# If $F\subseteq K$ are fields, $\alpha \in K$ Prove $\alpha$ is algebraic over $F$ [closed]

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

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.)

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

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• @Arnaud D. It's the extension field – Ilan Aizelman WS May 22 at 9:20
• Its equivalent to showing $F(\alpha) = F[\alpha]$ and $|F(\alpha) :F|$ is finite. – Ilan Aizelman WS May 22 at 9:22
• Note that $\Bbb Q\subset \Bbb Q(\pi)$, but $\pi$ is not algebraic over $\Bbb Q$. – Thomas Shelby May 22 at 9:24
• You need that the extension is finite. – Wuestenfux May 22 at 9:42
• @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. – Ilan Aizelman WS May 22 at 9:51

$$(1)\implies(2)$$ 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 …

• 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? – Ilan Aizelman WS Jun 10 at 13:03
• @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? – egreg Jun 10 at 13:05
• Can I conclude from that, that there exists some non-zero polynomial $g(x)$ with coefficients in $K$ such that $g(a) = 0$? – Ilan Aizelman WS Jun 10 at 13:08
• @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 $c_0+c_1x+\dots+c_nx^n\in F[x]$. – egreg Jun 10 at 13:14
• I see, thank you. (: – Ilan Aizelman WS Jun 10 at 13:20

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 $$p(x)$$.