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If $F\subseteq K$ are fields, $\alpha \in K$, and $K$ is an extension field of $F$.

Prove the following are equivalent:

  1. $\alpha$ is algebraic over $F$
  2. $F(\alpha)=F[\alpha]$
  3. $|F(\alpha):F|$ is finite

I'm trying to prove the properties from here: https://en.wikipedia.org/wiki/Algebraic_element enter image description here

Edited the question because people voted to close for not understanding what I mean in the original question:

Intuition: I did prove (2.) and (3.) independently, but I'm not sure how to show (1.)

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closed as off-topic by Magdiragdag, Yanior Weg, Adrian Keister, Xander Henderson, José Carlos Santos May 23 at 15:04

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  • $\begingroup$ @Arnaud D. It's the extension field $\endgroup$ – Ilan Aizelman WS May 22 at 9:20
  • $\begingroup$ Its equivalent to showing $F(\alpha) = F[\alpha]$ and $|F(\alpha) :F|$ is finite. $\endgroup$ – Ilan Aizelman WS May 22 at 9:22
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    $\begingroup$ Note that $\Bbb Q\subset \Bbb Q(\pi)$, but $\pi$ is not algebraic over $\Bbb Q$. $\endgroup$ – Thomas Shelby May 22 at 9:24
  • $\begingroup$ You need that the extension is finite. $\endgroup$ – Wuestenfux May 22 at 9:42
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    $\begingroup$ @ThomasShelby I want to show that (1), (2), (3) are all equivalent. so I want basically to show that $(1)$ iff $(3)$ in this sense. $\endgroup$ – Ilan Aizelman WS May 22 at 9:51
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$(1)\implies(2)$ Let $f(x)$ be the minimal polynomial of $\alpha$, which is irreducible. Take a nonzero element $g(\alpha)\in F[\alpha]$, where $g(x)\in F[x]$. Since $g(\alpha)\ne0$, we have that $f(x)$ doesn't divide $g(x)$; hence a greatest common divisor of $f(x)$ and $g(x)$ is $1$, so by Bézout's identity there exist $p(x)$ and $q(x)$ such that $f(x)p(x)+g(x)q(x)=1$. Can you conclude $g(\alpha)^{-1}\in F[\alpha]$?

$(2)\implies(3)$ It should be easy to show that $F[\alpha]$ is a finite dimensional vector space over $F$.

$(3)\implies(1)$ If $|F(\alpha):F|=n$, then $1,\alpha,\dots,\alpha^n$ are …

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  • $\begingroup$ Hi egreg, from $(3)$ to $(1)$, I can say that $1,\alpha, ..., \alpha^n$ are the linearly dependent elements over $F$ and thus $\alpha$ is algebraic? $\endgroup$ – Ilan Aizelman WS Jun 10 at 13:03
  • $\begingroup$ @IlanAizelmanWS No, they aren't all the elements; but they're $n+1$ elements in a vector space of dimension $n$, so they're linearly dependent. What does this give? $\endgroup$ – egreg Jun 10 at 13:05
  • $\begingroup$ Can I conclude from that, that there exists some non-zero polynomial $g(x)$ with coefficients in $K$ such that $g(a) = 0$? $\endgroup$ – Ilan Aizelman WS Jun 10 at 13:08
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    $\begingroup$ @IlanAizelmanWS Yes: the elements are linearly dependent, so there are $c_0,c_1,\dots,c_n\in F$, not all zero, such that $c_0+c_1\alpha+\dots+c_n\alpha^n=0$, so $\alpha$ is a root of $c_0+c_1x+\dots+c_nx^n\in F[x]$. $\endgroup$ – egreg Jun 10 at 13:14
  • $\begingroup$ I see, thank you. (: $\endgroup$ – Ilan Aizelman WS Jun 10 at 13:20
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Hint:$(3)\implies (1)$ : Suppose degree of $\alpha$ over $F$ is $n$, that is, $[F(\alpha):F]=n$. Then consider these $n+1$ elements: $1,\alpha,\alpha^2,\ldots \alpha^n$, which are linearly dependent over $F$.

$(1)\implies (3)$ : If $\alpha$ is algebraic over $F$, then $F(\alpha)\cong\dfrac{F[x]}{(p(x))}$, where $p(x)$ is the minimal polynomial of $\alpha$ over $F$. Dimension of $\dfrac{F[x]}{(p(x))}$ as a vector space over $F$ is same as the degree of $p(x)$.

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