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Let $\alpha\in K$ be algebraic over $F$ and $K$ be a field extension of $F$, and let $f(x)$ be the minimal polynomial for $F$ for $\alpha$ over $F$. Then how do I show that the ideal generated by $f$ is maximal? I is a maximal ideal in ring R if I is not properly contained in any other ideal in R. I do not want to use the fact that $F[\alpha]$ is a field. Thank you.

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    $\begingroup$ You do not want to use that $F[\alpha]$ is a field??? $\endgroup$
    – Bernard
    Apr 7, 2020 at 13:10
  • $\begingroup$ i dont want to use this fact $\endgroup$
    – maths
    Apr 7, 2020 at 13:17
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    $\begingroup$ The assertion is false if $F[\alpha]$ is not at least an integral domain. $\endgroup$
    – Bernard
    Apr 7, 2020 at 13:20
  • $\begingroup$ yes that true and also i can think about the case when $\alpha$ is not algebraic $\endgroup$
    – maths
    Apr 7, 2020 at 13:27
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    $\begingroup$ may i know reason of downvote $\endgroup$
    – maths
    Apr 8, 2020 at 12:02

1 Answer 1

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Let me clear the situation. I guess $F\subseteq K$ where $F$ and $K$ are fields and $\alpha\in K$ is algebraic over $F$.

If $f\in K[x]$ were not irreducible, then you could write $f=gh$ where $g,h\in F[x]$ are polynomials of positive degree. Since you have $f(\alpha)=0=g(\alpha)h(\alpha)$, you get that $g(\alpha)=0$ or $h(\alpha)=0$ where the degrees of $g$ and $h$ are strictly smaller than the degree of $f$: this would contradict the minimality of $f$. Thus $f$ is irreducible, hence $(f)$ is a maximal ideal of $F[x]$ (since the latter is a PID).

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  • $\begingroup$ May I know the reason of the downvote? $\endgroup$
    – Gaussian
    Apr 7, 2020 at 18:39

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