Integral Domains -- Abstract Algebra 18
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M_n( *R* )[X] and M_n( *R* [X] ) are isomorphic but not literally equal, because the elements of the former are polynomials while the elements of the latter are matrices
The notation used for defining the characteristic of a ring raises an interesting question: is there a special name, or special properties, to rings where the multiplicative operation is the same as repetition of the additive operation? Or, to include more than Z in this, where the multiplicative operation is sometimes the same as repetition of addition? For example, in the ring Q, we have that 3 * r = r + r + r, even if we can't expand 7/3 * r as adding r to itself multiple times. For the ring of matrices, however, multiplication and addition are "genuinely" different.
Seems to me that whether a ring is made up of 2 genuinely different operations, compared to just 1 operation and its iteration, ought to be important, but this never seems to be discussed anywhere.
Even for matrices, you can represent repeated addition as multiplication by a matrix with the same natural number repeatedly on the diagonal. These elements have to exist, because m+m+m=(I+I+I)m, but for matrices we just don't write (I+I+I) as "3". So I think the interesting distinction is whether there are any elements besides the ones that are the same as repeated addition, but the rings that don't have any elements that multiply specially are isomorphic to Z or Z_p.
Isn't Z_2[x] an infinite integral domain that doesn't contain a copy of the integers? It's the char(R)=0 ones that contain a copy of the integers, not the ones that may be infinite in ways only reached through multiplication.
can you give a timestamp for the part of the video you're talking about
@@schweinmachtbree1013 Starts at 7:44; he's discussing the integral domains that don't have to be fields because they're not finite (and are therefore not studied in finite field theory), and makes an off-hand comment that the name "integral domains" makes sense because the infinite ones contain a copy of the integers. He says that an infinite integral domain would have all the integers: 0 and 1, and 1+1=2, and 1+1+1=3, and so on, but in Z_2[x], 1+1=0, and it's also not a field because x doesn't have a multiplicative inverse (unlike in, say, Z_2[x]/(x^4+x+1), where (x)(x^3+1)=1).
@@iabervon Indeed, good catch
Ah. now i understand better what the course syllabus is like... i was surprised when we moved on from group theory so soon.
@23:40 technically the characteristic can seemingly also be 1, in the case of the trivial ring.
The trivial ring isn't considered an integral domain - one can rule out the trivial ring depending on one's precise definitions, which vary from author to author. Some authors put "non-trivial ring" directly into the definition of an integral domain. Others define an integral domain as "a ring in which the product of non-zero elements is non-zero", i.e. if a_1,a_2,...,a_n ≠ 0 then a_1 a_2 ... a_n ≠ 0 - taking n = 0 in here gives (empty product) ≠ 0, thus 1 ≠ 0.
Michael's definitions are incorrect, because he has defined zero-divisors to be nonzero elements and integral domains to be rings with no zero-divisors, and the trivial ring satisfies this definition - therefore in Michael's definition of integral domains, "ring with 1" should be changed to "ring with 1 ≠ 0".
The reason we don't want the trivial ring to be a domain is because we don't the trivial ring to be a field (although I'm not sure that Michael knows that the trivial ring is not a field judging by the recent video on his main channel. one easy argument for the trivial ring not being considered a field is that it would make dimension for vector spaces not well-defined). As we may see later in the course, every integral domain has a "field of fractions", and if {0} were an integral domain then its field of fractions would also be (isomorphic to) {0} - so since we don't want {0} to be a field, it shouldn't be an integral domain either.
@@schweinmachtbree1013 gotcha, but based on the definitions provided in the video, what I said seems correct but I guess those definitions are not the standard ones
@ 7:10 can we have *right* cancellation for integral domains ?
yes, reason explained at 5:13
22:13 I don't think it's immediately obvious that, for natural numbers a and b, the natural number product ab being n necessarily means that the ring product (a1)(b1) is n1. It's true by distributing on both sides.
(a1)(b1) = ab = n = n1
@@alegal695 If the ring doesn't have characteristic 0, it doesn't contain a copy of the integers, so it's potentially misleading to refer to elements as "a", "b", "ab", and "n". But a1 = (1+1+...+1) a times, and b1 = (1+1+...+1) b times, and the distributive rule then means that the product of these is (1+1+...+1) ab=n times, which is n1. (For example, in F_5, if a is 2 and b is 3, (a1)(b1)=1, but so is n1.)
@@iabervonI don't follow. Is there any step in the proof in the video that is not clear to you?
@@alegal695 Ah, right. Earlier today, I'd forgotten which proof I'd noticed this in.
My original comment was that, in a completely arbitrary ring, it's not (at this point) obvious how strangely ring multiplication can behave. So, when he assumes that ring multiplication agrees with integer multiplication, it's because an a-term sum times a b-term sum is an ab-term sum by the distributive property of ring multiplication. It's hard to write a comment using what he writes as "(1+...+1)" with "a times" under it; that's what I originally meant by "(a1)".
@@iabervon I see now. Yes I agree with you: at 21:55 Mike didn't really explained why n1 = a1*b1. This can be showed as you explained by repeated use of the distributive rule and associativity of addition
M_{2x2}( R[x] ) would be in a way a set of matrices elements which are polynomials of different independent variables
Please change your content to some complex topics
??? this is a very odd request to say the least. If you saw the author of a book you didn't like on the street, you wouldn't say "please change your writing to a different genre." Creators create what they want. Consume it only if you want to. This channel has no obligation to do what you want it to do.
@@noahprentice751 Agreed. I think it wasn't an actual comment. The real comment meant to be read as
"I am way smart guys... affirm me."
Figo , if you want to learn complex analysis there are many videos about that on the internet and many books about it .
Sorry to all of you if my comment heart you