane curve N in C^2, whose ring of polynomial functions equals the integral domain C[x,y]/(f). I.e. this is the ring of polynomial functions on the plane restricted to N, so that those which equal zero on N are set equal to zero. The field of fractions of this domain is called the field of rational functions on N. This field is isomorphic to the field of meromorphic functions on the Riemann surface of N, i.e. the compactified desingularization of N. It is a theorem that a compact non singular connected one dimensional complex manifold is determined up to isomorphism by its field of meromorphic functions.

The special case where f = y, gives the field C(x), rational functions of one variable. the associated Riemann surface is the Riemann sphere, hence a Riemann surface is isomorphic to the Riemann sphere if and only if its field of meromorphic functions is isomorphic to the field C(x). An irreducible plane curve N is called rational if its field of rational functions is isomorphic to C(x), if and only if its Riemann surface is isomorphic to the Riemann sphere.

So fields of rational functions are a fundamental invariant of Riemann surfaces, and the simplest example is C(x). Notice that for almost any function f as above, (except for f = x-a), the field C(x) is a subfield of the fraction field of C[x,y]/(f), corresponding to the fact that we can project a plane curve (other than a vertical line) surjectively onto the x axis. Thus it is natural to study any Riemann surface via such a projection, making the surface a finite cover of the Riemann sphere (the compactified x axis). Correspondingly, any rational function field of any plane curve is a finite algebraic extension of the basic field C(x). In this way the field C(x) is a very basic tool for studying all Riemann surfaces. Riemann himself for example proved that if a Riemann surface S has topological genus g, and if d is the smallest integer d ≥ (g/2)+1, then S is a finite cover of the Riemann sphere of degree d. Equivalently, the rational function field of S is an algebraic extension of C(x) of degree d.

E.g. if S is a torus of genus one, then S is a double cover of the Riemann sphere, (branched at 4 points). Equivalently a rational function field of genus one is a quadratic extension oif the field C(x). E.g. the elliptic curve y^2 = x(x^2-1), has rational function field generated by x and y, where y is quadratic over C(x), by virtue of this defining equation. But quadratic rational function fields can have any genus g, by considering curves with equations of form y^2 = x(x-1)(x-2).....(x-(2g+1)).

Remark: Thus some curves of genus g are 2-1 covers of the sphere, but the curve x^4+y^4 = 1 of genus 3, is a 3-1 cover but not a 2-1 cover of the Riemann sphere. I.e. the curves of genus g listed 2 or 3 lines above, are all double covers of the sphere, hence are not general curves of genus g, but special ones.

Last edited:
mathwonk
Homework Helper
one more remark about the usefulness of rational functions as opposed to everywhere regular functions (polynomials). Although rational mappings are not well defined at the zeroes of their denominators, this can actually be an advantage! Consider the polynomial mapping taking t to (t^2,t^3), from the t- line to the plane cubic curve y^2 = x^3. This curve is not smooth but has a "cusp" at the origin, where the curve comes to a point. Still the map is bijective from the smooth t-line to the curve. The cusp prevents the map from having a regular inverse, but nonetheless it has a rational inverse defined by sending (x,y) to y/x. The fact that this map is not well defined at (0,0) is essential, and it still defines an inverse to the original map, everywhere one exists. In general such rational maps are used to desingularize curves with singularities, by mapping them to curves without singularities, in a way which is as close as possible to an isomorphism, but of course necessarily not quite an isomorphism.

here is another similar one but with a small difference: map t to (t^2-1, t(t^2-1)), to parametrize the curve y^2 = x^2(x+1), also by the t-line. This curve is also singular but has a node at the origin, i.e. it crosses itself transversely there and there are two points of the t-line which map to the point (0,0), namely both t=1 and t= -1. The "inverse map" again takes (x,y) to y/x, and is not defined at (0,0) precisely because it is impossible, by a regular map, to send (0,0) back to both its preimages. Nonetheless this crucial feature allows us to desingularize this curve by this inverse map, the closure of whose image is the non singular t-line. The inverse map is a bijection from the curve minus the origin, to the open complement of the two points {1, -1} in the t-line. In general a desingularization of a singular curve is defined as a regular "proper" map from a smooth curve onto the singular curve, and that is isomorphic except over the singular points. But how to find such a map? I.e. how do we know which non - singular curve to use, since it is only the singular curve we have our hands on? We do this by constructing a sequence of rational maps from our singular curve, until we at last find a non singular image curve. This is called "blowing up" singularities, and the rational maps can be constructed by using projections from the singularities on a singular curve embedded in high dimensional space.

Notice that projection of the projective (x,y,z) plane to the (x,y) axis, a projective line, is not defined at the point (0,0,1) in the plane, since (0,0) is not a point in the projective line, so rational maps give an algebraic realization of the geometric process of projection.

E.g. the map of the nodal plane curve y^2 = x^2(x+1) to a line, can be realized geometrically by sending a point (x,y) away from (0,0) to the line through the origin spanned by the origin and the point (x,y). This maps the complement of the origin in the affine plane, to the projective line, and one can see that as we approach the origin along either branch of the node, the limiting value of the map, i.e. of the secant lines dfetermined by (0,0) and (x,y), is the corresponding tangent line to the nodal curve at the origin. Since there are two such tangent lines, we get in the limit a map sending the origin to two different points of the t- line, depending on which direction we approach the origin along the singular curve. Since t = y/x, the affine coordinate t of course only parametrizes the affine subset x≠0 of the projective (x,y) line. Note also that y/x, i.e. the slope, is a natural coordinate for the line through (x,y) and (0,0).

Note also that the two examples of singular plane curves had parametrizations which failed to have a regular inverse in different ways. The parametrization of the cusp was bijective but at t=0 the velocity vector was zero, so it was not locally invertible there. the parametrization of the node had two preimages of the origin, hence was not globally invertible, but at both of these preimages the velocity vector of the parametrization map was non zero. To have a regular inverse, a map must be both globally and locally invertible, but the rational inverses we found existed in both cases, i.e. rational maps solve both types of problems of non invertibility, which is rather wonderful!

Last edited:
mathwonk