# Importance and Intuition of Polynomial Rings

I know that a polynomial ring $$R[x]$$ is the ring with elements consisting of polynomials with coefficients in $$R$$. However, this definition leaves me confused when I try to really understand the concept of polynomial rings, rather than just accept that definition for what it is. Where did the polynomials come from? Why are polynomial rings important? Is there a way to “construct” the polynomials in the context of Abstract Algebra, so that we may see why studying rings made up of polynomials is important? Are polynomial rings actually just sets of polynomials, or are the polynomials just a concrete way to represent some more abstract idea? What does the variable $$x$$ mean in polynomial rings? I apologize for all the questions, I’m just having trouble seeing the importance or intuition behind them.

• Polynomial rings are the universal $R$-algebras, if that makes sense to you. See en.wikipedia.org/wiki/Polynomial_ring#Polynomial_evaluation
– lhf
Dec 14 '20 at 17:48
• I don't understand what specifically you are looking for here. Ignoring the ring-theory for now... you have seen and used polynomials before, yes? They occur naturally in many contexts and are important in many applications and have been in use in some form or another since the ancient Greeks or even before. So then... given the importance of polynomials, why not formalize it via group theory and ring theory? Dec 14 '20 at 17:50
• One could say that polynomials "come from" geometry: they arise as the global sections of affine space. Dec 14 '20 at 18:42
• I also don't really understand the question here. Polynomials are important, undeniably; we use them for all sorts of things, solving equations, representing geometric objects, etc. We can add and multiply polynomials. So we have a ring, and it's worth studying. Dec 15 '20 at 6:19
• Polynomial rings are exactly free commutative rings. In category theory "free" things are fundamentally important - any individual algebraic structure is an algebra of the appropriate monad. Here it's Set \rightarrow CommRing. Commutativity is fundamental because the \NN^d is commutative whereas free monoids in general aren't. \NN^d lives inside \RR^d which gives your algebraic objects spatial intuition via e.g. Cayley graphs. So finitely presented quotients of 3-generated Comm Rings are like the basic ways to fold 3d space onto itself in a way that respects some spatial structure... Dec 15 '20 at 21:57

Great question! Polynomial rings really get glossed over in my opinion, when they are actually quite complicated objects.

Super formally, we define the polynomial ring $$R[x]$$ as follows. Let $$S$$ be the set of all sequences $$(r_0, r_1, r_2,\ldots)$$ where the $$r_i$$ are elements of $$R$$ and only finitely many are nonzero. We explicitly define operations $$+$$ and $$\cdot$$ on this set by $$(a_0, a_1, a_2,\ldots) + (b_0, b_1, b_2, \ldots) = (a_0 + b_0, a_1 + b_1, a_2 + b_2, \ldots)$$ and $$(a_0, a_1, a_2, \ldots) \cdot (b_0, b_1, b_2, \ldots) = (a_0b_0, a_1b_0 + a_0b_1, a_2b_0 + a_1b_1 + a_0b_2, \ldots).$$ That is, if $$a = (a_i)$$ and $$b = (b_i)$$, then $$(a+b)_i = a_i + b_i$$ and $$(ab)_i = \sum_{k=0}^i a_kb_{i-k}$$.

It is easy to check that $$(S, +, \cdot)$$ is a ring with zero $$(0, 0, 0, \ldots)$$ and identity $$(1, 0, 0, \ldots)$$.

Furthermore, as an $$R$$-algebra, $$S$$ is generated by the element $$x = (0, 1, 0, 0, \ldots)$$ (if you don't know what an $$R$$-algebra is, the point is that we can "get to" every element of $$S$$ using just the element $$x$$ and the elements of $$R$$). To see this, note that for each $$n$$, the element $$x^n$$ is the sequence $$(0, \ldots, 0, 1, 0, \ldots)$$ with a $$1$$ in the $$n^\text{th}$$ entry and zeros elsewhere. The element $$(a_0, a_1, \ldots)$$ of $$S$$ is equal to $$a_0 + a_1x + a_2x^2 + \ldots.$$ Thus we can "get to" $$(a_0, a_1, \ldots)$$ just by raising $$x$$ to various powers and multiplying by elements of $$R$$.

We denote this ring by $$R[x]$$. The intuition is that we are "adjoining" an "indeterminate" $$x$$ to the ring $$R$$, which means that we are adding some element that has no constraints on it. In some sense, the element $$x$$ is "free". In practice, this is how we always think about polynomials.

Polynomial rings are extremely important. Without them, mathematics would basically not be possible. The most ubiquitous example I can think of is in Linear Algebra, where the theory of polynomial rings allows us to prove results like the Cayley-Hamilton Theorem. Another example is in the theory of probability generating functions. Polynomials are also vital to number theory and geometry. Every field of mathematics I can think of is built at least implicitly on the theory of polynomials.

Edit I got a bit carried away with advanced topics in my examples. For a more basic one, we need the theory of polynomial rings to understand even basic things like factorising quadratics. To prove that factorisation exists and is unique, you need to understand the ring $$\mathbb{R}[x]$$. Factorising such polynomials is very useful, as any high school student will tell you.

The important thing to understand about polynomials is that they are templates for maps between rings. If you know a bit about coding, you'll probably know the concept that a template is something that can take different types of objects as inputs in a way that you don't have to specify what the objects are beforehand. For a polynomial, this means that it's a template for maps which take objects for which the following operations are meaningful:

• multiplication of two or more of the objects
• multiplication of an object with an element of the underlying ring
• addition of two or more objects

For instance, take the polynomial $$f=2X^2+4X-1\in\mathbb R[X]$$. This polynomial defines a map \begin{align}f_{\mathbb R}:&\mathbb R\longrightarrow\mathbb R,\\&x\mapsto 2x^2+4x-1. \end{align} But it also defines a map \begin{align} f_\mathbb C:&\mathbb C\longrightarrow\mathbb C,\\ &z\mapsto 2z^2+4z-1 \end{align} Or even more exotic, it defines a map \begin{align} f_{\operatorname{Mat}_{n\times n}(\mathbb R)}:&\operatorname{Mat}_{n\times n}(\mathbb R)\longrightarrow \operatorname{Mat}_{n\times n}(\mathbb R)\\ &A\mapsto2A^2+4A-1 E_n. \end{align} This last one is important to formulate the Cayley-Hamilton theorem, for instance, which says that if $$f$$ is the characteristic polynomial of the matrix $$A$$, then $$f_{\operatorname{Mat}_{n\times n}(\mathbb R)}(A)$$ is the zero matrix.

More formally, this means that if $$R$$ is a commutative ring and $$R\subseteq S$$ a commutative ring extension, then for every $$s\in S$$ there is a unique evaluation map $$R[X]\longrightarrow S$$ which maps $$X\mapsto s$$ and $$r\mapsto r$$ for all $$r\in R$$, and is a homomorphism. The map evaluates each polynomial at $$s$$ (it "plugs in" $$s$$ for $$X$$). This property of the polynomial ring even uniquely characterizes it (up to isomorphism), and some will take this property as the definition of the polynomial ring. The construction via finite sequences in $$R$$ can then be used as proof of existence of such a ring.

In our examples from before, $$\mathbb R\subseteq\mathbb R$$ is a trivial ring extension. $$\mathbb R\subseteq\mathbb C$$ is another obvious ring extension which also gives us an evaluation map. And $$\operatorname{Mat}_{n\times n}(\mathbb R)$$ can also be viewed as a ring extension of $$\mathbb R$$ by identifying $$r\in R$$ with $$rE_n\in\operatorname{Mat}_{n\times n}(\mathbb R)$$, the matrix whose diagonal contains only $$r$$. This gives us a way to evaluate polynomials at matrices.

To summarize: Polynomials are there as templates for maps between ring extensions of the underlying ring - but only those special maps which are the result of applying these basic operations to an element $$x$$: addition, multiplication by an element of $$R$$, or multiplication by itself.

• Indeed. This ultimately is the best way to view most polynomials. Whether or not we encode them as finite sequences of coefficients is actually irrelevant to the whole point of polynomial functions as functional objects with a generic type parameter. Now of course there are extra things one can do with the syntactic encoding, such as polynomial rings with infinitely many indeterminates, but that kind of stuff is not typically what we use polynomials for. Dec 15 '20 at 16:26

$$R[x]$$ can be thought of as the simplest way to add one element to the ring $$R$$ and still have a ring.

Suppose I have a ring $$R$$ and I want to add one extra element which is not already in $$R$$. Call that extra element $$x$$. Consider the set $$A = R\cup \{x\}$$. To make $$A$$ into a ring, we need to add in all sums of terms that look like $$r x^k$$ where $$r\in R$$ and $$k\in \mathbb{N}$$. The simplest way to define multiplication in this ring is just to let multiplication of two members of $$R$$ stay the same and let all the products $$r x^k$$ be unique elements of the new ring. This is just the polynomial ring $$R[x]$$.

(This is actually just a restatement of the comment by @lhf, but it is likely that OP does not know what a universal $$R$$-algebra is.)

The philosophy behind polynomial rings is that we are not committing ourselves to saying exactly what $$x$$ represents. It is just an "indeterminate" and we can manipulate it according to formal rules like $$x^{a} x^{b} = x^{a+b}$$. If at some point we wish to specify a value for $$x$$, we can do so using a ring homomorphism $$R[x] \rightarrow R$$ that replaces every instance of $$x$$ with that particular value.

Note also we can define an homomorphism from $$R[x]$$ to the ring of all functions from $$R$$ to $$R$$ with pointwise operations. Just send $$r \in R$$ to the constant function $$r$$ and send $$x$$ to the identity map on $$R$$. In the case $$R = \mathbb{R}$$, this map is an isomorphism onto its image. This is how we get the correspondence between polynomial rings and polynomials as we know them from grade school algebra.

I will elaborate my answer on how to see the polynomial ring without the indeterminate $$x,$$ and on the universal property of polynomial rings. I will show you two alternative ways to understand the definition of polynomial rings. Those interpretations will allow you to see this construction without wondering about the role of indeterminate(s). Let $$R$$ be a commutative ring with unity $$1_R.$$

1. Monoid Ring on the Natural Numbers $$R[\mathbb{N}]$$:
This is the monoidal version of the group ring construction over the monoid of natural numbers. Here objects are formal finite linear combinations of natural numbers with coefficients in $$R.$$ For example $$r_1n_1+r_2n_2+\cdots+r_jn_j,\qquad r_1, r_2, \cdots, r_j\in R,\quad n_1, n_2,\cdots, n_j\in\mathbb{N}.$$ It is not difficult to see that this is another way to look at the polynomial ring by interpreting above element as the polynomial $$r_1x^{n_1}+r_2x^{n_2}+\cdots+r_jx^{n_j}\in R[x].$$ Furthermore, if you use $$\mathbb{N}^n$$ (ordered $$n$$-tuples of natural numbers) in this construction, you would get the polynomial ring $$R[x_1, x_2,\cdots, x_n]$$ with $$n$$ variables.

2. Ring of $$R$$-sequences of finite support $$R^{(\mathbb{N})}$$:
Another way to look at the linear combination $$r_1n_1+r_2n_2+\cdots+r_jn_j$$ is as a function $$f:\mathbb{N}\to R$$ whose values are $$f(n_i)=r_i$$ for all $$i=1, 2, \cdots, j$$ and zero everywhere else. Since the domain of this function is non-negative integers, this is nothing but a sequence. Also, the sequence is zero for all but finitely many terms, therefore we say it is finitely supported. On the other hand if you are given a such sequence we can easily create a unique polynomial (and a linear combinaion). For example: Say $$R=\mathbb{Z}/3\mathbb{Z},$$ then the finitely supported sequence $$(1, 1, 1, 0, 0, 0, \cdots)$$ represents the polynomial $$1+x+x^2=(2+x)^2$$ and vise versa. Note that even this factorization carries all the information about the associated sequence.

Above two equivalent (ring isomorphic) constructions shows us that powers of indeterminate $$x$$ are nothing but placeholders. We can identify $$x$$ with the linear combination $$1_R1$$ or with the sequence $$(0, 1_R, 0, 0, \cdots).$$ Also, if we accept any of them as the polynomial ring, say the first construction, then we have an injective ring homomorphism (embedding) $$\iota : R \to R[\mathbb{N}]$$ given by $$r\mapsto r0$$ (or in the other construction $$r\mapsto (r, 0, 0, 0, \cdots)\in R^{(\mathbb{N})}$$). This map is universal in the sence that any other such map factor through $$\iota$$ while sending $$x$$ to a priscribe value. This is the universal property of polynomial ring (of a single variable) over $$R$$:

For every commutative unital ring $$S$$ that has an injective ring homomorphism $$R\xrightarrow{\phi}S$$ and for any chosen $$a\in S,$$ there is a unique extension (evaluation homomorphism) $$R[\mathbb{N}]\xrightarrow{\hat\phi_a}S$$ such that $$\hat\phi_a\circ\iota=\phi$$ and $$\hat\phi_a(x)=a.$$ $$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && R[\mathbb{N}]\\ & \diaguparrow{\iota} @VV\hat\phi_a V \\ R @>>\phi> S \end{CD}$$

One can easily formulate similar universal property for polynomial ring with any number of variables.