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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.

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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 $K[t]$.

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  • $\begingroup$ So $t$ is a variable for a constant polynomial? But then $\phi(t)$ is also a variable in $F$, right? $\endgroup$ – Gogis Aug 25 '17 at 11:01
  • $\begingroup$ @Gogis no! no constant polynomials here (though I am not sure what you mean by "constant polynomial" - but it is usually understood that a constant polynomial is one of the form $a$, $a\in K$, i.e. $(a,0,0,\ldots)$. If you mean something else, please clarify). I don't understand your question. Could you please find another way to state it? $\endgroup$ – Cauchy Aug 25 '17 at 11:07
  • $\begingroup$ Sorry, I misunderstood your answer. What do you mean by saying that $t$ is identified with $(0, 1, 0, 0, ...)$? $\endgroup$ – Gogis Aug 25 '17 at 11:10
  • $\begingroup$ @Gogis I shouldn't have said this. Actually, $t$ is equal to $(0,1,0,0,\ldots)$ . The idea of $(a_0,a_1,\ldots)$ is that the $a_i$ represent coefficients of a polynomial, so $(a_0,a_1,\ldots)$ is to be thought of as $\sum_{i=0}^{\infty} a_i t^i$. In this view $(0,1,0,0,\ldots) = 0 + 1 \cdot t + 0\cdot t^2 + \cdots=t$ $\endgroup$ – Cauchy Aug 25 '17 at 11:15
  • $\begingroup$ Oh, I see. I think I just got it. The root of $\pi(x)$ in $F$ is an image of the polynomial $t$ (i. e. $\phi(t)$), correct? $\endgroup$ – Gogis Aug 25 '17 at 11:23

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