At 16:28, how is A_1 a subring of A? If the polynomials p(x) are the elements of the ring, and the constant term of p(x) is 0, then there's no multiplicative identity. Yes, the additive identity is p(x) = 0, but the multiplicative identity would have to be p(x)=1, correct? Is A_1 a subRNG and not a subRING?
This lecture is a quick review of the notion of rings, ideals, quotient ideals, homomorphisms of rings, kernel of a ring, the first isomorphism of rings, zero-divisors, nilpotent elements and units. Examples mentioned include: R[x], the ring of all polynomials in one variable, Z/nZ.
around 35:05, they refer to the Rank-Nullity Theorem of linear algebra. I'm not very familiar with Indian accents, so I had a hard time figuring that out.
minor typo at 9:56 should be A[x] = { a_0 + a_1.x + .... + a_n.x^n: n\member \mathbb{N}, a_i \in A} ie the coefficients are in A not in \mathbb{R}. What he wrote is \mathbb{R}[x].
the camera moves too much and does not give time to read the blackboard. camera man is trying to center the teacher at the expense of the blackboard content. very annoying.
help me please. 21:40 he says the kernel f must be a subring of A, which means kernel f must have an identity 1' even if it is not equal to 1 (identity of A). Now i have a ring homomorphism from Z to Z/2Z, kernel f is obviously 2Z which has no identity. So he made a mistake here, as kernel f may not be a subring, right?
See 14:08 - for this series of lectures, he does not require a subring to contain an identity element. This allows all ideals to be considered as subrings. This is fairly atypical for commutative algebra, but since that's the definition he's using, he is correct in saying that kernels are subrings. If you require a subring to have multiplicative identity (or more common: have the _same_ multiplicative identity as the ring itself), then most ideals (and hence kernels) are not subrings.
Our ring contains multiplicative identity (definition). When you say subring is itself is a ring under the same binary operation. Than subring has to contain some multiplicative identity. Isn't it?
At 16:28, how is A_1 a subring of A? If the polynomials p(x) are the elements of the ring, and the constant term of p(x) is 0, then there's no multiplicative identity. Yes, the additive identity is p(x) = 0, but the multiplicative identity would have to be p(x)=1, correct? Is A_1 a subRNG and not a subRING?
This lecture is a quick review of the notion of rings, ideals, quotient ideals, homomorphisms of rings, kernel of a ring, the first isomorphism of rings, zero-divisors, nilpotent elements and units.
Examples mentioned include: R[x], the ring of all polynomials in one variable, Z/nZ.
around 35:05, they refer to the Rank-Nullity Theorem of linear algebra. I'm not very familiar with Indian accents, so I had a hard time figuring that out.
Same here. Thanks bro.
Very nice. Good profesor
minor typo at 9:56 should be A[x] = { a_0 + a_1.x + .... + a_n.x^n: n\member \mathbb{N}, a_i \in A} ie the coefficients are in A not in \mathbb{R}. What he wrote is \mathbb{R}[x].
Never mind - he fixed it at 11:05
You use \member instead of \in? Wild
Did someone happen to transcribe these lectures? Would you be willing to share?
Anybody proved exercise given at 28:55?
35:08 what is ragionality theorem?
Rank-Nullity Theorem.
the camera moves too much and does not give time to read the blackboard. camera man is trying to center the teacher at the expense of the blackboard content. very annoying.
exactly!
help me please. 21:40 he says the kernel f must be a subring of A, which means kernel f must have an identity 1' even if it is not equal to 1 (identity of A). Now i have a ring homomorphism from Z to Z/2Z, kernel f is obviously 2Z which has no identity. So he made a mistake here, as kernel f may not be a subring, right?
See 14:08 - for this series of lectures, he does not require a subring to contain an identity element. This allows all ideals to be considered as subrings. This is fairly atypical for commutative algebra, but since that's the definition he's using, he is correct in saying that kernels are subrings.
If you require a subring to have multiplicative identity (or more common: have the _same_ multiplicative identity as the ring itself), then most ideals (and hence kernels) are not subrings.
Our ring contains multiplicative identity (definition). When you say subring is itself is a ring under the same binary operation. Than subring has to contain some multiplicative identity. Isn't it?
Yes, but it doesn't have to be the same as the identity of our original ring
As an example the zero ring is a subring of all rings, but has a different identity element.
Eyes are running from corner to corners
Sir Please Made Videos on A Course of Abstract Algebra from Basic to end, Please Sir.......
There are extraordinary lectures by prof Ben Gross on abstract algebra: ruclips.net/video/VdLhQs_y_E8/видео.html
JAZZAKALLAH SIR
Stupid Cameraman
there is no cameraman in the room