Eisenstein criterion and Problems 7

Lecture 30

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[sachinpasarge]

Minor Error: 26:22 ( According to hypothesis "p divides a0, p^2 doesnot divide a0") ( Check it out at 02:40 ) PS: It should have been p^2 instead of p, other things are fine.

[Sayan_Ghosh]

At 29:36 the idea seems not very clear to me because here we're going to show f is irreducible in Z[X] via contradiction by taking f is not irreducible in Z[X] right? But again we're assuming f is irreducible here in 29:36. If we've settled with f is irreducible then we're done, right? So all we need to do is to assume f is not irreducible first and then we'll get a contradiction!! Or if I'm missing something idk

[sripad]

Thanks a lot Mr Rahul for uploading the videos.

[hemraj]

Here we are given that f is primitive and we are assuming that f is reducible in Z[X].And we will prove by contradiction .so f can only be written as a product of some constant and a integer polynomial.f can not be written as product two non constant integer polynomial.because if we could write f as a product of two non constant Integer we could also write f as a product of two non constant rational polynomial because any integer polynomial is a rational polynomial but sir already proved that f is irreducible in Q[X] so the only way we can write f is as a product of some constant and integer polynomial .let f=cf0 that mean c divide f that mean c divide every coefficient of f but since f is primitive gcd of coefficient of f is one in other words all coefficient are not divisible by any integer other than 1 and we know that c is not 1 because we have written f as reducible factor. So c has to be 1.If c is one we are done then f =cf0 is irreducible factorisation.So for f to be irreducible in Z[X] it is important that f is primitive.