# ideal generated by minimal polynomial is maximal in $F[\alpha]$

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.

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

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).