For simplicity, all rings considered in this post will be supposed commutative and unitary.

Suppose we have a ring $R$ and we somehow want to add in it an element $x \notin R$ having no other relations with the other elements of $R$ excepted those from the defining axioms for a ring. How should we proceed? Well, one useful thing to do would be to try and pinpoint exactly what we are looking for since the problem is quite vague.

First, since we want the resulting thing to be a ring it does not suffice to just add $x$ in the underlying set of $R.$ On the other hand, it would certainly be neat to construct this new ring with as few new elements as possible. What we are looking for is thus the smallest ring containing $x$ and all of the elements of $R.$

Since the result must be a ring, it will contain all powers of $x$ and all linear combinations of these powers with coefficients in $R.$ In other words, we are looking for the ring of polynomials $R[x].$ We could stop here and be happy about it but let’s think of this more conceptually:

We would like to consider for our new ring not only rings which have $R$ as a subring but work more abstractly and demand only that it contains an isomorphic copy of $R.$ Let’s denote by $X$ the ring we are trying to construct. In this context, the condition that $X$ contains $R$ becomes:

that there exists an injective ring homomorphism $\iota : R \to X.$

Moreover, in this more general context, the order relation to consider on rings is naturally injective ring homomorphisms. i.e. the ring $A$ is smaller than $B$ iff there exists an injective morphism $A \to B.$ So the condition that $X$ be the smallest ring containing $R$ possessing an element $x$ such that $x$ satisfies no relations with the elements of $R$ excepted the defining axioms becomes:

that for every ring $B$ possessing such an element $x_B$ and such that there exists an injective ring homomorphism $\varphi : R \to B,$ there exists an injective ring homomorphism $\phi : X \to B$ such that $\phi (\iota (R)) = \varphi (R)$ and $\phi(x) = x_B.$

This is summarized with this commutative diagram:

It is easy to see that if such a ring exists, it is uniquely defined up to isomorphism. Indeed, let $X_1$ and $X_2$ be two candidates. Then there exist injective morphisms $\phi_1 : X_1 \to X_2$ and $\phi_2 : X_2 \to X_1.$ Since they are both injective, $\phi_2 \circ \phi_1 : X_1 \to X_1$ is injective and so its image is the whole of $X_1.$ From this follows that $\phi_2$ is surjective and therefore is an isomorphism.

But how can we be sure that such a ring even exists? Well, the ring $R[x]$ of polynomials in one variable with coefficients in $R$ does satisfy this property. Indeed, let’s denote by $\iota$ the map sending an element of $R$ to the corresponding constant polynomial in $R[x].$ Given an injective morphism $\varphi : R \to B,$ the map $\phi : R[x] \to B$ defined by letting

$\phi(\iota(r)) = \varphi(r)$ and $\phi(x) = x_B$

can be extended by linearity to an injective morphism defined on $R[x]$ sending $\sum r_ix^i$ to $\sum \varphi(r_i)x_B^i.$ It is easily seen to satisfy the universal property designed above.

So basically, we have shown that if you adjoin to a ring $R$ a new element $x$ having no more relations with other elements of $R$ other than the defining axioms for rings, i.e. the smallest ring containing $R$ and $x,$ then not only does the polynomial ring $R[x]$ gets the job done, there is no other sensible way of doing it. This might be a reason why polynomials are so often considered when working in algebraic settings. More generally, if want you want is to adjoin n such new elements to your ring, what you get is $R[x_1, \ldots, x_n].$

In fact, the universal property satisfied by polynomials can be stated more strongly:

Universal property for polynomials: For every ring homomorphism $\varphi : R \to B$ and every $b \in B,$ there exists a unique ring homomorphism $\phi : R[x] \to B$ such that $\phi|_R = \varphi$ and $\phi(x) = b.$

With this in mind, you can somehow view the polynomial ring $R[x]$ as the best possible ring containing $R$ but with a “wild card” $x$ which can take the form you want. I’ll elaborate on this in my next post when I’ll talk about adjunction of new elements in more detail.