(Re-edited) Let $k$ be a field and $a,b$ algebraic over $k$ but not inside $k$ and $a \neq b$. Suppose that $[k(a,b) : k]=[k(a):k] [k(b):k]$. What does this tell us about the relation between $a,b$?

By the multiplicativity of the degree, we must have $[k(a)(b):k(a)]=[k(b):k]$ and $[k(a)(b):k(b)]=[k(a):k]$.

Can we say anything about the inseparability of e.g. the extension $k \subseteq k(a)$? Can we deduce any other interesting fact?

Edited: Conversely, if $k \subseteq k(a),k \subseteq k(b)$ are algebraic and purely inseparable, can we conclude that $[k(a)(b):k(a)]=[k(b):k]$ and $[k(a)(b):k(b)]=[k(a):k]$?

  • $\begingroup$ A simple condition is that $[k(a):k]$ and $[k(b):k]$ are coprime. $\endgroup$ – Dan Petersen Oct 13 '12 at 19:49
  • $\begingroup$ @DanPetersen: Thanks, i edited the question, as i am interested in necessary conditions. $\endgroup$ – Manos Oct 13 '12 at 19:57
  • $\begingroup$ Why do you think separability has something to do with it ? $\endgroup$ – Belgi Oct 14 '12 at 15:38
  • $\begingroup$ @Belgi: I am working on a rather hard and somehow unclear homework problem which is about proving an equivalence of two facts. If i start from the other side of the equivalence i am getting something like $k \subseteq k(b),k \subseteq k(a)$ being purely inseparable, and i am supposed to conclude that $[k(b):k] = [k(b)(a):k(a)]$. But again, the problem is unclear and i am not yet certain of the validity of my interpretation. I will not post the actual homework problem here until i submit my solutions. $\endgroup$ – Manos Oct 14 '12 at 15:45

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