I think another nice way to see understand fields it to notice they are commutative Rings such that everything has a multiplicative inverse except zero. Just because we are able to connect the concept to Rings too.
i love these videos! Abstract Algebra never been this easy to understand, you make it to the right way of teaching this! Please make some more about ideal and quotionent ring. i will be waiting^^. hehe btw thanks a lot you really help me a bunch!
It is mentioned in this video that in an algebraic field you can add, subtract, multiply, and divide (and show those operators), but then it is said that "everything is commutative," which I found a bit confusing. Saying "you can subtract and divide" and showing the subtraction and division operators seems to imply that those operators are defined in a field, which is not the case. Subtraction and division are defined such that you only invert the right-hand operand. So a / b = a * (b^-1) [^ means "raised to the power of"], and if divide is to be commutative then it would have to be true that a / b = b / a which is saying a * (b^-1) = b * (a^-1) which is not true, nor is it true for subtraction: a - b = b - a is the same as saying a + (-b) = b + (-a) which is not true. So you can divide and subtract with a * (b^-1) = (b^-1) * a and a + (-b) = (-b) + a which was mentioned a few times in previous videos in this series. In fact there is no division or subtraction operator in a field, it is simply that those operations are possible when you use inverses. I think it would be better if the subtract and division operators were omitted from the graphics in the presentation. But anyway, I absolutely love this series, and I intend to watch all of the content on this channel.
I find the constructive definition of a field from a commutative ring (with no zero-devisors) via adding in fractions (localization) to be a better starting point which underlines deeper theoretical implications, for example it shows why fractions with denominator (or any other zero-divisor) can't happen...
In general take a commutative ring with 1, localize by a prime ideal, get a local ring. Localize one more time by the image of the same prime ideal, we get a residue field.
is it correct to say that polynomial is not a field because it doesnt "close" under the operation of division? i think matrix multiplication is not a field either coz it's not commutatitive
so if addition is the same as subtraction and multiplication the same as division in a field, does that mean an element x in a field, will also have its inverse x^-1 in the same field? and also by that result a field would contain an multiplicative identity?
A ring is a set of elements equipped with two binary operations that satisfy certain conditions. The first operation is associative and commutative. With respect to the first operation there is an identity element in the set and every element has an inverse in the set. The second operation is associative but not necessarily commutative. With respect to the second operation there is an identity element in the set but not necessarily an inverse for every element. The distributive property also holds among the two operations. If the first operation is addition, then there is an additive inverse for every element in the set. So sums between positive and negative values are equivalent to subtraction.
why must you omit 0 from being commutative with multiplication? considering 0 in R, (anything * 0) = 0 = (0 * anything): in that way 0 commutes with respect to multiplication. zero's multiplicative "inverse" wouldn't be commutative, but it does not necessarily exist. am i missing something?
Omitting 0 is not because of the commutative part; it's because of the "group" part. There is no multiplicative inverse for 0, so you don't get a group if you include 0 (recall that every element in a group has an inverse).
Despite the name, a vector field is not a field. In this case I believe the word "field" was brought over from physics, where they talk about electric fields, etc. We're actually working on a video about Vector Fields this week. It should be out in a few days.
in Brazil we call fields as bodies and fields are just related to vectors, so we have vector fields and bodies. That is why we never did this confusion and it took some time to me to understand that a field is a body but a vector field is not a body hehe, strange right?
Whoops: this is the field of rational expressions, not functions. You used fractions with polynomial terms, not polynomial function terms, and in any case the functions need not be defined on the reals (wherever the denominator vanishes). I think this is muddled and likely to confuse anyone who wants to study algebra further.
There's a tiny mistake in the introduction (0:14). Real numbers and decimal numbers aren't the same. While π is real, it isn't decimal as it cannot be written as an integer over a power of 10. The sets are very different (cardinality, least upper bound property, etc.).
i love these videos! Abstract Algebra never been this easy to understand, you make it to the right way of teaching this! Please make some more about ideal and quotionent ring. i will be waiting^^. hehe btw thanks a lot you really help me a bunch!
+Laura Estwin We're so glad you enjoy them! We're filming our next Abstract Algebra video in a few days. And we will definitely cover ideals & quotient rings eventually.
Yeay, i'm so happy to hear that! Unfortunately my exam about those topics is in 3 days ahead :'( hiks.. but it's okay, you have done a good job Socratica, a lot of people will be helped with your videos :D (y)
Sign up to our email list to be notified when we release more Abstract Algebra content: snu.socratica.com/abstract-algebra
"There's even fields with a finite number of elements. We call those... finite fields." 😂😂😂
She took us on a field trip
It really hurts that we didn't think of this pun for the video. 💜🦉
Lol
@@Socratica am a maths teacher and am learning the face expression n confident from yu thanks mate ,it was indeed a great lecture n demonstration
I definitely appreciate how Socratica fields questions and comments on their videos, even if they came out a long time ago!
You make this sound more epic and dramatic than it really is. XD
+Aegix Drakan i was thinking the same thing....
just making sure we dont get bored easy
These really stand the test of time. It's hard to find such concision these days.
I really love this video. Mathematics needs to be presented in this dramatic way because mathematics, is dramatic.
Wished I ran across your video in the beginning of the semester! Please make more abstract algebra video. These few video really helped! thanks
Fantastic Job! Thanks for providing a solid introduction to Abstract Algebra.
The best way of explaining and dramatizing I have ever seen..! XD... Loved it
Love this person's explanations. Wish I knew what a polynomial was.
I think another nice way to see understand fields it to notice they are commutative Rings such that everything has a multiplicative inverse except zero. Just because we are able to connect the concept to Rings too.
i love these videos! Abstract Algebra never been this easy to understand, you make it to the right way of teaching this! Please make some more about ideal and quotionent ring. i will be waiting^^. hehe btw thanks a lot you really help me a bunch!
Hello, what does commutative mean?
Man I’m having a field day with these videos!
I love the way you explain!
Amazing series, thanks a lot! When I finish Algebra II, I will review these videos! see ya
Artfully explained 👍🏾
My best lecturer
It is mentioned in this video that in an algebraic field you can add, subtract, multiply, and divide (and show those operators), but then it is said that "everything is commutative," which I found a bit confusing. Saying "you can subtract and divide" and showing the subtraction and division operators seems to imply that those operators are defined in a field, which is not the case. Subtraction and division are defined such that you only invert the right-hand operand. So a / b = a * (b^-1) [^ means "raised to the power of"], and if divide is to be commutative then it would have to be true that a / b = b / a which is saying a * (b^-1) = b * (a^-1) which is not true, nor is it true for subtraction: a - b = b - a is the same as saying a + (-b) = b + (-a) which is not true.
So you can divide and subtract with a * (b^-1) = (b^-1) * a and a + (-b) = (-b) + a which was mentioned a few times in previous videos in this series. In fact there is no division or subtraction operator in a field, it is simply that those operations are possible when you use inverses. I think it would be better if the subtract and division operators were omitted from the graphics in the presentation.
But anyway, I absolutely love this series, and I intend to watch all of the content on this channel.
love the way Liliana de Castro teaches. Plz more math videos with her 🤩
socratica made me mathematics too interesting..😍😍
Superb videos!! Many thanks. Please upload more. ☺
Wow that was such a great and simple explanation THANKS
Very well explanation !
Nice class it's very helpful
I find the constructive definition of a field from a commutative ring (with no zero-devisors) via adding in fractions (localization) to be a better starting point which underlines deeper theoretical implications, for example it shows why fractions with denominator (or any other zero-divisor) can't happen...
In general take a commutative ring with 1, localize by a prime ideal, get a local ring. Localize one more time by the image of the same prime ideal, we get a residue field.
Amazing presentation.
Mam i am from pakistan your learning style is very good👌👌👌👌👌👌
Greetings to Pakistan kind friend! 💜🦉
is it correct to say that polynomial is not a field because it doesnt "close" under the operation of division? i think matrix multiplication is not a field either coz it's not commutatitive
sure, it's a ring
Thanks!
Thank you so much for your kind support! It makes a huge difference for our work. 💜🦉
thanks for the nice explanation
so if addition is the same as subtraction and multiplication the same as division in a field, does that mean an element x in a field, will also have its inverse x^-1 in the same field? and also by that result a field would contain an multiplicative identity?
Thank you so much.
Now , I got it. Thanks!
please make videos about modules...😊
Why is that good to know!
1:21 you said, Ring is only under addition and multiplication, why there's substraction?
A ring is a set of elements equipped with two binary operations that satisfy certain conditions. The first operation is associative and commutative. With respect to the first operation there is an identity element in the set and every element has an inverse in the set. The second operation is associative but not necessarily commutative. With respect to the second operation there is an identity element in the set but not necessarily an inverse for every element. The distributive property also holds among the two operations.
If the first operation is addition, then there is an additive inverse for every element in the set. So sums between positive and negative values are equivalent to subtraction.
Prove or disprove that in Z[x], the ideal
< x > + < 3 > is a principaL ideal
Anyone pl help
Thanks alot Miss❤️❤️
why must you omit 0 from being commutative with multiplication? considering 0 in R,
(anything * 0) = 0 = (0 * anything): in that way 0 commutes with respect to multiplication. zero's multiplicative "inverse" wouldn't be commutative, but it does not necessarily exist.
am i missing something?
Omitting 0 is not because of the commutative part; it's because of the "group" part. There is no multiplicative inverse for 0, so you don't get a group if you include 0 (recall that every element in a group has an inverse).
MuffinsAPlenty smacking my forehead! I should’ve paid attention to “group” and not “commutative” thanks !
What is Borel's field ?
Very nice and easily explained
link?????
Removed the abstraction out of abstract algebra .. thx
Very nice mam
how find z(a) this group??
interger not field y?
cool math
Make Algebra Great Again
why can't i mulitply by zero? i am not dividing by zero.
Recall that the operations of division and subtraction are defined in terms of multiplication and addition.
Is a vector field a field?
Despite the name, a vector field is not a field. In this case I believe the word "field" was brought over from physics, where they talk about electric fields, etc. We're actually working on a video about Vector Fields this week. It should be out in a few days.
in Brazil we call fields as bodies and fields are just related to vectors, so we have vector fields and bodies. That is why we never did this confusion and it took some time to me to understand that a field is a body but a vector field is not a body hehe, strange right?
same in Germany, they call it body (Körper)
1:30 why do we have to expand polynomials into field ?
0:26 🤣 🤣🤣
amzinggggggggggggg!!!!
Finite fields are also called Galois fields, in honor of the short lived temperamental genius Evariste Galois.
thank you madam........
So, all commutative rings are fields?
No, not at all. The ring of integers is a commutative ring which is not a field.
Whoops: this is the field of rational expressions, not functions. You used fractions with polynomial terms, not polynomial function terms, and in any case the functions need not be defined on the reals (wherever the denominator vanishes). I think this is muddled and likely to confuse anyone who wants to study algebra further.
There's a tiny mistake in the introduction (0:14). Real numbers and decimal numbers aren't the same. While π is real, it isn't decimal as it cannot be written as an integer over a power of 10. The sets are very different (cardinality, least upper bound property, etc.).
So humans also form Rings but what's the Identity LOL
i think your definition of a field is incomplete because you didn't mention anything about inverse and identity
actually these things are a part of definition of a group under multiplication
The real numbers are not 'decimals', this is an insult! :-)
i love these videos! Abstract Algebra never been this easy to understand, you make it to the right way of teaching this! Please make some more about ideal and quotionent ring. i will be waiting^^. hehe btw thanks a lot you really help me a bunch!
+Laura Estwin We're so glad you enjoy them! We're filming our next Abstract Algebra video in a few days. And we will definitely cover ideals & quotient rings eventually.
Yeay, i'm so happy to hear that! Unfortunately my exam about those topics is in 3 days ahead :'( hiks.. but it's okay, you have done a good job Socratica, a lot of people will be helped with your videos :D (y)
You are a great teacher.