Integral Domains (Abstract Algebra)

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Integral Domains are essentially rings without any zero divisors. These are useful structures because zero divisors can cause all sorts of problems. They complicate the process of solving equations, prevent you from cancelling common factors in an equation, etc.
In this lesson we introduce the idea of an integral domain, talk about solving an equation over rings with and without zero divisors, and show how the cancellation property does hold in an integral domain.
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We recommend the following textbooks:
Dummit & Foote, Abstract Algebra 3rd Edition
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Milne, Algebra Course Notes (available free online)
www.jmilne.org/math/CourseNote...
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Teaching Assistant: Liliana de Castro
Written & Directed by Michael Harrison
Produced by Kimberly Hatch Harrison
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Опубликовано:

3 июн 2017

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Комментарии : 187

@Socratica 2 года назад

Sign up to our email list to be notified when we release more Abstract Algebra content: snu.socratica.com/abstract-algebra

@mathematicalexpert208 Год назад

ples madm you sport me

@derendohoda3891 4 года назад

Whoever wrote the scripts for this playlist did an absolutely outstanding job.

@MuffinsAPlenty 4 года назад

@Hugh Jones Most RU-vidrs have scripts, even if they write the video themselves. But in this case, it's a "well-known fact" that Liliana de Castro is an actor and that someone else writes the scripts for these videos. You can even look in the video description: "Teaching Assistant: Liliana de Castro Written & Directed by Michael Harrison Produced by Kimberly Hatch Harrison"

@Socratica 3 года назад

@@MuffinsAPlenty Thanks for sharing our credits! It's kind of funny how few people read the video descriptions - it's all in there! Lots of important info! 💜🦉

@giuliocasa1304 2 года назад

@@Socratica Indeed, IMO another important info is "Milne, Algebra Course Notes (available free online)": I've looked at the notes e.g. about Galois theory and at first sight this looks quite well explained, thank you also for it and of course for this superb video about integral domain!

@damnstupidoldidiot8776 4 года назад

I came here thinking it is about domains of functions you integrate. I learned something new instead.

@duckymomo7935 3 года назад

That’s just a measurable space

@sididia 3 года назад

same

@PunmasterSTP 3 года назад

I think this lecture is very well grounded; it’s got a lot of roots! In all seriousness, this is another phenomenal video, and I do plan on binge-watching the rest of the abstract algebra playlist today.

@johnfisher2495 3 года назад

Why does abstract algebra seem so much simpler this way? Love this. Maybe you could add one on the definition of an algebra.

@codatheseus5060 Год назад

They were talking about zero divisors in my linear algebra course and it makes so much more sense now after this video I'd love to have a socratica video series about dual numbers

@jonathanpopham5483 2 года назад

I was reading a book on abstract algebra that explained integral domains and it completely left out the part about the usage of modular arithmetic. Totally didn't understand. This makes it very clear, thank you.

@PearlKhurana Год назад

I always revisit the series before any of my abstract algebra papers for basic definition clarity! Would love to see more being uploaded. Kudos to the author!

@danielholme398 Год назад

Liliana, these videos are so good, compressing down the succinct core knowledge on a topic into a few minutes, just what's needed to dip in and update as required.

@jordanweir7187 4 года назад

These are all wonderful videos, brilliantly narrated, keep up the good work

@ChaudharyAteeq440 7 лет назад

Thanks for adding a new video about Abstract Algebra...Plz upload more videos on Pure Mathematics... Your explanation technique is great...

@tomhu7917 6 лет назад

Really best video for abstract algebra: the instructor explains clearly in reasoning

@somadityasantra5572 4 года назад

Your channel deserves a lot more attention.

@turalsadigov3602 7 лет назад

Just great! I just discovered your videos, and it is fun to watch them!

@Socratica 7 лет назад

We're so glad you've found us! Thanks for watching! :)

@Lycheeee11 7 лет назад

Love your channel! Glad to know you are uploading new videos. Could you please talk more about commutative rings?

@rajkumarmukherjee268 4 года назад

You are my all time favourite Teacher.

@ThePrimebrook 7 лет назад

Your videos are very well done and helpful. Keep it up!

@GabrielConstantinides 7 лет назад

thanks for yet another video Socratica

@kassaadane8660 6 лет назад

Always I am interested when I am watching videos u present! I like it.

@theboombody 2 года назад

The video is good throughout, but the end part alone is worth a thumbs up.

@Dannnniel Год назад

this is like the 5th of your vids so far, thanks for helping me pass my algebra exam :D

@ndubuisiuguru3239 Год назад

This is quite clear and easy to understand.

@padraiggluck2980 2 года назад

Great series 👍. Thank you.

@avishekadhikari1793 4 года назад

Innovative lectures I liked it so much

@akarshroy4461 2 года назад

Love your way of explanation from India Ma'am . 👍🏻

@kunslipper 6 лет назад

Thank you so much.

@adesina1 3 года назад

You are a wonderful teacher... Thank you very much

@yashikajain643 3 года назад

You've made this interesting. Thanks.

@Cabajal 7 лет назад

Un excelente video, me gustaría que hablaran un poco más de los diferentes tipos de Dominios como los de factorización única o los euclídeanos.

@alphalunamare 4 года назад

Best stand up Maths I ever understood! :-)

@frozenstrawbs 7 лет назад

love this channel

@jennamorabito2526 3 года назад

Big help, thank you!

@boradmay 4 года назад

This is so much better!

@akshettrj 4 года назад

I was taught about integral domains in our college, but this really made me understand the idea behind these.... Great Video

@gauravsinha6060 6 лет назад

Great Video 👍👍

@Eternal_stardust 5 лет назад

what a nice explanation ma'am. Lucid one ..

@benterrell9139 4 года назад

Fantastic. If only my University could make videos like this

@priyakataria8027 6 лет назад

these videos are very helpful,May I get more videos in ring theory by socratica,these are pretty particular abt topic..i need them

@Yoyimbo01 7 лет назад

Beautifully explained and laid out. I wish I had access to this material when I studied abstract algebra!

@chin6796 3 года назад

Thanks for amazing video

@ismaelmonsegur9581 7 лет назад

Good videos. Can you upload Topology videos?

@nkuniqueave8914 3 года назад

Good lecture

@abdelrahmangamalmahdy 7 лет назад

never studied abstract algebra, but I think this can be easily solved by setting the right hand side of the equation to 12n where n is an integer. so we can solve 2 equations, first when n=0 and then when n=1 so we get multiple solutions to the modular equation.

@monian2877 6 лет назад

awesome video

@ACZ29 7 лет назад

nice explanation mam.....i have studied before but never understood......this example makes me clear all doubt about integral domain ......thnxx mam

@nikolacvejic523 7 месяцев назад

Thanks a lot!!!

@nowornever5598 5 лет назад

Thanks

@andresvalera1430 4 года назад

Good explanation !, The book my university gave me just says "It's a ring in which a.b=0 implies that either a or b is zero" which is true but misses the whole point of defining a structure smh Also wanted to add that in some books "The integers mod n" are noted as Zn (The n being a subscript)

@hybmnzz2658 3 года назад

Zn is different from Z/nZ but the difference is very subtle and those groups are isomorphic to each other anyway. The elements of Z/nZ are sets themselves from the power set of Z and the emphasis is that it is constructed by cosets. The elements of Zn are equivalence classes which you could say are the same as the sets above but are constructed with Z x Z and a relation.

@thewalkingcrow8946 7 лет назад

Integral Domains were interesting to me because it's the minimum requirement for greater than or less than to be defined. Beneath that everything gets wibbly wobbly and is just there. Like colors, there's no way to define which one is greater or less than, but once you define color by it's frequency of light, you get the ROYGBIV we're familiar with red being the lower end and violet being the upper.

@standup4you 6 лет назад

I'm impressed

@mathmaths8380 4 года назад

Very interesting

@nsorrichmond2746 Год назад

😂😂, I love this woman, you are superb. God bless you very much for helping some of us

@MuhammadUsman-uu6rc 7 лет назад

Its very very good

@MuffinsAPlenty 7 лет назад

We generally also require an integral domain's 1 element to be distinct from 0. In other words, we don't want the 0 ring to be classified as an integral domain, since we want an ideal I to be prime if and only if R/I is an integral domain.

@Mizraab2912 4 года назад

product of 2 terms is 0 so one must at least be 0...ya had to do that in LinAl...saw how beautiful Math can be....you are a wonderful teacher.

@shivanidixit7395 6 лет назад

awesm video

@somadityasantra5572 4 года назад

Please make a video on unique factorisation domain too

@aakashlone4654 5 лет назад

It ix humble request to u ....plx make vedio 1 hour ....even I can watch your your vedio 10 sec even 10 hours also....best of best explanation...

@poomalaip2620 6 лет назад

Good teaching. Every one need example better

@saurabhsingh-ow7ue 3 года назад

thank you madam...................

@ronycb7168 9 месяцев назад

I use these videos as motivation before I read my text THE SECOND TIME 😂 ❤ 😅

@hashimm5860 2 года назад

Mam please share class about unique factorisation domain

@hashimm5860 2 года назад

Mam excellent class

@Socratica 2 года назад

Thank you, kind Socratica Friend! 💜🦉

@9erik1 5 лет назад

loving the videos here -- are there any notable groups/rings/etc with zero divisors other than groups of modular arithmetic? it seems that most of the caveat examples in this series' videos up to now come from modular arithmetic

@MuffinsAPlenty 5 лет назад

Consider the (noncommutative) ring R of nxn matrices whose entries are real numbers. Every matrix with determinant 0 is a zero divisor in the ring. If you happen to have an idempotent element in your ring (an element e so that e^2 = e) which is not 0 or 1, then that element is guaranteed to be a zero divisor (e*(1-e) = 0). This is true in Boolean rings, for example, since the algebraic structure resembles that of Boolean logic. A lot of other examples come from performing operations on rings. For example, if you take the direct product of two rings, elements of the form (a,0) and (0,b) are zero divisors. Or if you take any ring R and form the polynomial ring R[x], then if R has zero divisors, it is possible that R[x] could have zero divisors too. Or if you have a ring R with zero divisors, then taking the ring of nxn matrices whose entries in R will have zero divisors as well (all matrices whose determinants are zero divisors in R). Or taking a quotient ring of a ring if you're modding out by an ideal that isn't prime (but this is like modular arithmetic).

@9erik1 5 лет назад

@@MuffinsAPlenty wow this is awesome, thank you!

@chounoki 3 года назад

At 4:46, do you mean "1" represents the multiplicative identity and "0" represents the additive identity? Because the members of the ring may not be numbers at all.

@mathebryan 7 лет назад

Very informative and easy to follow. 9/10.

@lekhapratap1652 5 лет назад

Bryan Villegas what is the deduction of 1 for?

@awiggin4367 7 лет назад

These people are GREAT! Love the vids.

@mixbaal0 4 года назад

Thank you. You helped me giving sense to such a kind of concept. I am a self learning wanna be mathematician.

@santhoshwagle9857 4 года назад

I enjoyed that parody of "stand-up comedy" part....

@umairtech43 Год назад

Finally, if a is divisible by 6, then the equation reduces to 6x^2 + bx + c ≡ 0 (mod 12), which has the same number of solutions as 3x^2 + bx + c ≡ 0 (mod 6), which we have already counted. Therefore, the total number of distinct quadratic equations modulo 12 is 144 + 864 + 72 = 1,080.

@Zenene-ok5el 6 лет назад

(5:42) Why doesn't the factoring and cancelling method require division? Isn't cancelling a number in both sides of an equation the product of doing multiplication of the original number by it's multiplicative inverse (and isn't this the essence of division in the first place)? Edit: Never mind, the last part of the video solved my doubts. I didn't realize that, being a common property of all rings, distribution could be used to cancel a common factor from both sides of any equation.

@MuffinsAPlenty 6 лет назад

Although you figured it out, your intuition isn't bad. From any integral domain R, you can construct its field of fractions F, which is a field containing R as a subring and the inverses of every nonzero element of R (this works because R has no (nonzero) zero-divisors). If you have an equation xy = xz in R (with x not equal to 0), then it must still be valid in F (since R is a subring of F). Hence, you can multiply by x^-1 in F to get the equation y = z. But this is an equation consisting of elements in R, so it is also valid in R (since R is a subring of F). In general, given any commutative ring R in which x is a non-zero-divisor, if you know xy = xz, then you can conclude y = z, just from the properties of non-zero-divisors. But, since x is a non-zero-divisor, you can also find a ring S containing R as a subring in which x is a unit. Hence, you can use a "division" argument like you suggest to get the same result. So there is an interesting connection there anyway.

@information2949 6 лет назад

Plzzz ma'am tell me how to find zero divisor and unit in Qotient ring e.g R/i

@CosmicButterfly2 Месяц назад

If anybody wants a trick to do modular arithmetic fast, try using the Chinese Remainder Theorem

@Tracks777 7 лет назад

Keep it up! Looking forward for more videos from you, don't stop!

@kyleshim361 5 лет назад

1. change the question to "how many unique solutions". And we do not have to worry about the "zero divisors" because it's quadratic. zero divisors like accidental solutions. So it gives 12C1 + 12C2 = 78 there should be 78 sets of different solutions, which gives 78 unique equations. 2. 12 = 2x2x3 2x6, 3x4, 3x8, 4x6, 4x9, 6x8, 6x10

@edward_albergaria 7 лет назад

Representando Brasil

@Macieks300 5 лет назад

So what's the answer to the first puzzle? I only could come up with an upper bound of 11*12*12.

@Macieks300 5 лет назад

Here's the answer: math.stackexchange.com/questions/3185444/number-of-different-quadratic-functions-mod-12

@durgeshsharma4379 3 года назад

What is mod 12?

@anon2152 3 года назад

Theses videos are so much better. Indians are just obsessed with Definitions.

@pumpkin7889 7 лет назад

I really liked the video, but it would be much better if it was little bit slower :) I had to stop n play again lots of times cuz it was too fast for my brain lol

@StanislavBashkirtsev 4 года назад

mod 11 is also a field. It has multiplicative inverses which allows cancellation. Is there an example where no-zero-devisors alone makes cancellation possible?

@MuffinsAPlenty 4 года назад

I'm not sure if I'm understanding your question. Are you asking for an example of an integral domain which is not a field? If so, one example is the ring of integers, Z. The only units are 1 and -1, so it's not a field. But Z is an integral domain. In the ring of integers, if you have 2a = 0 for some element a in Z, it must be the case that a = 0, even though 2 is not invertible. You cannot find a finite example, though. In a finite commutative ring, every nonzero element must either be a unit or a zero-divisor. Therefore, all finite integral domains must be fields.

@StanislavBashkirtsev 4 года назад

@@MuffinsAPlenty huh, make sense. Thanks!

@alexsoutienscolaire719 3 года назад

Morning coffee, in the warm sun of May, lost nowhere in a village of french hilly countryside, surrounded by birds singing and soft wind in the trees, I was watching randomly your video, listening from one ear only. At 0:37 then you say "in abstract algebra, this isn't always true". I instantly stopped breathing for a second, then my heart went fast, starting to feel dizzy, sweating, thrills... I almost fell off my coffee mug on the ground. Then I said to myself, no, it's wrong, she must be wrong, how possible ? Then I watched the rest of the video, and now my whole world collapsed. I don't know what to do with this now. I feel anger cause I wasn't prepared for this. Shouldn't you have made a warning sign ? Hopefully I was sitting. I kind of understand how Pythagorician could feel at the time Hippase de Metaponte came to demonstrate the irrationality of square root of 2 xD Then hopefully the little stand-up comedy funny end was making all of my sadness disappear :D And now the birds sings sounds different, the wind seems way more mysterious and the sun tans my face with a new kind of light. You definitely change the world, at least mine :) Thanks so much for your videos Keep on !

@Socratica 2 года назад

The world is more beautiful because you are in it, Socratica Friend! 💜🦉

@Tracks777 7 лет назад

I enjoyed the video! Keep it up!

@Tracks777 7 лет назад

Brilliant work! Keep it up!

@adolfninh23 5 лет назад

ok from now my brain starts burning !

@majestic7768 4 года назад

What is a mod 12?

@lovemavi1827 3 года назад

I already watched this before our class in Abstract Algebra started but it turns out that her explanation and examples are similar to what is given here so I left the meet and watched this video again. Srlsly, I don't understand my prof's explanation.

@hoangtudaden1304 5 лет назад

Integral domain: If ab=0, then either a or b is 0 where a,b in R.

@WestOfEarth 4 года назад

When is mod used? It's a new concept to me.

@isaiahmrman6564 4 года назад

It's used all throughout number theory. It also has applications in cryptography

@zorro20010 4 года назад

wat is mod 12

@Tracks777 7 лет назад

Nice! Keep it up!

@Tracks777 7 лет назад

Fantastic video! Keep it up!

@hallgowrt 3 года назад

Watch socratica instead of sleeping on professor's notes... Time saved

@rogerwilcoshirley2270 4 года назад

Wow, very interesting ! Bet this is at the core of some cryptographic/decryption methods. Regardless, no way to really understand w/o trying a bunch of related basic problems in these topics, bet it won't take long to get totally stuck/ confused ;-)

@sss-ol3dl 7 лет назад

There are 11*12*12 different quadratics, if you mean just different coefficients and not the same functions x^2 + 2x + 4 is an example of a quadratic with only 2 solutions mod 12 (2 and 8)

@gundamlh 6 лет назад

hi, I'm confused, could you be kind to explain your solution a little bit?

@GabrielConstantinides 6 лет назад

What they did is this: A quadratic will be of the form ax^2 + bx + c, and we require a to be non-zero so that it is a quadratic. Now, this means that for a there are 11 choices *mod 12* (1,2,3,4,5,6,7,8,9,10,11). b and c can be zero if we want, so there are 12 choices for them mod 12 (0,1,2,3,4,5,6,7,8,9,10,11). Now, 11 choices for a, 12 for b, 12 for c gives a total of 11*12*12 possible quadratic equations mod 12. Now I am not sure if this is the correct answer, I wouldn't be able to come up with anything else. I suppose that we might consider two equations different mod 12 if there is an input (x value) which gives a different output for each equation.

@gundamlh 6 лет назад

11*12*12 : [939 (no solution), 144 (unique solution), 324 (just 2 solutions), 8 (have 3 solutions), 138, 0, 18, 0, 12, 0, 0, 0]

@Macieks300 5 лет назад

Yeah but some of those could be the same quadratic congruence.

@Macieks300 5 лет назад

I asked the question on Mathematics Stack Exchange and here's the answer: ( it's (12^3)/2 ) math.stackexchange.com/questions/3185444/number-of-different-quadratic-functions-mod-12

@wiggles7976 2 года назад

Should we make the distinction between trivial zero divisors, and non-trivial zero divisors? Since 0*0=0, or 0*2=0, etc., , 0 divides 0, and we had nonzero things times 0 giving 0, so 0 is a zero divisor. Also, any nonzero number like 1 satisfies 1*0=0, so 1 divides 0. However wikipedia says 1 needs to be multiplied by something nonzero and equaling 0 to be called a zero divisor, so 1 is not a zero divisor. So, 0 is a trivial zero divisor, and any nonzero element a dividing 0 with something nonzero (ie, ax=0 with x != 0) is a non-trivial zero divisor. When you say "zero divisor" it makes sense to mean "non-trivial zero divisor".

@MuffinsAPlenty 2 года назад

Different textbooks have different conventions on what "zero divisor" means. In areas of study where the focus is on integral domains (such as number theory), they like to exclude 0 from the definition of zero divisor because it makes it easier to define an integral domain (an integral domain is a commutative unital ring with 0 ≠ 1 and having no zero divisors). However, if you are working in a branch of mathematics which allows rings which are not integral domains, the typical definition is what Wikipedia says, and 0 is considered a zero divisor in nonzero rings. (The theory involving zero divisors is much easier to state if one considers 0 to be a zero divisor in such settings.)

@gregoryfenn1462 6 лет назад

Fields (and integral domains) have one (and only one) zero-divisor: zero. The video is good but misleading as says that domains (like Z_{11}) have no zero divisors, which is false.

@Socratica 6 лет назад

There are different conventions for handling the trivial case of 0. For example, Serge Lang defines them as "Elements x, y of A are said to be zero divisors if x!=0, y!=0, and xy = 0." Richard Dean at Caltech used the same convention. So you'll frequently hear mathematicians say certain rings do not have zero divisors. Thanks for watching!

@MuffinsAPlenty 6 лет назад

Socratica, do you know if there is a reason why Lang and Dean prefer that 0 not be considered a zero-divisor? I understand why in commutative algebra we consider 0 to be a zero-divisor (except in the 0 ring): the set of zero divisors is the union of the associated prime ideals of the ring. If you don't want to consider 0 as a zero-divisor, then you have to remove 0 from this union. So it makes things slightly easier to consider 0 as a zero-divisor in nonzero rings if you care about associated prime ideals. Is there something similar for excluding 0 from the set of zero-divisors, or is it just to make the definition of an integral domain simpler?

@Taricus 5 лет назад

@@MuffinsAPlenty I think it ties to making the definition for cancellation more clear. You don't want x to be zero and neither y, so that when xy=0 then you can definitely say whether you have zero divisors. If you include zero, then it becomes trivial and muddles it up a bit. If you count zero, then you lose some detail in the definitions. I mean, I'm here trying to figure out how to do my homework though LOL! So I may be completely wrong, but that's how I interpreted it.... take my comment with a grain of salt :P

@MuffinsAPlenty 3 года назад

@@Taricus Very delayed response to you, but I think you're pretty much right! In number theory, most of the rings they care about are integral domains. So, if you exclude 0 from the classification of zero-divisor, it makes it easier to define integral domains, which is pretty much all they care about anyway. In some sense, they are like, "Hey, in arbitrary ring theory, this bad thing will happen. But we won't consider any such situation." However, more general ring theory (such as commutative algebra) really does want to consider rings with zero-divisors. The theory of zero-divisors is cleaner when 0 is considered to be zero-divisor (except in the 0 ring). So in this context, it makes sense to consider 0 to be a zero-divisor. It's sort of a matter of convenience. If you want your study to include rings with zero-divisors, it's easier to consider 0 to be a zero-divisor (except in the 0 ring); but if your study only includes integral domains, then the only reason you care about zero-divisors is to say "bad thing doesn't happen", and it gives a cleaner definition of integral domain.