Units in a Ring (Abstract Algebra) 

Please add more videos on vector space or on linear algebra.. I will be very thankful..

So you can also write a "circle of sevens"?

@@ilguerrierodragone129 Basically, yeah! Instead of a "Circle of Fifths" you could call it a "Circle of Seven Half-step Jumps". The interval of seven half-steps is called a Fifth by musicians because it's the fifth note in the major scale. (and it doesn't help that each of the 12 equal steps in an octave are called "half-steps") It doesn't really make sense from a math perspective.

​@@nomukun1138I'm sorry i didn't understand very well. I've read your comment again and I think it's not correct that -1 is in Z/12Z, also there is 11 which is a unit too in Z/12Z. Said that i want to understand better your concept, I don't know very well what is the cricle of fifths in music, could you explain it to me? I think that 7 is not the inverse of 5 in Z/12Z. What is the perfect fourth? And how i organize the signatures based on the circle of fifths?

@@ilguerrierodragone129 Music terminology is extremely confusing. -1 in the context of Z/12Z is the same as 11. I thought calling it -1 emphasized that it steps backward through the ring. It's the same as negative one. 7 is the additive inverse of 5 in Z/12Z. The word "inverse" usually means the "multiplicative inverse" but I meant "additive inverse". In music theory, the Perfect Fourth Interval (5 half-steps) is called the inverse of the Perfect Fifth Interval (7 half-steps), that's the reason I used the word "inverse" without considering that it would be confusing. Let me try to explain the circle of fifths... Every melody or piece of music (in this traditional European system of music) has a Key Signature describing which notes are used most frequently. A melody with the Key Signature of C would have the notes C D E F G A B, and a melody in the Key Signature of G would have the notes G A B C D E F#. The Key Signature of C# includes the notes C# D# E# F# G# A# B#, which has no overlap with the Key of C, even though the root notes are only one space apart. Playing a melody in C# at the same time as a melody in C sounds horrible because none of the notes match. However playing a melody in G at the same time as a melody in C will often sound very good, because many of the notes in the Key Signature are the same. (and because 7 generates Z/12Z, every possible key signature is some precise number of Fifths away from every other key signature, and this number of Fifths is a good measure of similarity of the key signatures) The Key of C is "closer" in sound to the Key of G even though the root notes are a Fifth apart. The Key of C is "distant" in sound from the Key of C# even though the root notes are only one half-step apart. We say that C and G are "neighbors on the Circle of Fifths". Changing a piece of music by seven half-steps, corresponding to a Fifth, sounds pleasant and changing it by one half-step sounds rough.

@@nomukun1138 thank you very much, I didn't know that

im a bit confused on matrices because as someone who learned about linear algebra and it was my introduction to matrices i dont see how they are used in any different fields. You said "if you view matrices as linear transformations" so there are numerous ways to view them? What are some of them? And what are some resources i can see to learn about matrices in the general sense and not just linear transformations?

@@atrophysicistyou can also view them as just ”lists of numbers”. One thing about matrices that’s used a lot in other fields are the linear algebraic groups, which are basically algebraic groups that are sub-algebraic groups of GL(n,R) for some ring R, i.e. the group of invertible n x n matrices over R. Here the connection with lienar transformations becomes weaker, in particular when R is a ”bad” ring. These groups show up in many different places.

Same. Once I get enough money I’m going to give back to the open source community (this includes freely available RU-vid video and free software I used during my education)

I would also like to watch these

"all would be ideal" PUN! Ha!

*modules*

hi soctatica madam how are u.today i watched "units in rings (abstrct algebra)it is very i teresting.can u publish these subjects in text books written by u?it is not possible tu grasp certain points irealy appreciate your calibre.bye madam.

@@s.k.potdarpotdar8377 very desperate aloo masala.

Neato! Thanks for sharing! 💜🦉

I really need it, now for upcoming semester exam

Seems Z/nZ , with n being a multiple of 6, has significantly lower percentage of units. Check Z/24Z, Z/36Z etc..

Min phi(n)/n for n up to 50.

My proof was as follows: The number of co-primes to n is Euler(n), therefore the percentage of units is Euler(n)/n. Euler(n) is given by n*Pi(1-1/p) aka the product of the distinct primes that divide n. So the percentage will just be Pi(1-1/p), so we reduced our problem to find which number between 1 and 50 has the minimum product. We can see from the formula that the more distinct prime factors the lesser the percentage (since trivially 1-1/p < 1) therefore we need to find the number with the most distinct prime factors. This is still not an easy task but not insurmountable. My continuation from here was to look at the largest p(n)! that satisfies p(n)!

@@YougottacryforthisThat suggests that Z/210Z should also have a small number of units. (Edited for skipping nonprime coprimes) There are some coprime values that aren't actually primes, for example 11*11=121 is coprime with 210. There aren't any natural numbers coprime with 30 and less than 30 that aren't also primes, but that's not the case for larger primorials. Wolfram Alpha says there are 46 primes less than 210, but composite numbers with all prime factors greater than 7 will also be coprime. These are all two-factor composite numbers (because the smallest three-factor composite with factors greater than 7, 11^3=1331, is too large) and have the factors between 11 and 19 (because anything greater than 11*19=209 is too large) 11*11 = 121 11*13 = 143 11*17 = 187 11*19 = 209 13*13 = 169 13*17 = 221 (too large) (all other combinations too large) 46 primes less than 210 = 2*3*5*7 +1 one is not coprime -4 primes that are factors of 210, so not coprime +5 composite numbers that are coprime Therefore Z/210Z should have 48 units for a percentage of 22.9% Z/2310Z should also have a lower percentage of units than any smaller modulo ring.

That is quite the list: the aforementioned ring is the ring of all polynomials evaluated at root 2. A more intuitive presentation is a+b✓2 for integers a and b. This has a lot of units like (✓2 +1)^n for all positive integers n

Hello to our Socratica Friends in Bangladesh! We're so glad you're watching. 💜🦉

It's not a convention - it's a fact. Say you have x.y. Then if you multiply on the left by y^-1.x^-1, you get y^-1.x^-1.x.y. See how we have x^-1.x in there? That can become the identity element, leaving us with y^-1.y, which also becomes the identity element. So y^-1.x^-1.x.y = identity. Similarly, multiplying x.y on the right by y^-1.x^-1 leaves us with x.y.y^-1.x^-1. Then the y.y^-1 is the identity, leaving x.x^-1, which is also the identity. On the other hand, if you were to multiply x.y on the right by x^-1.y^-1, you get x.y.x^-1.y^-1. Can we simplify this at all? It's not clear. If multiplication is _commutative,_ then x^-1.y^-1 = y^-1.x^-1. But if multiplication is not commutative, then these two expressions are not equal, and (x.y)^-1 = y^-1.x^-1.

If the ring R does not have an identity element then it does not make sense to even question which elements have inverses. Thus it does not even make sense to speak about R^x

@@hybmnzz2658 OK. That makes sense. But it is NOT a requirement for a ring to exist 1, known as *unity*, or *multiplicative identity*,( at least according to some authors) as it is a requirement, nevertheless, to exist a 0, or zero *additive identity*... Therefore, should not change the slide at 2:23 for saying "for any ring with unity"?

The concept of a unit doesn't make sense unless 1 is in R. After all, a unit is an element that has a multiplicative inverse, which is an element that multiplies something to 1.

@@MuffinsAPlenty This is a reasonable argument. But there are definitions of a ring, which only demand the set with multiplication to be a semigroup, which does not necessarily have a neutral element. That's why I think it is sensible to mention it.

Thank you for your kind support! We so appreciate it!! 💜🦉

No. The real Hamilton quaternions are a division ring which are not a field.

Every 'Commutative' Division Ring is a Field

You are correct, and indeed, Bézout's Lemma is the way to go. But for the other direction of the proof, you need something a tiny bit stronger than Bézout's Lemma. For the other direction of your proof, the thing to realize is the following: if there exist integers x and y with ax+ny=c, then c is a multiple of gcd(a,n). Using this fact, we can conclude that if there exist integers x and y with ax+ny=1, then gcd(a,n) = 1. I'll let you try to do the rest of the proof from there. But feel free to comment back if you want an additional hint.

Hulk used to say to Captain America in Age of Ultron :D :D

40298 tesuya factorization is unique up to order means that if we consider the prime factorization of 30, 30=2×3×5, but can also be written as 30=3×5×2 or 30=5×2×3, these products are all the same and so we say the factorization is unique "up to" order, so we don't consider writing the product with it's terms scrambled up as counting as another factorization. 😃

I understand your explanation. thank you

It’s nothing to do with intelligence, don’t put yourself down. She is going fast, these videos are more like reviews. Keep your notes and textbooks at hand to go through at your own pace. Also, feel free to pause the video and rewind it until you understand, I believe in you 😁

Yeah and that is so because numbers that have a common factor with n are zero divisors

UMU-i-D you should try watching from the start of the Abstract Algebra playlist You’ll have a better chance of understanding what she’s talking about :D

Hello! You can turn off subtitles on your end (click the cc or gear icon).

No, it isn't inaccurate. By definition, units are not considered to be prime elements of a ring. Now, the reasons why 1 isn't considered a prime number (in the natural numbers) are pretty much the same reason why units are not considered prime elements (in an arbitrary commutative ring), but that doesn't make the statement inaccurate.

Ah, I see. It's been a while since I studies abstract algebra. "Prime element" has a precise definition with respect to rings.

Perhaps he meant Hindi subtitles haha