Why isn't $p(x)$ identically zero in $F[x]/(p(x))$

Question

Suppose $F$ is a field and $p(x)\in F[x]$ is an irreducible polynomial. Then $K:=F(x)/(p(x))$ is a field extension of $F$.

I am confused: Why in this field $p$ has the root $x$, and why it isn't the case that $p$ is identically zero in this field?

Answer 1

$p(x)$ is zero but the polynomial $p(T)$ in $(F[x]/(p(x)))[T]$ is not zero.

For example, let $p(T) = T^2 + 1$. Then in $\mathbf{Q}[x]/(p(x))$, the polynomial $p(T)$ splits as $p(T) = (T + x)(T - x)$. So $p(x) = 0$ but the polynomial is not identically zero.

If we write our polynomial ring as $F[x]$ then we want $x$ to be transcendental, meaning it is not the solution of any polynomial equation. Then in $F[x]$, two polynomials are equal if and only if they have the same coefficients—which would not be true if $p(x) = 0$. If we introduce a new relation: $p(x) \equiv 0$, then we can no longer use $x$ as the indeterminate of our polynomial ring. If $K = F[x]/(p(x))$ then $K[x] = K$, which is not the same as the polynomial ring $K[T]$.