Steinitz's Theorem
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Theorem
Let $E / F$ be a finite field extension.
The following statements are equivalent:
- $(1): \quad E / F$ is simple: there exists $\alpha \in E$ such that $E = \map F \alpha$
- $(2): \quad$ there are only finitely many intermediate fields between $E$ and $F$.
Proof
$1$ implies $2$
Suppose $E = F \sqbrk \alpha$ for some $\alpha \in F$.
Let $M$ be any intermediate field between $E$ and $F$.
Let $g$ be the minimal polynomial of $\alpha$ over $M$. Write
- $\ds g = x^ d + \sum _{i \mathop = 0}^{d \mathop - 1} a_ i x^ i$
Let $M'$ be the field extension of $F$ generated by all the coefficients of $g$.
- $M' = F \paren {a_0, \dots, a_{d-1}}$
Then $M' \subseteq M$ by definition of the minimal polynomial over $M$.
Since $g \in M' \sqbrk x$,
$g$ is the minimal polynomial of $\alpha$ over $M'$,
then the degree of $\alpha$ over $M'$ is $d$, but $E = M' \sqbrk \alpha$, so
- $\index E {M'} = d$
similarly, we have $\index E M = d$,
we conclude by Tower Law,
- $\index M {M'} = 1$
hence
- $M = M' = F \paren {a_0, \dots, a_{d-1}}$
Thus it suffices to show there are at most finitely many possibilities for the polynomial $g$.
Let $f$ be the minimal polynomial of $\alpha$ over $F$,
since $f \in M \sqbrk x$, and $\map f \alpha = 0$,
by definition of $g$,
- $g \divides f$
By unique factorization in $E \sqbrk x$,
the polynomial $f$ has only finitely many divisors in $E \sqbrk x$,
so there are only finitely many intermediate fields between $E$ and $F$.
$\Box$
$2$ implies $1$
Suppose there are only finitely many intermediate fields between $E$ and $F$.
If $F$ is a finite field (equivalently $E$ is a finite field), this follows from Finite Extension of $\F_p$ is Generated By a Single Element, since the generator of $E / \F_p$ also generates $E / F$.
If $F$ is infinite, then each intermediate field between $E$ and $F$ is a proper $F$-subspace of $E$,
by Vector Space over an Infinite Field is not equal to the Union of Proper Subspaces their union can't be all of $E$.
Thus any element outside this union must generate $E$, since it is not contained in any proper subfield of $E$.
$\blacksquare$
Also see
Source of Name
This entry was named for Ernst Steinitz.
Sources
- 1910: Ernst Steinitz: Algebraische Theorie der Körper (J. Reine Angew. Math. Vol. 137: pp. 167 – 309)
- 1996: Patrick Morandi: Field and Galois Theory: Chapter 1 Galois Theory, $\S5$ The Fundamental Theorem of Galois Theory, Proposition $5.6$
- 2002: Serge Lang: Algebra (Revised 3rd ed.): Chapter V, $\S4$ Separable Extensions, first part of Theorem $4.6$