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4 Correct typo: deg(P) = n+3, not n+2 as I wrote

[Edited to add the final field-theoretic paragraph]

Just noticed this in a list of "Related" problems. I've seen applications to the computation of Galois groups $G$ of explicit polynomials $P \in k[X]$. See for example Abhyankar's survey paper

[A] Abhyankar, Shreeram S., with an appendix by J.-P. Serre: Galois theory on the line in nonzero characteristic, *Bull. AMS (N.S.) 27 #1 (July 1992), 68–133.

(Note the dedication to "Walter Feit, J-P. Serre, and e-mail"!)

Let $N = \deg P$. If $G = S_N$ or $A_N$ then this can be proved by showing that $G$ is $n$-transitive for $n$ large enough (depending on $N$), at which point $G=A_N$ if ${\rm disc}(P\phantom.)$ is square in $F$, and $G=S_N$ if not. [This last assumes $2 \neq 0$ in $k$; there's a pseudo-discriminant criterion that works in characteristic 2.] As T. Sauvaget notes in his question, $n>5$ is always enough, but in fact $n=4$ suffices except for $N=11,12,23,24$ (Mathieu groups), and even $n=3$ brings it down to a usually-manageable list of possibilities (see [A, pages 86–87]).

This approach can be useful because showing (say) 4-transitivity amounts to proving that a few polynomials are irreducible. Indeed $P$ itself is irreducible iff $G$ is 1-transitive; in this case, we may adjoin to $k$ a root $X_0$ of $P$, and then the point stabilizer in $G$ is the Galois group of the degree-$(N-1)$ polynomial $P_1 := P(X)/(X-X_0)$ over $k_1 := k(X_0)$, so $G$ is 2-transitive iff $P_1$ is irreducible over $k_1$, in which case we can adjoin a second root, etc.; if $P_3$ is irreducible then $G$ is 4-transitive, and you're done (except in the four Mathieu cases where you must go one or two steps further). Again see [A], in particular Section 4 "Throwing away roots" (p.69).

This also has the following amusing consequence. Let $P\phantom.$ be an irreducible separable polynomial over $k$, and define $P_1, P_2, \ldots, P_n$ as before, as long as $P_m$ is irreducible for each $m<n$. Then:

i) If each of $P_1,P_2,P_3$ is irreducible but $P_4$ is reducible then $\deg P \phantom. \in \lbrace 6, 11, 23 \rbrace$.

ii) If each of $P_1,P_2,P_3,P_4$ is irreducible but $P_5$ is reducible then $\deg P \phantom. \in \lbrace 7, 12, 24 \rbrace$.

iii) Suppose $n \geq 5$. If $P_m$ is irreducible for each $m \leq n$, but $P_{n+1}$ is reducible, then $\deg P \phantom. = n + 2$3$. Moreover, each of the allowed possibilities for$\deg P\phantom.$in (i), (ii), and (iii) occurs for suitable$k$and$P$. This statement does not explicitly mention finite groups at all, but given Galois theory it is equivalent to the classification of 4-transitive permutation groups. 3 Add field-theoretic consequence [Edited to add the final field-theoretic paragraph] This also has the following amusing consequence. Let$P\phantom.$be an irreducible separable polynomial over$k$, and define$P_1, P_2, \ldots, P_n$as before, as long as$P_m$is irreducible for each $m<n$. Then: i) If each of$P_1,P_2,P_3$is irreducible but$P_4$is reducible then$\deg P \phantom. \in \lbrace 6, 11, 23 \rbrace$. ii) If each of$P_1,P_2,P_3,P_4$is irreducible but$P_5$is reducible then$\deg P \phantom. \in \lbrace 7, 12, 24 \rbrace$. iii) Suppose$n \geq 5$. If$P_m$is irreducible for each$m \leq n$, but$P_{n+1}$is reducible, then$\deg P \phantom. = n + 2$. Moreover, each of the allowed possibilities for$\deg P\phantom.$in (i), (ii), and (iii) occurs for suitable$k$and$P$. This statement does not explicitly mention finite groups at all, but given Galois theory it is equivalent to the classification of 4-transitive permutation groups. 2 Spelled out the "etc." in the last paragraph. Just noticed this in a list of "Related" problems. I've seen applications to the computation of Galois groups$G$of explicit polynomials$P \in k[X]$. See for example Abhyankar's survey paper [A] Abhyankar, Shreeram S., with an appendix by J.-P. Serre: Galois theory on the line in nonzero characteristic, *Bull. AMS (N.S.) 27 #1 (July 1992), 68–133. (Note the dedication to "Walter Feit, J-P. Serre, and e-mail"!) Let$N = \deg P$. If$G = S_N$or$A_N$then this can be proved by showing that$G$is$n$-transitive for$n$large enough (depending on$N$), at which point$G=A_N$if${\rm disc}(P\phantom.)$is square in$F$, and$G=S_N$if not. [This last assumes$2 \neq 0$in$k$; there's a pseudo-discriminant criterion that works in characteristic 2.] As T. Sauvaget notes in his question,$n>5$is always enough, but in fact$n=4$suffices except for$N=11,12,23,24$(Mathieu groups), and even$n=3$brings it down to a usually-manageable list of possibilities (see [A, pages 86–87]). This approach can be useful because showing (say) 4-transitivity amounts to proving that a few polynomials are irreducible. Indeed$P$itself is irreducible iff$G$is 1-transitive; in this case, we may adjoin to$k$a root$X_0$of$P$, and then the point stabilizer in$G$is the Galois group of the degree-$(N-1)$polynomial$P_1 := P(X)/(X-X_0)$over$k_1 := k(X_0)$, so$G$is 2-transitive iff$P_1$is irreducible over$k_1$, in which case we can adjoin a second root, etcetc.; if$P_3$is irreducible then$G\$ is 4-transitive, and you're done (except in the four Mathieu cases where you must go one or two steps further). Again see [A], in particular Section 4 "Throwing away roots" (p.69).

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