Extending Quotient Rings

April 9, 2026 · Luciano Muratore

It is well-known that we can construct $\mathbb{C}$ by just “adding” the value $i$ to $\mathbb{R}$. By the way, $i$ is the root of the polynomial $X^2+1$.

In other words, we can extend a field just by adding the root of a polynomial. But, the previous idea can also be applicable to Rings. For example, we can create the ring $\mathbb{Z}[i]$ by again considering $X^2+1$. Just by ‘adjoining’ a root of a polynomial $f(X) \in R[X]$ to the ring $R$, we would like to extend the ring into, maybe, a field.

The extension of the ring is defined as the quotient ring given by $R[X]/(f(X))$.

The first natural question to postulate is: what is the structure of this quotient ring?.

Since we want the quotient ring to be an extension of the ring $R$ we would expect $R$ to be embedded inside the quotient ring. In other words, we should be able to define an injective morphism $\varphi:R\to R[X]/(f(X))$. As previously mentioned, we want to construct the elements of the quotient by using the root of $f(X)$. ($i$ is a root of $X^2+1$ and the elements of $\mathbb{C}$ are $a+ib$). That would imply that $\alpha \equiv X \mod (f(X))$, which would mean that $f(\alpha) \equiv 0 \mod (f(X))$.

In the end, we would like any element $z \in R[X]/(f(X))$ to be writable in a unique way as $z=r_0+r_1\alpha+\cdots+r_{n-1}\alpha^{n-1}$ with $r_i \in R$, in such a way that $\{1,\alpha,\ldots,\alpha^{n-1}\}$ would act as a finite basis.

I need to make a pause. There is something that I have not mentioned. The polynomial $f$ must be monic; if not, there is a risk of not having a unique linear combination.

To make this clear, let’s consider the example $R[X]/(2X-1)$. Then, $2\alpha=1 \in R[X]/(2X-1)$. Now, take the element $\beta=\alpha$. And, if $\alpha=2\alpha^2$ then $\beta=2\alpha^2$. That shows that the expression of $\beta$ as a linear combination of powers of $\alpha$ is not unique.

Hope the example was enough to convince the relevance of the polynomial being monic. There are situations when the polynomial cannot be monic; I would let the readers find them by themselves.

Continuing with the narrative, I want to make clear the uniqueness of the linear combination for any element in the quotient ring, and for that it is necessary to present some kind of proof.

Let $z \in R[X]/(f(X))$. Then, $z \equiv g(X) \mod (f(X))$ for some polynomial $g(X) \in R[X]$. By the Euclidean division of $g(X)$ by $f(X)$, there exist unique polynomials $q(X)$ and $r(X)$ such that $g(X)=f(X)q(X) + r(X)$ with $\deg r < n$. Then, $z \equiv r(X) \mod (f(X))$ because $f(X)q(X) \in (f(X))$. $r$ is unique, giving $z=r_0+r_1\alpha+\cdots+r_{n-1}\alpha^{n-1}$.

Great, we have worked through the structure of $R[X]/(f(X))$. So far, we have been concerned with rings, but what if we operate with fields?.

Let $K$ be a field, and $K[X]$ its polynomial ring. But, what if we would like to infer what the quotient $K[X]/(f(X))$ looks like from the polynomial $f$? I mean: “would the extension of a field still be a field?”. Would it be possible? For example, $\mathbb{Q}[X]/(X^2+X+1)$ is a quotient with a polynomial whose root is $\frac{-1+i\sqrt{3}}{2}$, which is not in $\mathbb{Q}$. And, what can we say of $\mathbb{Q}[X]/(X^4-X^2-2) = \mathbb{Q}[X]/((X^2+1)(X^2-2))$?.

If the polynomial is irreducible in $K$ and $K[X]$ is a PID (Principal Ideal Domain), then the quotient is a field, as in the first example where $\mathbb{Q}[X]/(X^2+X+1) \simeq \mathbb{Q}[i\sqrt{3}]$.

In the second example, the polynomial is reducible over the rational field into $(X^2+1)$ and $(X^2-2)$. Both polynomials have roots outside of $\mathbb{Q}$. And, we express $\mathbb{Q}[X]/(X^4-X^2-2)$ as a product of two fields, the ones made with the roots of $(X^2+1)$ and $(X^2-2)$, i.e. $\mathbb{Q}[i]$ and $\mathbb{Q}[\sqrt{2}]$.