Separable Degree is At Most Equal To Degree
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Theorem
Let $E / F$ be a tower of fields.
Let $\index E F$ be finite.
Then $\index E F_s$ is finite, and:
- $\index E F_s \le \index E F$
where:
- $\index E F$ denotes the degree of $E / F$
- $\index E F_s$ denotes the separable degree of $E / F$.
Proof
This theorem requires a proof. In particular: Theorem $4.1$ of Lang's Algebra You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 2002: Serge Lang: Algebra (Revised 3rd ed.): Chapter $\text V$: $\S 4$: Separable Extensions: Theorem $4.1$