0
$\begingroup$

Let $f \in K[X]$ with $deg(f)\geq 1$. Then there exists an algebraic field extension $L/K$, such that $f$ has a root in $L$.

Proof: WLOG we can assume that $f \in K[X]$ is irreducible. Since $L=K[X]/(f)$ is a field and $K[X]/(f)\cong K(a)$, where $a$ is a root of $f$, it is clear that $L/K$ is an algebraic field extension. It is apparent that $x=X+(f)$ is a root of $f$ in $L$.

The lemma showing $K[X]/(f)\cong K(a)$ was shown by assuming that $f=\mu_{a,K}$, and then showing that $(f)$ is the kernel of the evaluation map at $a$ and it is only cited in the proof, not explicitly written out.

This lemma is used in the proof of existence of algebraically closed fields and is proved before it.

  1. Why can we assume that there exists a root of $f$? Does this assumption not already prove the lemma by itself? Is this not circular?
  2. Is the class of $f$ not $0 \in L$, since $f \in (f)$? Why even look for roots if everything suffices?
  3. Why would $x=X+(f)$ be proposed as the root?
$\endgroup$
  • 1
    $\begingroup$ Please correct "where $a$ is a root of $K$" which makes no sense; fields do not have roots. It is not clear to me why that part is needed in the first place, all that matters is that $L$ is a field generated by the algebraic element $x$ of the next sentence. $\endgroup$ – Marc van Leeuwen Mar 4 at 8:44
  • $\begingroup$ Corrected, thank you. $\endgroup$ – B.Swan Mar 4 at 11:35
  • $\begingroup$ @MarcvanLeeuwen I think you answered the only open question, namely why $L/K$ is algebraic. If you want to write out an answer I will accept it. $\endgroup$ – B.Swan Mar 4 at 11:46
1
$\begingroup$

Well, the situation is quite simple. Given the quotient ring $K(x)/\langle f(x)\rangle$, the residue class of $x$, $\bar x = x+\langle f(x)\rangle$, fulfills $f(\bar x)=f(x)+\langle f(x)\rangle = \bar 0$ and so $\bar x$ is a zero.

If $f(x)$ is irreducible over $K$, then $K[x]/\langle f(x)\rangle$ is a field extension of $K$ whose degree is given by the degree of $f(x)$, and $f(x)$ is the minimal polynomial of $\bar x$.

$\endgroup$
  • $\begingroup$ Okay, I understand. But why can we assume $K[X]/(f)$ as an algebraic field extension, if for that we need to assume that $f$ is the minimal polynomial of some $a$? $\endgroup$ – B.Swan Feb 28 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.