Talk:Separable polynomial

In mathematics, a polynomial P(X) is separable over a field K if its roots in an algebraic closure of K are distinct

I thought this definition was only valid for an irreducible polynomial; and a general polynomial is separable if its irreducible factors are separable? Or is this the same (if it is, I don't think this is obvious)?

mazi 12:04, 20 February 2006 (UTC)[reply]

It's not the same. For example, it is not the same as P and its derivative having no common factor. Charles Matthews 15:13, 20 February 2006 (UTC)[reply]

Statement about connection to Galois Theory

The article defines separability in terms of the irreducible factors of the polynomial. Then it says:

Irreducible polynomials over perfect fields are separable, which includes in particular all fields of characteristic 0, and all finite fields. This criterion is of technical importance in Galois theory. In this connection, the concept of separability is of lesser importance if P is not assumed irreducible, since repeated roots may then just reflect that P is not square-free.

The last claim makes no sense given the definition: if P is not square-free, this does not affect separability: either the irreducible factors (repeated or not) have multiple roots or they do not. Squaring a separable irreducible factor does not give the factor multiple roots. Magidin (talk) 05:31, 31 October 2010 (UTC)[reply]

The article gives two distinct definitions of separability. Squaring a polynomial preserves separability under the first definition, but breaks it under the second one, and that's presumably the one which was referred to in this sentence. Nevertheless, I have removed it as it only serves to add to the confusion, it does not convey any useful information.—Emil J. 13:36, 1 November 2010 (UTC)[reply]

I think the statement was meant to refer to the criterion that an irreducible polynomial is separable if and only if it is relatively prime to its formal derivative. But it made no sense after the comment about Galois theory. Thanks for the prompt action. Magidin (talk) 14:00, 1 November 2010 (UTC)[reply]

Delete the second definition??

The second (field-dependent) definition is not standard. I propose to delete it and to keep only one definition: separable over K means no multiple roots over Kbar. MvH Feb 6 2014. — Preceding unsigned comment added by 71.229.28.197 (talk) 01:06, 7 February 2014 (UTC)[reply]

I agree that nowadays only the first definition is standard. However, as the second one is sourced by a reference to a notable book, we cannot remove it completely. I suggest to give the first definition, and then to introduce the second one by "some authors have used a slightly different definition, which is no more in use". D.Lazard (talk) 10:08, 7 February 2014 (UTC)[reply]

Lets present the modern definition, and make the old definition less prominent (just like the definition of prime numbers, which used to include 1, but not anymore). MvH Feb 7 2014. —Preceding undated comment added 16:14, 7 February 2014 (UTC)[reply]

Do you know if the alternative definition has a name, other than "product of separable polynomials"? --nBarto (talk) 20:47, 17 February 2018 (UTC)[reply]