Given a commutative ring $R$, the polynomial ring in one variable $R[x]$ can be defined as the set of all the formal expressions $a_0+a_1x+\cdots+a_nx^n$ with 'obvious' rules of addition and multiplication.

What exactly do we mean by a 'variable' here is not very clear though. My main question is the following:

Qustion 1. Does it mean anything to say that $ax=xa$ in $R[x]$ for all $a\in R$?

When we talk about multivariable polynomial rings, this approach becomes cumbersome. Even when talking about $R[x, y]$, the multilpication seems rather artificial.

Further, we also have an isomorphism $R[x][y]\cong R[x, y]$. This is making me a bit uncomfortable:

Question 2. In $R[x][y]$, it seems a bit odd to write $xy=yx$ (See Question 1) but we do certainly want to write this.

I know these questions are rather vague. So finally I can ask this: Is there a better way to think about polynomial rings? Also, can we intrinsically define what a variable is?


  • $\begingroup$ $ax = xa$ clarifies that the operation is commutative. $\endgroup$ Jun 9, 2015 at 22:26
  • $\begingroup$ Ah yes. Let me edit. $\endgroup$ Jun 9, 2015 at 22:27
  • $\begingroup$ @ Question 1: Yes, as long as you regard both $a$ and $x$ as elements of $R\left[x\right]$. (The element $a$ of $R$ can be canonically viewed as an element of $R\left[x\right]$ via the canonical embedding $R \to R\left[x\right], a \mapsto \operatorname{pol}_{R, x}\left(a,0,0,0,\ldots\right)$, where $\operatorname{pol}_{R, x}\left(u_0, u_1, u_2, \ldots\right)$ means the polynomial in $R\left[x\right]$ with coefficients $u_0, u_1, u_2, \ldots$. The indeterminate $x$ is identified with the polynomial $\operatorname{pol}_{R, x}\left(0, 1, 0, 0, 0, \ldots\right)$ as usual.) $\endgroup$ Jun 9, 2015 at 22:28
  • $\begingroup$ @ Question 2: This is perfectly valid. As you said, it is just a particular case of Question 1. $\endgroup$ Jun 9, 2015 at 22:29
  • $\begingroup$ @darijgrinberg Hmm. It does make sense. I see. Can you also comment on what is meant by a variable? $\endgroup$ Jun 9, 2015 at 22:31

5 Answers 5


The most satisfactory (by that I mean that it seems rigorous) way I've seen of defining polynomial rings is by defining $R[X]=\bigoplus_{n=0}^\infty R$. That is, elements in $R[X]$ are of the form $(a_0,a_1, a_2, \dots)$ with all but finitely many $a_i=0$. Then one defines a product on $R[X]$ by

$(a_0,a_1,\dots, a_n, \dots)(b_0,b_1,\dots, b_n, \dots)=(a_0b_0, a_0b_1+a_1b_0, \dots , \sum_{i+j=n}a_ib_j, \dots )$.

Then after proving that this makes $R[X]$ into a ring, one can define the standard $i^{th}$ basis vector $(0, \dots, 0, 1, 0, \dots):= X^i$, with the convention that $X^0=1$.

Then $ax=xa$ means

$(a,0,0, \dots )(0,1,0, \dots)=(0,1,0,\dots)(a,0,\dots)$

which makes sense and is true; both are equal to $(0,a,0,\dots)$.

Next, for polynomials of severable variables, people usually inductively define $R[X_1, \dots , X_n] := R[X_1, \dots X_{n-1}][X_n]$, so that in particular $R[X,Y]=R[X][Y]$. Now elements in this ring are really sequences of sequences, but with the way we've defined $X^i$, we can instead just write them as sequences of things that actually look like polynomials, and then define $Y^i$ in the same way we defined $X^i$. Your question 2 is then the exact same as question one: What does it mean to say $(X,0,\dots)(0,1,0,\dots)=(0,1,0,\dots)(X,0,\dots)$?

  • $\begingroup$ Thanks. I guess it is a mater of time then. The definitions don't seem elegant right now. $\endgroup$ Jun 9, 2015 at 22:49
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    $\begingroup$ I suppose they're not. I mean when you're actually working with polynomials, you just forget about the formalities and do all the same algebraic manipulations with them that you did in high school. However, if you want to see that the object $R[X]$ can be formally defined for peace of mind, this is how I would do it. $\endgroup$ Jun 9, 2015 at 22:57

The notation $R[x]$ may be read as "the commutative ring created by taking $R$ and adding a new element $x$." The fact that this new $x$ should commute with the existing elements is merely a part of the definition - when we're working with polynomial rings, we're generally interested in commutative rings, so we don't venture out of that territory when we add a new element.

Think of particular instances like $\mathbb Q[\sqrt[3]{2}]$. This is the set of reals generated by the rationals and $\sqrt[3]{2}$. It happens that every such rational may be written as a polynomial in $\sqrt[3]{2}$ - i.e. a sum of its powers multiplied by rational coefficients - and this follows just by distributing any expression you could come up with using rationals and $\sqrt[3]{2}$ with multiplication and addition. Of course $\sqrt[3]{2}$ commutes with every other element, because everything involved are real numbers. In a more general sense, we could consider $\mathbb Q[x]$ with $x$ standing in for some unknown complex or real number - it definitely commutes with the other elements, so everything is a polynomial in $x$ - but we can't make any reductions beyond that without knowing more about $x$. So, as far as commutative algebra is concerned, letting $x$ commute with things is the best way to represent our intuition about adding new elements to commutative rings. Abstracting a little more, we see that $R[x]$ is just the commutative ring that occurs when we stick a new element in, call it $x$, and see what happens. $x$ isn't really a variable - it's a new object we're sticking into the ring. So, if we stick both $x$ and $y$ in, given that we still want things to commute, we will end up with $xy=yx$.

That said, we certainly could consider the structure of a ring $R$ adjoined with a new element $x$ which we do not assume to commute with anything - the reason we don't is likely because structure of this is far harder to work with (each monomial has to be of the form $axbxcx\ldots$ rather than $ax^n$, though we can still always distribute into such monomials) and isn't particularly useful to algebraic number theory or Galois theory, where $R[x]$ is likely to appear.


When we write $K[X]$, the variable $X$ merely stands for the sequence $(0, 1, 0, 0, \dots)$. The product of polynomials is so defined that $X^i$ can be identified with the sequence $(0, 0, \dots, 0, 1, 0, \dots)$ with $1$ on the $i$-th position. It is natural to identify the constant $1$ with the sequence $(1, 0, 0, \dots)$. Thus, the polynomial $f = a_0 + a_1 X + \dots + a_n X^n$ can be identified with the sequence $(a_0, a_1, \dots, a_n, 0, 0, 0, \dots)$. This identification allows us to identify each polynomial with a sequence of finite support of elements from $K$. In other words, $K[X] = \{f : \Bbb N \to K \Big| \space \#|\text{supp} f| < \infty \}$.

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    $\begingroup$ i.e. the subring of $\,{\rm End}(R^{\Bbb N})\,$ generated by constant scalings $\, (x_i)\mapsto (rx_i)\,$ and the shift map $\, (x_0,x_1,\ldots)\mapsto (0,x_0,x_1,\ldots).\ \ $ $\endgroup$ Jun 9, 2015 at 22:43

I like the idea of using the concept of Monoid ring $R[M]$ where $R$ is a ring and $M$ is a monoid. The elements of $R[M]$ are finite formal sums

$$r_1m_1 + \cdots + r_p m_p $$ with the obvious addition and multiplication by using distributivity.

If $M=\mathbb N$, the nonnegative natural numbers under addition, we get the polynomial ring $R[x]$. Here an element of $M$ is written $x^i$ with $x^ix^j = x^{i+j}$.

But also if $M= \mathbb Z$, the integers under addition, and working as above, we get the ring of Laurent polynomials.

Note that we get polynomials in many variables using $R[x,y]= (R[x])[y]$.

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    $\begingroup$ Polynomials in two indeterminates correspond to $M=\mathbb{N}\times\mathbb{N}$ (the free commutative monoid on two generators). $\endgroup$
    – egreg
    Nov 4, 2017 at 12:15

One way to think about the space of degree $m$ terms $R[x_1,\cdots,x_n]^m$ is as a coordinate representation of $\operatorname{Sym}^m R^n$, where $\operatorname{Sym}$ is symmetric power. I think this is a more geometric view, compared with the more formal views implicit in the other answers here. Then the answer to the question "what are the $x_i$?" is: a choice of basis.

In this view the whole coordinate ring $R[x_1,\cdots,x_n]$ is $\operatorname{Sym}R^n=\bigoplus_{i=0}^\infty\operatorname{Sym}^iR^n$

Now question 1 is understood as an expression of commutativity of $R$. and question 2 is a statement of the linear algebra identity $\operatorname{Sym}^n\operatorname{Sym}^m=\operatorname{Sym}^{n+m}$

  • $\begingroup$ This is actually very nice. Thanks. Could you give me a reference where polynomial rings are looked at as symmetric powers? $\endgroup$ Sep 26, 2015 at 14:18
  • $\begingroup$ This approach seems to have some analogies with your questions about invariant descriptions about exterior algebras, so I thought it might appeal to you. Unfortunately I don't know a reference for this point of view, though I suppose there must be some. Probably look more toward algebraic geometry side, than pure algebra. $\endgroup$
    – ziggurism
    Sep 26, 2015 at 15:05

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