1
$\begingroup$

Let $R$ is a ring and $x_1, x_2, \cdots x_n$ are $n$ variables. Now let us define a polynomial ring $R[x_1,x_2\cdots,x_n]$ be a collections of multivariate polynomials where coefficients will come from ring $R$.

Is it a right way to define polynomial rings ? Is there a condition on variables also.

For example :

  1. Is $R[x_1,x_2]$ a polynomial ring ? where $x_1 = x_2^{2}$
  2. Is $R[x_1,x_2]$ a polynomial ring ? where sin$ x_1 = $ cos $x_2$
$\endgroup$
1
  • 1
    $\begingroup$ The moment you give algebraic relations between the variables, the ring ceases to be a polynomial ring, at least in general. $x_1=\cos x_2$ is not an algebraic relation, so there I don't know. $\endgroup$
    – Arthur
    Commented Aug 9, 2017 at 12:51

2 Answers 2

2
$\begingroup$

Yes, your original description is basically correct. You can describe the polynomial ring over $R$ having indeterminates $x_1,\ldots ,x_n$ as formal linear combinations of monomials with coefficients from the ring $R$.

But what you describe in 1) and 2) is different from that.

  1. $R[x_1,x_2]$ "where $x_1=x_2^2$" is better known as the quotient ring $R[x_1,x_2]/(x_1-x_2^2)$ of the polynomial ring $R[x_1,x_2]$. This is different from the full set of linear combinations because using the relation, lots of the monomials collapse.

  2. Is not even well defined, since $x_1$ is just a symbol for an indeterminate, not a real number, or any number $\sin()$ is defined on. A relation between $x_1$ and $x_2$ can only be established using the operation that are available in $R[x_1,x_2]$, and the only two operations available are addition and multiplication.

A defining feature of polynomial rings is that the lack extra relations like $x_1=x_2^2$. Of course, it is very useful to study such rings, and they are studied as quotients of polynomial rings, but the concept of the big thing that is totally free of any relations is the important thing to establish as a creature of its own. Because polynomial rings are free of relations, they are also called "free $R$ algebras in commuting indeterminates $x_1,\ldots, x_n$."

$\endgroup$
1
$\begingroup$

No, there are no relations among variables $x_i$. Polynomial ring is freely generated commutative $R$-algebra with generators $\{x_1,\ldots,x_n\}$. This is similar to basis of vector space, however there is difference since you will not get all polynomials as linear combinations of generators, but you have to multiply them as well. Still, if you consider this analogy, you have that basis for vector space must be linear independent; here, however, you have to consider multiplication as well, so there are no polynomial relations among generators, i.e. something like $x_1 = x_2^2$ can't hold by definition.

That said, your first example still makes sense, and I will get back to it, but your second does not. It seems that you are treating variables as if they were real/complex numbers. That is not the case. Expressions like $\sin x_1$ don't even make sense because $\sin$ is not a polynomial. It's like trying to multiply general vectors, there is no structure to support the notion.

But let us get back to your second example. What you described is actually quotient ring $R[x_1,x_2]/(x_1-x_2^2)$, which is no longer polynomial ring, because it is not free, i.e. there are relations among the generators.

If you are familiar with free groups, this behaves the same way. Free groups are those that do not have relations among their generators. As soon as you impose some relation, you get corresponding quotient group, which is no longer free.

$\endgroup$
2
  • $\begingroup$ +1 sorry, we gave almost the same answer in parallel, but I think it's good to see both versions :) $\endgroup$
    – rschwieb
    Commented Aug 9, 2017 at 12:58
  • $\begingroup$ @rschwieb, agreed, +1 to your answer as well. $\endgroup$
    – Ennar
    Commented Aug 9, 2017 at 13:00

You must log in to answer this question.