Abstract Algebra | Eisenstein's criterion

[@felipelopes3171]

Damn, these proofs were amazing. Keep these videos coming!

[@brandonk9299]

I'm struggling with a counter example to both approaches: x^2+4x+16. By Eisenstein, p=2 (or using 2^2), it should be reducible. If we use Z_3 where p=3 x^2+x+1 shows it should be reducible when x=1 a factor. The original roots are complex to boot. If I introduce x=x+1, then it changes to x^2+6x+21. Then using p=3 satisfies Eisenstein. Is there a way to know when to not trust the false positive given by Eisenstein and the Zp test and/or which element to use to test?

[@mizarimomochi4378]

Really late, but: - As 16 is divisible by 4 (2^2), Eisenstein's criteria can't be used with p = 2. - With x^2 + x + 1, it does indeed factor over Z_3[x] because it has that root. However, Eisenstein's is only used for Q[x] (and by proxy, Z[x]). x^2 + 6x + 21 is irreducible over Z[x] (no 2 integers multiply to 21 and add to 6), and so it is irreducible over Q[x]. If you take x^2 + 6x + 21 mod Z_3[x], you get x^2, which is reducible, so you can't conclude whether x^2 + 6x + 21 is reducible or irreducible from that. The statement is if f(x) mod Z_p[x] is irreducible and has degree n, then f is irreducible over Z[x].

[@cowworker53]

i found many proofs for eisenstein criterium, and this one is simple and convincing. But i still have a question, if you could help me. At 21:19, we see that degree of C(X) equals n. Thus degree of B(X) is 0. But if b_0 is different from 1 and -1, B(X)=b_0 is not inversible in Z[X], so b_0 * C(X) is a reduction i think (like, for example, f(X)=2X is factorised in 2 times X in Z[X]). Isn't it right? thanks

[@cowworker53]

Answer to myself : we want a reduction in Q[X] of f(X) with coefficient in Z[X]. So it is proved that there is no reduction in Q[X]. Great proof! thanks

[@digxx]

Can anyone help me out here at 15:10. p(x) is a rational polynomial. Let's name the coefficient of x^n as c_n/d_n. Who says that d_n does not contain the prime p in which case p(x) would not be defined over Z_p[x] or put differently d/d_n in d/c*p(x) does not allow to factor out the prime p.

[@navyasubba_4493]

Why did u multiply the gcd with a

[@42-94]

I believe that treating Gauss's Lemma as a lemma is really insulting. This fact after 30 seconds of meditation tells us that R[x] is an UFD for any UFD R. This is a very good result

[@sumitsingh7646]

Love from india...

[@thayanithirk1784]

Sumit Singh comment from india

[@AmanKumar-bz7pt]

@@thayanithirk1784 where the hell is your educational channel thaynithi

[@sumitsingh7646]

What's about this channel

[@ibrahimkachal6759]

what is integral domain

[@aweebthatlovesmath4220]

Integral domain is an ring with 1 and no zero divisor a and b are said to be zero divisor if a≠0≠b but ab=0 Z/4Z is not an integral domain because 2×2=4=0 but 2≠0

[@otakurocklee]

But Gauss' lemma (as you've proved it here) only applies to monic polynomials. How can you deduce that a non-monic polynomial irreducible over integers is also irreducible over the rationals?

[@filipbaciak4514]

I have thougth of the same problem