Abstract Algebra | Properties and examples of ring homomorphisms.

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Michael Penn

Michael Penn

Күн бұрын

We present some important properties of ring homomorphisms and give some examples. For instance we prove that 2Z and 3Z are isomorphic as groups but not rings.
www.michael-penn.net
www.researchgate.net/profile/...
www.randolphcollege.edu/mathem...

Пікірлер: 11
@darrenpeck156
@darrenpeck156 Жыл бұрын
Great examples!!!
@darrenpeck156
@darrenpeck156 Жыл бұрын
Don't we need to be in Q for n to have an inverse, but we are only in Z. It seems to be a big problem. Unless we are in Z mod 3?
@ashishKjr
@ashishKjr 4 жыл бұрын
Hello! Can you make some videos on Orbit Stabilizer theorem, Class equation, Sylow's theorem?
@MichaelPennMath
@MichaelPennMath 4 жыл бұрын
Yes, these are already in my overall plan.
@darrenpeck156
@darrenpeck156 Жыл бұрын
Please make these videos. I love the involved long examples; it's like story telling!!!
@darrenpeck156
@darrenpeck156 Жыл бұрын
How do we know n has an inverse for division?
@asht750
@asht750 3 жыл бұрын
Some textbooks use the additive group or subgroup as a condition to prove if a subset of a ring is ideal.
@carlosdominguez9420
@carlosdominguez9420 2 жыл бұрын
Thanks Bro
@JaredCai
@JaredCai Жыл бұрын
12:44 I think he meant there's no ring isomorphism(where he said homomorphism) from 2Z to 3Z.
@maxdickens9280
@maxdickens9280 10 ай бұрын
He didn't solve the equation correctly. In fact, n=0 is an additional solution for 6n = 9n^2. Nonetheless, this implies that the homomorphism simply maps every number to zero, resulting in a trivial homomorphism.
@maxdickens9280
@maxdickens9280 10 ай бұрын
At 12:44: Michael didn't solve the equation correctly. In fact, n=0 is an additional solution for 6n = 9n^2. Nonetheless, this implies that the homomorphism simply maps every number to zero, resulting in a trivial homomorphism.
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