# Why $K = F[x]/(f(x))$ contains a root of $f(x)$, an irreducible polynomial

A theorem states that given an irreducible polynomial $f(x) \in F[x]$, the field $K = F[x]/(f(x))$ contains a root $\alpha$ of f(x).

The proof says that if we let $\alpha = \overline{x} \in F[x]/(f(x))$ then $f(\alpha) = \overline{f(x)} = 0$ so $\alpha$ is indeed a root of $f(x)$ in K.

The proof is what I don't understand - how is it valid? How does setting $\alpha = \overline{x}$ work? It is clear that $\overline{f(x)} = 0$, but how does this show that there exists a root (which is a constant) of f?

• $\bar{x}$ means the coset of $x$ in the quotient ring $F[x]/(f)$ – Nick Mar 25 '17 at 22:14

It's the equivalence class of the monomial $x$ in the quotient ring $K=F[x]/(f)$ where basically the element $f$ will be equal to $0$.
So, $f$ had no roots in the field $F$, but we want a (necessarily bigger) field where it has at least one root. And we do it by a construction: first we adjoin a formal element to $F$ - that's how $F[x]$ enters the picture, and then we force this new element to be a root of $f$, by taking the quotient.
In formulas, suppose $f=a_0+a_1x+a_2x^2+\dots$, and taking $\alpha:=\bar x$, we will have $$f(\alpha) = a_0+a_1\bar x+a_2\bar x^2+\dots=\overline{a_0+a_1x+a_2x^2+\dots}=\overline{f}=0$$ where in the middle we used that taking the quotient respects the ring operations.
Can you catch where we use that $f$ is irreducible?
When you define the polynomial ring $K[x]$, you impose no relations on $x$: all the powers of $x$ are distinct and none of them can be expressed as a linear combination of other powers of $x$. However, there is no reason why we cannot define a different ring which is the same as $K[x]$ except for the additional condition that $x$ satisfies $f(x) = 0$: all the ring axioms are readily verified. This ring contains a root of $f$ by definition. Moving from $K[x]$ to this ring is exactly what you accomplish when you quotient by $(f(x))$.