Ring homomorphisms and isomorphisms -- Abstract Algebra Examples 21
Vložit
- čas přidán 7. 09. 2024
- ⭐Support the channel⭐
Patreon: / michaelpennmath
Merch: teespring.com/...
My amazon shop: www.amazon.com...
🟢 Discord: / discord
⭐my other channels⭐
Main Channel: / michaelpennmath
non-math podcast: / @thepennpavpodcast7878
⭐My Links⭐
Personal Website: www.michael-pen...
Instagram: / melp2718
Randolph College Math: www.randolphcol...
Research Gate profile: www.researchga...
Google Scholar profile: scholar.google...
Michael, for the proof of the "Second Ring Isomorphism Theorem" you should also prove that the function φ is well defined.
Let s1+i1=s2+i2. We should prove that φ(s1+i1)=φ(s2+i2).
But s1-s2=i2-i1, so s1-s2 ∈ S∩I, so φ(s1+i1)=s1+S∩I=s2+S∩I=φ(s2+i2)
pinned!
Thank you for commenting this. Very helpful!
In the last problem Michael uses proper divisors instead of just divisors by mistake - it should be ord φ(1) ϵ {1,2,3,5,6,10,15,30} and ord φ(1) ϵ {1,2,4,5,10,20}, and hence ord φ(1) is in the intersection of these two sets which is {1,2,5,10}. More efficiently we can argue that since ord φ(1) | 30 and ord φ(1) | 20 we have ord φ(1) | gcd(30, 20) = 10, and hence ord φ(1) ϵ {1,2,5,10}.
@24:25 we don’t need any number theory results. If ab=0 and a^2+3b^2=2 then from ab=0 we know that either a or b is 0. If b=0 this becomes a^2=2 which has no solution in the rationals and if a=0 this becomes 3b^2=2 which implies b^2=2/3 which also has no solution in the rationals. Thus a contradiction, and therefore no ring isomorphism.
@ 27:01 I guess you forgot to check ord [φ(1)] = 10. Then, by ord(m) = n / gcd(m, n) [equivalent to gcd(m, n) = n / ord(m)], you would get gcd(m, n) = gcd(φ(1), 20) = 20 / ord [φ(1)] = 20 / 10 = 2. All the numbers 0 ≤ m < 20 such that gcd(m, 20) = 2 are in the set {2, 6, 14, 18}. Now observe that:
1) 2^2 = 4 ≠ 2 [in Z_20], so φ(1) cannot be equal to 2
2) 6^2 = 36 = 16 [in Z_20] ≠ 6, so φ(1) cannot be equal to 6
3) 14^2 = 196 = 16 [in Z_20] ≠ 14, so φ(1) cannot be equal to 14
4) 18^2 = 324 = 4 [in Z_20] ≠ 18, so φ(1) cannot be equal to 18.
I wanted to point it out even though you didn't miss any solutions :) great job as always
Thank you for this!
The homomorphism options from Q[root 2] to itself reminds me of when someone pointed out that the choice of which square root of -1 is i is arbitrary, but making the other choice results in an indistinguishable structure for the complex numbers. If you don't have an order relation added to your field of Q[root 2], you can't tell which root you used.
You mean the example of Q[root 3] right?
Cuz you mentioned 2, but I want to make sure I understand what you’re referring to.
Are you saying similar to how both i and -i both square to -1?
@@Happy_Abe Yes, sorry, root 3. Since both roots square to 3 and otherwise can't be simplified, they've got to be the same up to isomorphism.
hello in video Ring Homomorphisms and Isomorphisms -- Abstract Algebra 21 we have used the fact that phi(1)=a with a that can be different to 1, but in the first exemple of this video we have directly set phi(1)=1 I'm confused if someone can clarify this
In the lecture, phi was only required to be a homomorphism. Here in the example, phi is an isomorphism.
1:42 based god lil b reference