You are here
Homepolynomials in algebraic systems
Primary tabs
polynomials in algebraic systems
Polynomials in an Algebraic System
Let $(A,O)$ be an algebraic system. Given any nonnegative integer $n$, let $S$ be a set of functions from $A^{n}$ to $A$, satisfying the following criteria:
1. the projection $\pi_{i}:A^{n}\to A$ is an element of $S$ for each $i$ between $1$ and $n$.
2. for every $0$ary operator symbol $c\in O$, the corresponding element $c_{A}\in A$ is an element of $S$.
3. for any $m$ary operator $\omega_{A}$ on $A$, and any $p_{1},\ldots,p_{m}\in S$, the function $\omega_{A}(p_{1},\ldots,p_{m}):A^{n}\to A$, given by
$\omega_{A}(p_{1},\ldots,p_{m})(a_{1},\ldots,a_{n}):=\omega_{A}(p_{1}(a_{1},% \ldots,a_{n}),\ldots,p_{m}(a_{1},\ldots,a_{n}))$ is an element of $S$.
Take the intersection of all such sets and call it $\textbf{P}_{n}(A)$. Then $\textbf{P}_{n}(A)$ is the smallest set satisfying the above two conditions. We call an element of $\textbf{P}_{n}(A)$ an $n$ary polynomial of the algebra $A$. A polynomial of $A$ is some $n$ary polynomial of $A$. Two polynomials $p,q$ are said to be equivalent if they have the same arity $n$, and for any $\textbf{a}\in A^{n}$, $p(\textbf{a})=q(\textbf{a})$.
Examples.

Let $G$ be a group. Then $f(x)=x$ is a unary polynomial. In general, a unary polynomial of $G$ has the form $x^{n}$, $n\geq 1$. For a binary polynomial, first note that $f(x,y)=x$ and $g(x,y)=y$ are both binary. For this, a binary polynomial of $G$ looks like
$x^{{n_{1}}}y^{{m_{1}}}\cdots x^{{n_{k}}}y^{{m_{k}}},$ where $n_{1},m_{k}\geq 0$ and the rest are positive integers. If $G$ is abelian, a binary polynomial has the simplified form $x^{n}y^{m}$, and more generally, an $n$ary polynomial of an abelian group has the form
$x_{1}^{{n_{1}}}\cdots x_{k}^{{n_{k}}}.$ 
Let $R$ be a ring. $f(x)=x$ is a unary polynomial, so are $g(x)=nx$ and $h(x)=x^{m}$ where $n,m$ are nonnegative integers. In general, a unary polynomial of $R$ looks like
$n_{m}x^{{m}}+n_{{m1}}x^{{m1}}+\cdots+n_{1}x+n_{0},$ where $n_{i}$ are nonnegative integers with $n_{m}>0$, and $m$ is any positive integer. This is the form of a polynomial that is familiar to most people.

Continuing from the example above, note that, however,
$a_{m}x^{{m}}+a_{{m1}}x^{{m1}}+\cdots+a_{1}x+a_{0}$ where $a_{i}\in R$ is in general not a polynomial under this definition. This is an example of an algebraic function in an algebraic system.
Remarks.

$\textbf{P}_{0}(A)$ is nonempty iff $O$ contains constant symbols (nullary function symbols).

By virtue of the definition, $\textbf{P}_{n}(A)$ is an algebraic system for each nonnegative integer $n$.
Polynomial Symbols
If we have two algebras $A,B$ of the same type, then, by the definition above, a polynomial of $A$ may be considered as a polynomial of $B$, and vice versa. Putting this formally, we introduce polynomial symbols for a class $K$ of algebras of the same type $O$. Let $X=\{x_{1},x_{2},\ldots\}$ be a countably infinite set of variables, or symbols. For any nonnegative integer $n$, consider a set $S$ consisting of the following elements:
1. $x_{1},\ldots,x_{n}$,
2. every nullary operator symbol,
3. if $\omega\in O$ is $m$ary and $p_{1},\ldots,p_{m}\in S$, then $\omega(p_{1},\ldots,p_{m})\in S$.
Take the intersection of all such sets and we end up with a set satisfying the two conditions again, call it $\textbf{P}_{n}(O)$. Any element of $\textbf{P}_{n}(O)$ is called an $n$ary polynomial symbol of $K$. A polynomial symbol of $K$ is just some $n$ary polynomial symbol. In model theory, a polynomial symbol is nothing more than a term over $X$ in the $O$language.
Now, suppose $A\in K$. For a given nonnegative integer $n$, consider the function $f_{A}:\textbf{P}_{n}(O)\to\textbf{P}_{n}(A)$, defined recursively by
1. $f_{A}(x_{i}):=\pi_{i}$,
2. $f_{A}(c):=c_{A}$,
3. $f_{A}(\omega(p_{1},\ldots,p_{n})):=\omega_{A}(f_{A}(p_{1}),\ldots,f_{A}(p_{n}))$.
Then by the constructions of $\textbf{P}_{n}(O)$ and $\textbf{P}_{n}(A)$, $f_{A}$ is a surjection. Any polynomial $p$ of $A$ is said to be induced by a polynomial symbol $q$ of $K$ if $f_{A}(q)=p$. We usually write $p=q_{A}$ to denote that the polynomial $p$ (of $A$) is induced by the polynomial symbol $q$ (of $K$). It is not hard to see that, between two algebras of the same type, there is a onetoone correspondence between their respective polynomials the same arities. However, equivalence of polynomials in one algebra does not translate to equivalence in another.
Example. $x\vee(y\wedge z)$ and $(x\vee y)\wedge(x\vee z)$ are both polynomials symbols in the class of lattices. In the subclass of distributive lattices, the induced polynomials are equivalent. However, in the subclass of modular lattices, the equivalence no longer holds.
Remark. Like $\textbf{P}_{n}(A)$, $\textbf{P}_{n}(O)$ is also an algebraic system, called the $n$ary polynomial algebra, or absolutely free algebra, of $O$. Furthermore, the function $f_{A}$ above is an algebra homomorphism.
References
 1 G. Grätzer: Universal Algebra, 2nd Edition, Springer, New York (1978).
 2 S. Burris, H.P. Sankappanavar: A Course in Universal Algebra, Springer, New York (1981).
Mathematics Subject Classification
08A40 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections