I have a question regarding the proof of the theorem that for any irreducible polynomial $f(x) \in F[x]$, where $F$ is a field, there is a field extension in which $f(x)$ has a root.
I think the proof is pretty standard and it is fully described here (theorem $2.1$).
Specifically, what concerns me is that we say (using the definitions from the proof above) the congruence class of $t$ is a root of $\pi(t)$ in $F$. However, I don't understand what the congruence class of $t$ is, since $t$ is indeterminate, i. e. a variable. So when $\phi:K[t] \rightarrow F = K[t]/(\pi(t))$ is a canonical homomorphism, how can we evaluate $\phi(t)$? How can it be even defined, since $t$ is not a member of $K[t]$, but merely a symbol?
There are some similar questions, though none of the answers provides sufficient explanation for me.
No, $t$ is a member of $K[t]$. Recall that $K[t] := \{(a_0,a_1,\ldots): \forall j \ a_j \in K \land \exists i, a_j = 0 \ \forall j \ge i\}$, but also $t$ is identified with $(0,1,0,0,\ldots)$ so it is a member of $K[t]$.