Download PDF

explicit definition of polynomial rings in
arbitrarly many variables∗
joking†
2013-03-22 3:39:53
Let R be a ring and let X be any set (possibly empty). We wish to give an
explicit and formal definition of the polynomial ring R[X].
We start with the set
F(X) = {f : X → N | f (x) = 0 for almost all x}.
If X = {X1 , . . . , Xn } then the elements of F(X) can be interpreted as monomials
X1α1 · · · Xnαn .
Now define
R[X] = {F : F(X) → R | F (f ) = 0 for almost all f }.
The addition in R[X] is defined as pointwise addition.
Now we will define multiplication. First note that we have a multiplication
on F(X). For any f, g : X → N put
(f g)(x) = f (x) + g(x).
This is the same as multiplying xa · xb = xa+b .
Now for any f ∈ F(X) define
M (h) = {(f, g) ∈ F(X)2 | h = f g},
Now if F, G ∈ R[X] then we define the multiplication
F G : F(X) → R
by putting
(F G)(h) =
X
F (f )G(g).
(f,g))∈M (h)
∗ hExplicitDefinitionOfPolynomialRingsInArbitrarlyManyVariablesi
created: h2013-03-2i
by: hjokingi version: h42238i Privacy setting: h1i hDefinitioni h12E05i h13P05i h11C08i
† This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
Note that all of this well-defined, since both F and G vanish almost everywhere.
It can be shown that R[X] with these operations is a ring, even an R-algebra.
This algebra is commutative if and only if R is. Furthermore we have an algebra
homomorphism
E : R → R[X]
which is defined as follows: for any r ∈ R let Fr : F(X) → R be the function
such that if f : X → N is such that f (x) = 0 for any x ∈ X, then put Fr (f ) = r
and for any other function f ∈ F(X) put Fr (f ) = 0. Then
E(r) = Fr
is our function, which is a monomorphism. Furthermore if R is unital with the
identity 1, then
E(1)
is the identity in R[X]. Anyway we can always interpret R as a subset of R[X]
if put r = Fr for r ∈ R.
Note, that if X = ∅, then R[∅] is still defined and E : R → R[X] is an
isomorphism of rings (it is ,,onto”). Actually these two conditions are equivalent.
Also note, that X itself can be interpreted as a subset of R[X]. Indeed, for
any x ∈ X define
fx : X → N
by fx (x) = 1 and fx (y) = 0 for any y 6= x. Then define
Fx : F(X) → R
by putting Fx (fx ) = 1 and Fx (f ) = 0 for any f 6= fx . It can be easily seen that
Fx = Fy if and only if x = y. Thus we will use convention x = Fx .
With these notations (i.e. R, X ⊆ R[X]) we have that elements of R[X] are
exactly polynomials in the set of variables X with coefficients in R.
2