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There is a theorem in my book which says the following:

Let $K$ be a field and let $f(X) \in K[X]$ be irreducible over $K$. Then there exists a field extension $L/K$, such that $\exists u \in L$: $f(u) = 0$.

Proof: Notice that $K\subseteq K[X]/(f(x))$. Clearly, $K[X]/(f(x))$ is a field since $(f(x))$ is maximum ideal. If we take $u = \overline{X}$, then $f(u) = f(\overline{X})= \overline{f(X)} = 0$.

Question: I really don't get if $u = ...$ in the proof, since I don't know what $\overline{X}$ means in this context, does anyone know?

Thanks in advance.

Unfortunately I cannot provide a link to the book as it is my dutch (non pdf) syllabus for my algebra course.

EDIT: Seems like I get it now: $f(\overline{X}) = \overline{f(X)} = f(x)+ (f(x)) = (f(x))$ which is $0$ for $K\subseteq K[X]/(f(x))$.

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    @Rankeya: Sorry if it sounded like I was discounting your comment, as I agree that it's an illuminating way to think about the result (and there's no need to defend yourself :) ). To compensate for issues I have when dealing with results normally taken for granted, I force myself to take a fine-tooth comb to anything approaching one of those results (ex: to this day I'm unable to explain the isomorphism theorems since I cannot disassociate "surjective homomorphism" from "projection onto quotient" or "quotient" in any meaningful way).2011-12-14

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$\overline{X}$ means the class of $X$ mod $f(X)$ in the quotient ring $K[X]/(f(x))$.

Now, given $g\in K[X]$, consider the associated polynomial function in $K[X]/(f(x))$. Then, by the definition of the ring operations in $K[X]/(f(x))$, we have $g(u) = g(\overline{X}) = \overline{g(X)}$. When $g=f$, we get $g(u)=\overline{0}$.

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    @Rankeya, thanks, completely forgot about how how multiplication is defined in quotient rings!2011-12-14