Arguing inductively, it suffices to prove that if 𝐿 =𝐾(𝛼,𝛽) with 𝛼,𝛽 separable elements over 𝐾, then 𝐿 is simple. We also assume that 𝐾 is infinite, since the finite case is covered by Finite extension of a Galois field.

Suppose we have distinct morphisms of field extensions 𝜄,𝜄′ ∈𝖥𝗅𝖽𝐾(𝐹,――𝐾). Then the polynomials

are distinct, for otherwise 𝜄(𝛼) =𝜄′(𝛼) and 𝜄(𝛽) =𝜄′(𝛽) implying 𝜄 =𝜄′. Therefore the polynomial

is not identically zero. Since 𝐾 is infinite, it follows that there exists a 𝑐 ∈𝐾 so that the evaluation 𝑓(𝑐) ≠0. This means distinct 𝜄 ∈𝖥𝗅𝖽𝐾(𝐹,――𝐾) map 𝛾 =𝑐𝛼 +𝛽 to distinct elements 𝜄(𝛾) =𝜄(𝛼)𝑐 +𝜄(𝛽). Since the cardinality of 𝖥𝗅𝖽𝐾(𝐹,――𝐾) is the separable degree [𝐹 :𝐾]s, and each 𝜄(𝛾) is a root of the minimal polynomial of 𝛾 over 𝐾, we have

But by hypothesis the extension is separable, so the upper and lower bounds are equal, squeezing [𝐾(𝛾) :𝐾] =[𝐹 :𝐾], whence 𝐾(𝛾) =𝐹.