Why Quotient by a Polynomial?

Central Question

Why does imposing the relation $p(x)=0$ create a root, and when does it create a field?

Why This Matters

Quotients turn algebraic equations into construction rules. Irreducibility explains why this construction sometimes gives a field and sometimes does not.

Mathematical Notes

What the quotient actually says

For a field $F$ and a nonconstant polynomial $p\in F[x]$, identify $f$ and $g$ when $p$ divides $f-g$. In $A=F[x]/(p)$, let $\alpha=[x]$. Then

Here $[x]$ is a whole equivalence class of polynomials. It becomes a root because the arithmetic has been changed so that multiples of $p$ are zero. Constants embed into $A$, since a nonzero constant is never divisible by a nonconstant polynomial.

Polynomial division gives a unique representative of degree less than $\deg p$ for every class. Consequently, $A$ has basis $1,\alpha,\ldots,\alpha^{n-1}$ over $F$, where $n=\deg p$, whether or not $A$ is a field.

Proof mechanism. If $p$ is irreducible and $[f]\ne0$, then $\gcd(f,p)=1$. Bézout gives $uf+vp=1$, so $[u][f]=1$. Conversely, a nontrivial factorization $p=ab$ produces nonzero classes $[a],[b]$ with $[a][b]=0$, which cannot happen in a field.

Three different outcomes

• $\mathbb Q[x]/(x^2-2)$ is a field: the polynomial is irreducible.
• $\mathbb Q[x]/(x^2-1)\cong\mathbb Q\times\mathbb Q$ by $[f]\mapsto(f(1),f(-1))$. Two distinct roots create two components, not a field.
• $\mathbb Q[x]/(x^2)$ contains a nonzero element $\varepsilon=[x]$ satisfying $\varepsilon^2=0$. Repeated factors can create nilpotents.

For example, in the first quotient $(1+\alpha)(\alpha-1)=1$. In the second, $(\alpha-1)(\alpha+1)=0$ even though neither factor is zero.

Why this construction is canonical

If $B$ is a commutative $F$-algebra containing $\beta$ with $p(\beta)=0$, evaluation factors uniquely through $A$:

Any arithmetic interpretation of the imposed relation must use this map. When $p$ is irreducible and $B$ is a field, it identifies $A$ with $F(\beta)$.

Learning Prompt

I understand quotient rings, but I want to deeply understand why $F[x]/(p(x))$ effectively creates a root of $p(x)$. Explain what the element $[x]$ actually is and why $p([x])=0$. Compare irreducible and reducible $p(x)$, and show why irreducibility gives a field.

Ideas

Examples

Questions

Connections

• Adjoining a Root
• When Different Constructions Give the Same Field
• Quotient Spaces
• Week overview

One Thing That Surprised Me

Further Rabbit Holes

Sources

Ian Stewart, Galois Theory, fifth edition, CRC Press, §5.3; §§17.1–17.2. Publisher and edition details.