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This is the exact wording of a proposition from a note in commutative algebra.

Let $E/F$ be a field extension such that $E=F(x)$ for some element $x \in E$ (meaning that $E$ is the smallest field containing $F$ and $x$.) TFAE:

(i) $E/F$ is a finite extension

(ii) $x$ is algebraic over $F$.

(iii) $E$ is generated by $x$ as an $F$-algebra.

My question:

  1. I am confused with what the differences between $F(x)$ and $F[x]$ is. It seems equality holds when $x$ is algebraic.

  2. How would we argue the equivalences? In particular, I couldn't get from (iii) back to (ii) or (i).

Below are my attempts for 1. (can ignore, but please check if you have time):

$(1) \Rightarrow (2) $ Suppose $\dim (E_F) =k < \infty$, then note that $1,x,\ldots ,x^k$ form a dependent set, hence $\sum f_i x^i = 0 $. (So this shows that any element of $E$ is algebraic over $F$)

$(2) \Rightarrow (1)$ If $x$ is algebraic over $F$, then we should have $F(x) = F[x]$. This could be obtained by considering the quotient of the minimal polynomial. By divison algorithm, we can show $|F[x]:F|=n$ where $n$ is degree of minimal polynomial.

$(2) \Rightarrow (3)$ As above $(2)$ implies $F(x) = F[x]$, and so is an $F$-algebra, and is clearly the smallest $F$-algebra containing $x$.

EDIT: After more research, I updated my attempts.

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    1. same thing no difference 2. dont 3. learn how 2 use google :P – 2017-02-03
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    For 1. If $K/F$ is a field extension, and let $x \in K$ be transcendental, then the smallest subfield of $K$ that contains both $x$ and $F$ is not necessarily $F[x]$. In fact, I think they are the rationals of $F[x]$ (?) – 2017-02-03
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    You are right that $F(x)=F[x]$ when $F$ is algebraic. But if $x$ is a transcendental element over $F$, then not only are they not the same, but $F[x]$ isn't even a field! (it doesn't contain $x^{-1}$) – 2017-02-03

1 Answers 1

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(i) $\implies$ (ii) is correct.

(ii) $\implies $(iii): all you have to prove is $P(x)^{-1}\in F[x]$, ($P(x)\ne 0$) if $x$ is algebraic over $F$. To see this, consider the (linear) map of multiplication by $P(x)$ in $F[x]$: it is an injective endomorphism of $F[x]$, hence it is surjective since $F[x]$ is finite-dimensional, so $1$ is attained, i.e. there exists a polynomial $Q(X)$ such that $P(x)Q(x)=1$: this means $P(x)^{-1}=Q(x)$.

(iii) $\implies$ (i): If $E=K[x]$, the ideal of polynomials in $K[X]$ which vanish at $x$ is a maximal ideal, generated by a polynomial $P(X)$ of minimal degree $d>0$.

$E\simeq K[X]/(P(X))$ is a finite dimensional $K$-vector space, generated by $1, x, \dots , x^{d-1}$. Indeed, any element in $E$ is a polynomial $Q(x)$ in $x$. By Euclidean division in $K[X]$, we have $$Q(X)=P(X)Q_1(X)+R(X), \quad\deg R

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    I am fine with (i)=>(ii)=>(iii). But I don't understand in (iii) that you can state the existence of a minimal polynomial. – 2017-02-03
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    Maybe I was a little too elliptic. I've added a small justification. Is that clearer? – 2017-02-03
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    Yes, that is clearer. But is it trivial that its the maximal ideal? I would have to argue as follows: Let S denote the ideal; then (i) K[X] is PID => S = <(p(X)) > (ii) p(x) = 0 so exists minimal polynomial by well ordering (iii) of which irreducible by Euclidean algorithm. OR (i) if f(x)g(x) =0, then f(x) =0 or g(x) =0 , so f in S or g in S, hence S is a prime ideal, as K[X] is integral domain. – 2017-02-03
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    It is a maximal ideal because the quotient is the field $E$. Now as $K[X]$ is not a field, a maximal ideal is $\neq 0$. – 2017-02-03
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    For (ii)$\implies$(iii) it is not sufficient to show $x^{-1}\in F[x]$. You have to do it for every $y\in F[x]$, $y\ne0$. – 2017-02-03
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    @egreg: Quite right. I'll fix it in a moment. – 2017-02-03