As far as I can tell, a definition of algebraic function field (see below) which is commonly used suggests the following definition for the polynomial ring $R[x]$ of a ring $R$:

A polynomial ring $R[x]$ is the "extension ring" (free ring?) of $R$ with respect to some element $x \not\in R$ which is transcendental over $R$.

Question: how can we rigorously show that for any ring $R$ there exists some element/object $x$ which is transcendental over it, allowing the formation of a polynomial ring?

My attempt: Consider the following definition of a polynomial ring $R$ in terms of a universal property (see e.g. Lemma 21.3 here) -- I will try to derive the result using this definition:

Given a (commutative) ring $R$ (with unit), a polynomial ring over $R$ is a triple $(T, x, \iota)$ with $T$ a (commutative) ring (with unit), $x \in T$, $\iota: R \to T$ a (commutative) ring (with unit) homomorphism such that the following holds: for every $(S, s, \varphi)$, $S$ again a suitable ring, $s\in S$, $\varphi$ a suitable homomorphism, there exists a unique suitable ring homomorphism $\psi: T \to S$ such that $\psi \circ \iota = \varphi$ and $\psi(x)=s$.

Two obvious conclusions from this definition are that (1) $\iota$ is a monomorphism (take $(S,s,\varphi)=(R,1_R,id_R)$ then $\psi$ is just the homomorphism generated by $\psi(x)=1_R$ and the fact that $\psi$ is surjective and $\psi \circ \iota =id_R$ implies that $\iota$ is injective) and that (2) $(T,x,\iota)$ is an initial object in some category, and therefore must be unique up to canonical isomorphism.

I want to show that this definition via universal property implies that $x$ is transcendental over $R$ (or more rigorously speaking that $x$ is transcendental over $\iota(R)$, the image of $R$ under $\iota$), because then any construction satisfying this universal property (which I already have) will imply the existence of an element $x$ which is transcendental over $\iota(R)$.

The obvious approach is to assume by means of contradiction that $x$ is algebraic over $\iota(R)$ and then show that this causes the universal property to fail for some appropriate choice of $(S,s,\varphi)$ -- my guess then would be to choose some $S$ which is an appropriate algebraic extension of $R$, but then how do you define an algebraic extension of $R$ without resorting to $T$ and using circular reasoning?

Anyway, if we get this to work, then we also get that any polynomial ring over $R$ is canonically isomorphic to the "transcendental extension" definition of polynomial ring.

Background: Consider the following definition of an algebraic function field (Stichtenoth: Algebraic Function Fields and Codes, p. 1):

An algebraic function field $F/K$ of one variable over the field $K$ is an extension field $F \supseteq K$ such taht $F$ is a finite algebraic extension of $K(x)$ for some element $x \in F$ which is transcendental over $K$... The simplest example of an algebraic function field is the rational function field; $F/K$ is called rational if $F=K(x)$ for some $x \in F$ which is transcendental over $K$. Each element $0 \not=z \in K(x)$ has a unique representation $$z = a \cdot \prod_i p_i(x)^{n_i} $$ in which $0 \not= a \in K$, the polynomials $p_i(x) \in K[x]$ are monic, pairwise distinct and irreducible and $n_i \in \mathbb{Z}$.

  • 2
    $\begingroup$ Just to clarify: what you are trying to do is to construct the polynomial ring over a given ring. Talking about "existence of element transcendental over R" already assumes an embedding in some larger ring. $\endgroup$ – xyzzyz Oct 18 '16 at 17:22
  • $\begingroup$ @xyzzyz Yes, I am fine with assuming the existence of a larger ring in which $R$ can be embedded (although now that I think about it that is probably just as problematic), I just want to show that one such ring has to exist that contains an element transcendental over $R$, and thus can be used to construct the polynomial ring over $R$ (in the sense of both definitions, I think). $\endgroup$ – Chill2Macht Oct 18 '16 at 17:26
  • $\begingroup$ To show that "one such ring" exists presumably requires a construction that's equivalent to your original question in the non-already-embedded case. $\endgroup$ – Greg Martin Oct 18 '16 at 17:34

Maybe the following paragraph will clear things up.

Let $R$ be a ring. Consider the set $R[x]$ of $\mathbb{N}$-indexed sequences of elements of $R$ with finite support, i.e. if $(a_n) \in R[x]$ then $a_n = 0$ for $n$ big enough. Endow this set with pointwise addition, and the multiplication of $(a_n)$ and $(b_n)$ is given by $(c_n)$ where $c_n = \sum_{i+j=n} a_i b_j$. Then this is a ring, and $R$ embeds in $R[x]$ by sending $a \in R$ to $(a, 0, 0, \dots)$. It is now rather elementary to show that the element $x = (0,1,0,0,\dots) \in R[x]$ is transcendental over $R$.

tl;dr: $R[x]$ is not defined as being "generated by a transcendental element"... You can just define $R[x]$ in the usual fashion, and then you can prove that $x \in R[x]$ is transcendental. Not the other way around.

What the sentence you quoted actually means is the following: given any $R$-algebra $A$ and any element $a \in A$ transcendental over $R$, there exists a unique morphism of $R$-algebras $R[x] \to A$ such that $x \mapsto a$, and moreover this morphism is an embedding. So in a sense you take $R$, "add a transcendental element", and you get $R[x]$. But it's just a heuristic.

In other words, $R[x]$ is initial in the category of pointed $R$-algebras. This seems to be what you had in mind when you wrote your definition of "polynomial rings over $R$". But as usual, when you define something as an initial object in some category, you have to show that it exists... And there's no better way to show that something exists than to construct it. Defining something as the initial object of a category is not a definition until you can prove that the initial object exists.

  • $\begingroup$ Yes this makes sense -- sorry about all of the confusion $\endgroup$ – Chill2Macht Oct 18 '16 at 18:03
  • 1
    $\begingroup$ @William Don't worry -- sometimes one gets lost in the depths of abstract nonsense and forgets that we're still talking about things that make sense! :) $\endgroup$ – Najib Idrissi Oct 19 '16 at 8:40

What you can do is simply this:

  1. Start with a ring $R$.
  2. Consider a free $R$-module with basis $x^n$, $n = 0, 1, \ldots$, call it $S$.
  3. Define a ring structure on $S$ in the obvious way.

You can now show that $S$ satisfies the universal property, and that $x$ as an element of $S$ is transcendental over $R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.