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In Algebra by Hungerford, page 237 the sketch of proof for Theorem 1.10:

Theorem 1.10: If $K$ is a field and $f\in K[x]$ polynomial of degree $n$, then there exists a simple extension field $F = K(u)$ of $K$ such that $u\in F$ is a root of $f$.

In the proof I could not understand one part:

Let $\pi:K[x]\to K[x]/(f) = F$ be the canonical projection, then $f(\pi(x)) = 0$

I understand all the theorems before this point and every sentence until this one. If I understood correctly, $\pi$ maps $g(x)$ to $r(x)$ where $g = fs+r$ by division algorithm. I can only see why $\pi(f) = 0$ but I cannot see why $f(\pi(x)) = 0$. Some explanation please. Thank you.


EDIT: Also I don't know how to prove that $F = K(\pi(x))$.

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  • $\begingroup$ The notation $f(\pi(x))$ means, by definition, the image of $f$ under $\pi$. $\endgroup$ – Keenan Kidwell Jan 7 '14 at 2:38
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The canonical $\,K$- homomorphism we're talking about is (check that the following stuff follows from the basic definition of operations in quotient rings!):

$$\pi(p(x))=\pi\left(\sum_{i=0}^m c_ix^i\right):=\left(\sum_{i=0}^mc_i\,x^i\right)+(f)=\sum_{i=0}^mc_i\left(x^i+(f)\right)=$$

$$=\sum_{i=0}^mc_i\left(x+(f)\right)^i=p(\pi(x))$$

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$\pi$ is a homomorphism, so $\pi(f(x)) = f(\pi(x))$.

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    $\begingroup$ Not only "a homomorphism" but a $\,K$-homomorphism, which allows us to do $\,\pi(f(x))=f(\pi(x))\,$ . $\endgroup$ – DonAntonio May 1 '13 at 22:28
  • $\begingroup$ Right, the important thing is that it fixes $K$. $\endgroup$ – Paul Gustafson May 1 '13 at 22:39

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