Suppose I have two different ring structures on the same domain $\langle R,+,\cdot,0,1\rangle$, $\langle R,\oplus,\otimes,\bar 0,\bar 1\rangle$ and I throw the structures together into a single common structure
$$\langle R,+,\cdot,0,1,\oplus,\otimes,\bar0,\bar1\rangle$$
where I have all four operations together. Let us assume there is no regular interaction at all between the two ring structures. In this common language, we may form generalized polynomials that mix together all four operations, forming expressions such as $((x\oplus y)\cdot x+y)\otimes 0$. Ultimately, we form the full term algebra consisting of all possible expressions in this language.


**Question 1.** Is there any kind of normal form for such generalized polynomials?


What I am looking for is any kind of systematic simplification, as is possible in the ordinary language of rings, where polynomials can be represented as finite summands of expressions $c\_0+c\_1x+\cdots+c\_nx^n$ and in several variables sums of expressions of the form $\pm x\_0^{k\_0}x\_1^{k\_1}\cdots x\_n^{k\_n}$. In the mixed-language case, we have all four operations, with two additions and two multiplications.


**Question 2.** Is there a normal form for expressions with only one free variable?


One might hope initially to find a normal form consisting of $+$ sums of expressions of the form $p(x)x^k$, where $x^k=x\cdot x\cdots x$ and $p(x)$ is a polynomial in the language with $\oplus,\otimes$. But these expressions do not seem in general to be closed under $\oplus$ or $\otimes$, and so I think it cannot be correct.


**Question 3.** Is there a normal form even for the constant expressions?


For example, the term
$$((((\bar 0+\bar 0)\cdot\bar 1)\oplus 0)\otimes 1)$$
does not seem to simplify in any way, since in each case we are using the wrong constant for the given operation.


I expect that all three questions will have negative answers — in general we shall have only the full term algebra of all mixed terms in this language. But I not really sure how such a thing could be proved, and so my final question is:


**Question 4.** How can one prove that there are no normal forms of the desired form?


In case it helps, in the situation in which I am interested, the rings are fields and indeed algebraically closed fields of characteristic zero. Indeed, each of the two rings is isomorphic to the complex field (and to each other).


Perhaps one might hope to answer question 4 by proving that the only identities that can be proved in the underlying theory, which we might call the theory of two rings, are the trivial identities that follow from direct applications of associativity, commutativity, and distributivity of the separate ring structures. Can one mount such an analysis to show in a robust way that no nontrivial identity is possible?



Normal form for terms in language with two ring structures
There is no normal form besides that any subterm using $+,\cdot,0,1$—treating its subterms whose topmost functions are $\oplus,\otimes,\bar0,\bar1$ as black boxes—can be written as a polynomial (i.e., a possibly empty commutative sum of possibly empty commutative products), and vice versa. As the question puts it, “we shall have only the full term algebra of all mixed terms in this language”. This holds even if each structure is individually isomorphic to the complex field, as requested in the question. The way to show this is to look at the term functions in the following structure $(F,+,\times,0,1,\oplus,\otimes,\bar0,\bar1)$.


Let $(F\_0,+,\cdot,0,1)$ be a copy of $\mathbb C$. Treating the elements of $F\_0$ as formal variables, let $(G\_0,\oplus,\otimes,\bar0,\bar1)$ be the algebraic closure of the rational function field $\mathbb Q(F\_0)$. Treating the elements of $G\_0\let\bez\smallsetminus\bez F\_0$ as formal variables, let $(F\_1,+,\cdot,0,1)$ be an algebraic closure of the rational function field $F\_0(G\_0\bez F\_0)$. Treating the elements of $F\_1\bez G\_0$ as formal variables, let $(G\_1,\oplus,\otimes,\bar0,\bar1)$ be an algebraic closure of $G\_0(F\_1\bez G\_0)$. Etc. Then let $F=\bigcup\_{n\in\omega}F\_n=\bigcup\_{n\in\omega}G\_n$ with the respective structure. (For good measure, we can also define a set $G\_{-1}$ as a fixed transcendence basis of $F\_0$ over $\mathbb Q$.)


Let me define the normal forms more formally. Let $V$ denote the set of variables. Fix a linear order $<$ on the set of all terms over $V$.


If $X$ is a finite set of terms, define $\sum X$ as follows: write $X=\{x\_i:i<n\}$, where $n=|X|$ and $x\_0<\dots<x\_{n-1}$; put $\sum X=0$ if $n=0$, and $\sum X=((\cdots(x\_0+x\_1)+\cdots)+x\_{n-1})$ if $n\ge1$. Similarly, define $\prod X$, $\oplus X$, and $\otimes X$.


Let $T\_+$, $T\_\cdot$, $T\_\oplus$, and $T\_\otimes$ be the smallest sets of terms over $V$ satisfying the following inductive conditions:


* if $X\subseteq V\cup T\_\oplus\cup T\_\otimes$ is finite and $|X|\ne1$, then $\prod X\in T\_\cdot$;
* if $X\subseteq V\cup T\_\cdot\cup T\_\oplus\cup T\_\otimes$ is finite and $|X|\ne1$, then $\sum X\in T\_+$;
* if $X\subseteq V\cup T\_+\cup T\_\cdot$ is finite and $|X|\ne1$, then $\otimes X\in T\_\otimes$;
* if $X\subseteq V\cup T\_\otimes\cup T\_+\cup T\_\cdot$ is finite and $|X|\ne1$, then $\oplus X\in T\_\oplus$.


The set of normal forms is defined as $N=V\cup T\_+\cup T\_\cdot\cup T\_\oplus\cup T\_\otimes$.



> 
> **Proposition:** Every term is equal to exactly one term in normal form.
> 
> 
> 


**Proof sketch:**


Existence: by induction on the complexity of the term. A term whose main function is $+,\cdot,0,1$ can be written as $p(t\_0,\dots,t\_{k-1})$, where $p$ is a polynomial with nonnegative integer coefficients (other than a single variable), and $t\_i$ are either variables or terms whose main functions are $\oplus,\otimes,\bar0,\bar1$. We can write each $t\_i$ as a normal form in $V\cup T\_\oplus\cup T\_\otimes$ by the induction hypothesis, and express $p$ as an appropriately bracketed sum of products in the right order to make it a normal form.


Uniqueness: let $v(t)$ denote a fixed evaluation of terms $t$ in $F$ above, where variables are evaluated by distinct elements of $G\_{-1}$. We want to show that $v(t)\ne v(s)$ if $t,s\in N$ are distinct.


Again, a term in $T\_+\cup T\_\cdot$ is of the form $p(t\_0,\dots,t\_{k-1})$, where $p$ is a polynomial with nonnegative integer coefficients other than a single variable, and $t\_i$ are subterms of $t$ from $V\cup T\_\oplus\cup T\_\otimes$. Using this, and the fact that elements of $\bigcup\_n(G\_n\smallsetminus F\_n)$ are algebraically independent over $\mathbb Q$ in $(F,+,\cdot,0,1)$, as well as the dual statements swapping the role of the two sets of operations, we can prove by induction on the complexity of $t\in N$ that $v(t)\in\bigcup\_n(F\_n\smallsetminus G\_{n-1})$ if $t\in T\_+\cup T\_\cdot$, $v(t)\in\bigcup\_n(G\_n\smallsetminus F\_n)$ if $t\in T\_\oplus\cup T\_\otimes$, and $v(s)\ne v(t)$ for all $s\ne t$ in $N$ whose complexity does not exceed that of $t$.


