While studying algebra, I came across with the following sentence: "The element $t \in F(t)$ is transcendental over $F$ since if $f(t) = 0$ for some $f \in F[X]$, then necessarily $f=0$".
Why is that?
Do not we have $g(X)=X-t \in F[X]$ so that $g(t) = 0$?
$t$ is transcendental over $F$ if and only if the map $$\varphi : F[x] \to F[t]$$ defined by $f(x) \to f(t)$ is isomorphism. This is because $\ker(\varphi) = (m_t(x))$, where $m_t(x)$ is the minimal polynomial of $t$ over $F$. So the $\ker(\varphi) = 0$ if and only if $t$ has no minimal polynomial over $F$, i.e. $t$ is transcendental.
In this case the isomorphism is clear because $F[x]$ and $F[t]$ are both the polynomial ring in one indeterminate over $F$, just with different names for the indeterminates.