Thanku so much sir , mei du se msc krri hu lekin hmare teacher jaise pdhate h tou field theory ko leke drr beth gya h dil mei bs apki videos hi dekh k exam krke aungi 9 Jan ko h thanku for videos ❤
Subfields of Q (root2, root3) are four i think, like Q (root2+root3), Q (root2×root 3), Q (root2/root3) , Q (root2--root3), aisa ek question dec2012 net exam may aaya hai please check, answer 4 subfields hii hai
Sir, in the last problem the degree of the extension is 4 but dimension is 2. But I have learned a definition regarding degree of extension in which it was written that " The dimension of of F as a vector space over K ks called the degree of the extension F/K". So, how the dimension can be different from from the degree of extension? Sir pls reply.
x^4+4=0 implies x=(4)^1/4 e^(2npii/4) =root2 ×(cosnpi/2 +isinnpi/2) n=0, 1, 2, 3, so i think splitting field over Q is Q (2^1/2, i) and [Q (root2, i):Q]=4, please check and reply me 🙏
Sir, could you please check your last worked out example. There seems to be some mistake in finding the splitting field. I think the splitting field is Q(√3i) and not Q(√3, i).
Sir, a small clarification please. In the problem "SF of x^4+x^2+1 over Q, the roots are w, -w, w^2,-w^2, where w is complex cube root of unity. Thus the SF for f(x) over Q is Q(w) ,degree is 2. It's not 4 as u said. Please correct if I am wrong
Sir ji, clearly Q(w) contains w,-w,w^2,-w^2 all the four roots. Minimal polynomial of w over Q is x^2+x+1. Just assign w to Q to get the SF, instead of assigning root3, i separately. Please reply sir
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Thanku so much sir , mei du se msc krri hu lekin hmare teacher jaise pdhate h tou field theory ko leke drr beth gya h dil mei bs apki videos hi dekh k exam krke aungi 9 Jan ko h thanku for videos ❤
Zabardast job. Very nice lecture
Thanku sir.....This was a confusing topic for me.....Bt now it's clear.....
Thank you so much Dear ❣️
Keep Watching
Subfields of Q (root2, root3) are four i think, like Q (root2+root3), Q (root2×root 3), Q (root2/root3) , Q (root2--root3), aisa ek question dec2012 net exam may aaya hai please check, answer 4 subfields hii hai
Please confirm any reply me 🙏, i think right or not
No .....jo maine karwaya wahi banegi .............
Jo apne banai hai wo sari same hai Sandeep ...ak daffa app check kare
@@parveenkumarmath ok sukriya, but please check csir-net question dec2012
Ok I will try
Nice explanation sir thanks for help us, please give one lecture on splitting field over finite field, like z_p
Lecture 6 is on this topic .......and also Our next Lecture is on same topic
Nicely explaine every topic sir
Thank you so much Dear . . . Keep watching
Nice explanation about splitting field...
Thank you so much 🙏🙏
Very much thank you professor
Always Welcome ...and Keep watching............🙂🙂🙂☺️☺️☺️
Sir, in the last problem the degree of the extension is 4 but dimension is 2. But I have learned a definition regarding degree of extension in which it was written that " The dimension of of F as a vector space over K ks called the degree of the extension F/K". So, how the dimension can be different from from the degree of extension? Sir pls reply.
Haa Bhai me bhi yahi likhne aaya tha ,
Lots of thanks sir...u make this so easy for us🙏🙏
Thank u sir
x^4+4=0 implies x=(4)^1/4 e^(2npii/4) =root2 ×(cosnpi/2 +isinnpi/2) n=0, 1, 2, 3, so i think splitting field over Q is Q (2^1/2, i) and [Q (root2, i):Q]=4, please check and reply me 🙏
Yes , Bilkul sahi kiya hai Apne ... Correct...
Sir, could you please check your last worked out example. There seems to be some mistake in finding the splitting field.
I think the splitting field is Q(√3i) and not Q(√3, i).
Yes!!😊
thank u so much sir🙏
*_Thank you so much Dear....for appreciation_*
x^n -a=0 spliting field kya hoga
Thank youuuuuu sir.........
Your Most Welcome Dear 🙂
How we came to know about complex is smallest field over R
Sir in the video last example [Q(√3,i):Q]=4
Why Basis=(i, √3)
I think Basis =(1,√3,i,i√3)
Becouse Degree=elements in Basis
Yes ..............right
Recommended Book please?
Thank you so much sir
Most welcome ☺️
Thanku sir
Please share link how to find automorphisms of field by conjugate roots method.
Sir, a small clarification please. In the problem "SF of x^4+x^2+1 over Q, the roots are w, -w, w^2,-w^2, where w is complex cube root of unity. Thus the SF for f(x) over Q is Q(w) ,degree is 2. It's not 4 as u said. Please correct if I am wrong
It's 4 , because w is made by root 3 and i
Sir ji, clearly Q(w) contains w,-w,w^2,-w^2 all the four roots. Minimal polynomial of w over Q is x^2+x+1. Just assign w to Q to get the SF, instead of assigning root3, i separately. Please reply sir
Nice explanation ❤❤❤
Thank you so much Dear ❣️
Keep Watching
Splitting field of x^4+4, x^7--1, x^5+1 over R=R (i)=C?? Please answer, splitting field of x^5+1 over Q is Q (e^(ipi/5)? ?please answer
Sir, plz tell the spliting field of x^5+1 over Q.
You can easily do it .....by calculating roots of this equation
Thanku sir 😊
Your Most Welcome Dear 😊😊
basis 2 ni 4 banen gi bcz basis is the cardinality of dimensions to basis {1,sqrt2,sqrt2i} banen gi check kren plz
Sir, we know {0} is a trivial subfield of every field.
Okey
Bhai {0} trivial kaise ko Sakta h because cardinality of finite field is prime power so at least cardinality is 2
Sir kya various universities me Phd entrance ke liye enough hai ye video ?
Yes ....Bilkul
Sir agr question
Find[Q(√3,5√7):Q]
Ho to kya find krn gy??
Degree, splitting field???
4
Sir please make more videos on fixed field and galois extension .My university exams are very near.please make it as soon as possible
*_Sure_*
Last example degree 2 ayegi
Sir apka number send kar do
9813155942