how to make a field

[@laurendoe168]

"If you build it, they will come."

[@pierreabbat6157]

You could also use the p-adic numbers as the base field.

[@JosBergervoet]

That's too simple, Andrew Wiles did that already, on this channel we need a certain level of sophistication.

[@henrikljungstrand2036]

​@@JosBergervoetLol! 😂😂😂

[@henrikljungstrand2036]

Yeah, algebraic extensions over p-adic numbers are pretty cool! Just like extending real numbers, except you get way more dimensions of the extensions.

[@driksarkar6675]

18:25 I don’t think it’s on screen. For anyone wondering, I believe x^4+1 reduces to (x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1).

[@05degrees]

23:41 I’d use what’s already established in this case: as we know α (α² + 1) = 1, then squaring both sides and commuting we get α² (α² + 1)² = 1, and finally we calculate (α² + 1)² = α⁴ + 1 = α (α + 1) + 1 = α² + α + 1, so α² and α² + α + 1 are inverses.

[@ScientiaDuxVitaeCertissimus]

Thanks!

[@martinnyberg9295]

3:29 Michael, could you do a video, or series of videos, on your favourite ways of defining the reals? And explore things like how the naïve picture of reals as infinite decimals is equivalent to the axiom of choice and such things? 😃

[@matematicacommarcospaulo]

I also want to see it

[@05degrees]

“Eudoxus reals” are pretty underrepresented (this construction starts right from ℤ and “almost additive” functions on it; pretty clever and neat IMO).

[@JosBergervoet]

At 18:26, "that should be on the scren right now," priceless, Michael! (Probably someone told you, "please use some modern animations", and this is how you silence them. Perfect job...)

[@ramziabbyad8816]

Michael I am so grateful for everything you do. I love elementary maths.

[@MrAboy97]

How do I translate this equation: (x^2 + 5x + 6 = 0) to Math Field. Plz reply. 🤔😩😫

[@BlackEyedGhost0]

18:20 That was not on the screen right now

[@ed.puckett]

(using a for alpha): a^4 = a^2 + a, not a^2 + 1, so if you are using the prior result for a^4 on the board you won't be able to verify the last inverse pair. Just in case someone else ran into that. Love the videos!!

[@skylardeslypere9909]

You can still verify the last inverse pair like this. α⁴ = α²+α and α²+α ≠ 1 because then α would we the root of a degree 2 polynomial with coëfficiënts in Z/2Z, contradicting the irreducibility of x³+x+1.

[@zadsar3406]

At 8:55, it's obviously unclear why such an α would always exist. I guess it's fine to take this for granted, but there's no need, as a simple and very illuminating construction shows that it exists. Let F be a field and f an irreducible polynomial with coefficients in F. We denote by F[x] the set of all polynomials with coefficients in F (so this is a ring) and by F[x]/f(x) this set "modulo" f. It's really just a set of all the remainders of polynomials from F[x] when divided by f. Now the operations of modular addition and modular multiplication of polynomials in this set give us a field, which can be checked without too much difficulty. The great thing is that F is a subfield of F[x]/f(x) (it's just the set of constant polynomials in F[x]/f(x)) and so this is somehow an "extension" of the field F. However, in this field, the element x satisfies f(x) = 0. So we know that a root exists in some extension of F. This is actually a very powerful idea. It allows us to construct extensions of F in which the polynomial f splits, that is, in which it can be factored as a product of linear polynomials - this is very useful in algebra. Additionally, with some tools from set theory or logic, we can show that there exists an extension of F in which *every* polynomial splits - this is called the algebraic closure of F. For example, the algebraic closure of R is C, the algebraic closure of Q are exactly the algebraic numbers, and so on.

[@manateepink9100]

multiplicative inverses exist because of Bezout's identity, right?

[@karl131058]

Fine, shortest explanation ever. I have a minor nitpick though: the algebraic closure of Q is the Algebraic Numbers in C, NOT C itself! (No transcendentals like e or π are needed for algebraically closing Q)

[@zadsar3406]

@@karl131058 Ah, true. Sorry. Will fix that now. I think it's pretty short for such an important construction.

[@karl131058]

@@zadsar3406 no problem, typos happen to the best of us.

[@elevedesfb]

Thanks, this explanation clearly was missing in the video

[@zh84]

23:58 I make α(α+1) out to be α²+α, not α²+1?

[@azavier-a]

it confuses me too, but either way it would show that a^2 is not it's own inverse

[@iabervon]

In much the same way that you can define Z_p as the remainders from integer division, you can define these fields as the remainders from polynomial long division. You can make it a lot like doing calculations in Z_101 while writing numbers in decimal, where having too many digits means you should subtract off a multiple of the modulus. (The main difference being that you don't carry between coefficients like you would in decimal arithmetic.)

[@Anna-dg9kp]

this vid makes things so easy, really strong didactics, I wish they existed back in the days when I was learning about splitting fields, thank to Mr Penn and his team

[@hxc7273]

Since you’ve already done an abstract algebra course on your second channel, it would be great if you could do like a 2nd semester abstract algebra course where you go into field and Galois theory.

[@kristianwichmann9996]

Any plans on doing some videos on Galois theory?

[@samwalko]

9:12 This is not true of binomials (in particular, x - alpha). They don't factor, but can have real roots. Do we need to explicitly exclude binomials (and maybe monomials, to ensure roots exist), or is there some workaround here? Edit: after watching the actual construction, I now realize that would simply result in the original field (or multiples of it), as b_(1-1)*alpha^(1-1) = b_0 is the only term.

[@2712animefreak]

For the task: The field has 9 elements: {0,1,2,a,1+a,2+a,2a,1+2a,2+2a}. 1 and 2 are their own inverses, the other pairs are: (a,2a), (1+a,2+a), (1+2a,2+2a).

[@epennrogers]

Great video! It’s probably good to note that this is specifically how to build fields that are algebraic extensions of some smaller field. For a transcendental extension, say a is transcendental over F, then F(a) is isomorphic to the entire field of fractions of the polynomial ring F[x]. In general we can say that a generic field extension of F is always isomorphic to the field of fractions of some quotient ring of F[x], but this is probably overkill. Cheers!

[@mathtiger_]

Simple but important examples that gives a nice intuition for Kronecker's theorem.

[@PeterBarnes2]

Multiplication of powers of α in Z_2(α) is addition in Z_7. o_O (This can be seen by continuing to multiply by α from α^4 in the video, which gets more counterintuitive all the way until you wrap around. The cool part isn't the inverses, it's that the powers wrap around in the first place, but this is probably evident from one of those theories about coloring a sequence of integers, or some similar idea in number theory in general.)

[@Tehom1]

That is known as the multiplicative group of a field. Be aware that it is only order 7 when the field is of order 8. In general it's every element of the field except 0.

[@goodplacetostop2973]

25:12 Michael The Builder

[@primenumberbuster404]

:)

[@MrRyanroberson1]

8:13 i found a VERY interesting way to construct the "3d" geometric algebra field R[x,y,z: x^2 = y^2 = z^2 = 1; xy+yx = xz+zx = yz+zy = 0]: take the imaginary quaternion units i,j,k to be xy, yz, xz, and apply the SAME complex number extension to the quaternions with i=xyz using the multiplication rule you would expect from extending the reals to the complex: (a+bi)(c+di) = (ac-bd)+(ac+bd)i. so by using the complex numbers in this way with the quaternions and having their imaginary units be mutually transparent and independent, it's isomorphic to GEO-3 under a trivial renaming transformation

[@not_vinkami]

9:30 are we assuming f(x) is not with order 1 in the first place? Order 1 functions like f(x)=x+1 of course has roots that are still in the field

[@Tehom1]

5:10 For prime p, which is not news to you since you say "we divide by the prime p" a few moments later but some watchers might need it.

[@cmilkau]

It's sometimes useful to have two additional elements ∞,¿, serving the role of 1/0 and 0/0, with the rules: x + ¿ = x·¿ = -¿ = 1/¿ = ¿ ∞ + ∞ = 0·∞ = ¿ x + ∞ = -∞ = 1/0 = ∞ when x≠∞,¿ x·∞ = ∞ when x≠0,¿ 1/∞ = 0 These rules follow common sense but also are recovered by defining a/b as module { (x,y) in R² | ay = bx } over some integral domain R and transporting the normal rules for fractions the same way, e.g. q + r = { (x,y) | (ad + bc)y = cdx for some (a,b) in q, (c,d) in r } (this strictly isn't even a ring but it's sometimes more useful).

[@MacHooolahan]

@Michael Penn and team, this video includes many concepts I've seen addressed usually used in proofs of Galois' insight into the quintic formula, and its general unsolvability. But I'll be honest - I've never quite *got* it - it never quite lodges in my head in a way I could explain back. Any chance of a fan request(!) where this could be tackled in a dedicated video, all in the (inimitable) MP way? I'd absolutely love to understand that one...! :D x

[@padgames8179]

Does your definition of a field preclude that 1=0? Or would the zero ring also be a field according to your definition?

[@antoniusnies-komponistpian2172]

It would be a highly trivial field because the only invertible element is excluded by the rule ^^

[@GlenMacDonald]

[7:23] "...because 3 times 5 is 1, because it's one more than 15". (He meant 14.)

[@xizar0rg]

Penn Fact: If he says it should be on the screen right now, it isn't. If he says there's a link in the description, there isn't.

[@bigchungus7327]

LOL

[@schweinmachtbree1013]

This construction is an example of a quotient field (i.e. a quotient of a ring by a maximal ideal) for those interested :)

[@TheEternalVortex42]

I think it's usually called a "quotient ring" (quotient field is something else). Although technically the construction in the video is only isomorphic to a quotient ring, since that acts on polynomial rings :)

[@schweinmachtbree1013]

​@@TheEternalVortex42 Unlike quotient group, quotient space, and quotient ring, "quotient field" doesn't mean "quotient of a field" because fields don't have any non-trivial quotients, so considering their quotients is useless. The construction in the video is isomorphic to a quotient field in the same way that *Z* _n is isomorphic to *Z* /n *Z* . In abstract algebra, equality of algebraic structures is never used; isomorphism of structures is the correct concept, and therefore e.g. a sentence of the form "the ring is the ring ..." should be taken to mean "the ring is isomorphic to the ring ..." - for example when we say "there are 5 groups of order 8" we mean that there are up to isomorphism 5 groups of order 8 (if we want to clarify this we can say "there are five nonisomorphic groups of order 8", but once one has understood isomorphism it no longer needs clarification)

[@CatchyCauchy]

​​@@schweinmachtbree1013wait i thought a quotient field Q of a Ring R follows that every injective function from R to a Field K has a function from Q to K such that the standard universal property diagram commutes Thats the definition i know

[@schweinmachtbree1013]

​@@CatchyCauchy Ah EternalVortex was right, "quotient field" does mean something else - I'm just so used to saying "field of fractions", which is what you're thinking of (but is defined for integral domains R, not all rings). The more common definition -- because undergraduates don't learn any category theory -- is in terms of formal fractions a/b, however the universal property is an equivalent definition (but should say "injective homomorphism" and "unique homomorphism" instead of "injective function" and "function"). Note however that definitions by universal properties do not prove that the object being defined exists! (it only proves that _if_ it exists then it is unique up to canonical isomorphism) - to prove existence you have to actually construct the object and show that it satisfies the universal property. A few examples of this (there are many more) are: - the polynomial ring R[X] can be defined by a universal property, but to show it exists one has to construct it in terms of formal polynomials - the free group on n generators F_n can be defined by a universal property, but to show it exists one has to construct it in terms of (equivalence classes of) words in n "letters" and their inverses - the tensor product of K-vectorspaces (or more generally, of a right R-module and a left R-module) can be defined by a universal property, but to show it exists one has to construct it in terms of a huge quotient of a huge free K-vectorspace (more generally, of a huge free abelian group) - the product and disjoint union of two sets can be defined by (dual) universal properties, but to show they exist one has to construct them in terms of pairs and labels.

[@ktlikes]

at 25:47 should be alpha squared plus alpha.

[@krisbrandenberger544]

That is at 24:07.

[@wagnerrodrigues6440]

Yup, gotta be to check if we are paying attention kkk

[@johannmeier6707]

Is it "valid" to justify the rational numbers Q to be a field by using a notation/symbol/sign (the "division line") that is not part of the general defintion of the field together with a (yet) "undefined" set (natural numbers)? And then saying that getting from Q to R is hard? I'd say, the other way round, that justifying R to be a field is "easily" valid and then from there proof that Q is also a field is the actual hard way.

[@schweinmachtbree1013]

*Q* is the field of fractions of ( *Z* , +, ×) while *R* is the order completion of ( *Q* , +, ×,

[@TheEternalVortex42]

Q is easy to define as ordered pairs of integer (a, b) which we write a/b, and s.t. (a, b) = (c, d) iff ad = bc. Then we define (a, b) * (c, d) = (ac, bd) and (a, b)^1 = (b, a) follows immediately.

[@schweinmachtbree1013]

@@TheEternalVortex42 ordered pairs of integers (a, b) with b ≠ 0*, (a, b)^{-1} = (b, a) for a ≠ 0*, (a, b) ~ (c, d) iff ad = bc*, and equivalence classes of ordered pairs*.

[@travisporco]

have you ever done a video on tropical stuff? What's up with that?

[@stephenhamer8192]

Re definition of a field. Don't we also have to have 0 =/= 1?

[@deepjoshi356]

Don't clean the desk before you die.

[@paosusuu]

build-a-field time

[@matematicacommarcospaulo]

As Brazilian, I thought interesting how Michael pronounces F[x]. The equivalent in Brazilian Portuguese is "f of x" just like a function f(x)

[@Mosux2007]

Would F = {0, 1, -1, sin x, -sin x, cos x, - cos x, csc x, - csc x, sec x, -sec x} be a field? It seems to satisfy all the requirements. If is it, would it have any use?

[@antoniusnies-komponistpian2172]

How is sinx*cosx in the field?

[@HideyukiWatanabe]

The object ⊥ is like the floating-point value NaN; Any operation(+-*/) including NaN is NaN and so is ⊥. Especially ⊥-⊥ = ⊥ and NaN-NaN produces NaN. (I didn't say NaN-NaN=Nan. In the floating-point system NaN is not equal to ANYTHING including NaN -- you can check e.g. x=float("NaN"); x==x produces False in python; you'll have simillar result in C/C++, java, javascript, perl, ...)

[@HideyukiWatanabe]

The new system including ⊥ has one inconvenience: a-b=0 ⇔ a=b ∧ {a,b}∩{⊥,∞}=emptyset

[@louisreinitz5642]

Shades of Galois

[@alielhajj7769]

First I thoight so we get Z8 then it poped out that Z8 is not a field and that we actually got a whole new structure of a finite group with 8 elements

[@theflaggeddragon9472]

Michael! .... Michael... you cannot use Z_p for the field with p elements :(

[@ludo-ge9fb]

why

[@theflaggeddragon9472]

@@ludo-ge9fb Z_p denotes the p-adic integers. You should use C_p for the cyclic group with p elements, Z/pZ for the ring, and F_p for the finite field with p elements (note Z/pZ and F_p are the same thing!)

[@Packerfan130]

the factorization of x^4 + 1 is not on the board :(

[@antoniodamianvargasmoreno]

Muy interesante. 👍

[@jplikesmaths]

I guess this makes us farmers now

[@Victor_Gonzalez98]

I just took the exam a few days ago for a subject on algebraic structures, a pre Gaolis Theory subject. Many definitions and abstract constructions without examples, without motivations in the ideas. If I had watched this video before studying the subject, or simply if it were explained from the point of view from which you do it, everything would have been easier. You know how to prepare the way for the student, motivating ideas and making what is seen make sense, thank you very much, professor. (sorry for my english, I don't know if I have made any mistakes or inconsistencies in my writing, but I think it is understandable)

[@OlivierGeorg]

It looks like un french It is called a "corps", but some assume that multiplication is commutative, other that it is not required!

[@antoniusnies-komponistpian2172]

Do non-commutative fields exist?

[@cmilkau]

you may want to request the degree of f be at least 2.

[@nunoalexandre6408]

Love it!!!!!!!!!!!!

[@Dshado]

Why do the exponents on alpha not simplify in Z2? E.g. alpha^2=alpha^0=1

[@Alex_Deam]

Because when working in a field like Z2, you're working congruent mod 2. Congruency mod 2 means you're asking for the remainder when divided by 2. The remainder of 2*alpha when divided by 2 is zero, since 2*alpha is divisible by 2. Meanwhile, alpha^2 is just shorthand for alpha*alpha. It's not divisible by 2.

[@landsgevaer]

alpha itself isn't in Z2, so it doesn't follow the rules of Z2. Apart from that, even if it were, then that wouldn't mean alpha*alpha = 0. (You mean alpha+alpha = 0?)

[@thierrypauwels]

When you write alpha^2, the 2 really is a natural number, not a number in Z2. alpha^2 is just a shorthand for alpha multiplied by alpha.

[@antoniusnies-komponistpian2172]

Because the result of powers is not explicitly defined in Z2, but derived from the multiplication results

[@bobh6728]

At the end, can you just assume that the remaining pair are an inverse pair? I thought that would have to be proven.

[@TheEternalVortex42]

We can given that we proved that this construction produces a field, which I guess didn't happen ;)

[@ethanbartiromo2888]

I’m on break at work rn, so I don’t have the time until I get home, but I’ve always been interested in thinking of cool operations and checking if they make a ring or field. So please nobody respond with an answer until I reply to my own comment, but my idea of operations is the plus operation being the remainder of dividing the numbers and the multiplication being the quotient of the numbers over the integets

[@ethanbartiromo2888]

Now that I think about it, the addition wouldn’t be commutative

[@ethanbartiromo2888]

@jojijoestar4762 I actually already watched that video

[@asappia]

what is 6^(-1) in Z7 ? 6*6 = 36 mod 7 = 1 Does it mean that 6 is inverse of itself in Z7?

[@mattikemppinen6750]

yes, or I guess it would be more correct to say that the residue class of 6 mod 7 is it's own multiplicative inverse

[@schweinmachtbree1013]

Yes :)

[@Alan-zf2tt]

For small fields like this it may help to think of a clock with zero at the top and for Z7, number 6 left of zero. The clock only uses integers 0, 1, ... 6 on clock face Sooo in Z7 (-1) = 6 as it is one step to left on clock (-2) = 5 as it is two steps to left on Z7 clock face. Does it help? well if you are asked what is (-3) in Z53 it should be trivially easy to work it out (on Z53 clock face 0 ~ 53, 1 ~ 1, (-1) ~ 52, (-3) ~ ....... (?) Also in Z53 (-2) x 53 ~ -106 ~ as ??? and (-2) x 53 - 10 ~ -116 ~ ????

[@TheEternalVortex42]

6 = -1 (mod 7) and -1 is always its own inverse in Z_p

[@IngamNogard]

Sooooooooo, the invers of all real numbers larger than 1 is contained within the open interval between 0 and 1......

[@antoniusnies-komponistpian2172]

Yes, which already proves that these intervals have the same cardinality.

[@rocky171986]

Lol clearly your editor is sleeping, the request d factorisation of x^4+1 did not come up on the screen

[@wojteksocha2002]

By the way Z(alpha) is isomorfic with (Z_2)^3

[@tracyh5751]

Only as groups under addition. As rings, the two have different multiplications (the latter has nonzero numbers which multiply together to become zero).

[@albertogarcia4177]

They are isomorphic as 3-dimensional Z_2-vector spaces

[@Anonymous-zp4hb]

Pretty cool. 𝛼 is like your 'imaginary constant' of Z_2. Does it follow that { a+b𝛼 | a,b ∈ Z_2 } is algebraically closed? Or would that require further work?

[@thierrypauwels]

No. A finite field can never be algebraically closed. The proof is quite simple. There is an infinite number of polynomials, since the degrees can extend to infinity. But the number of functions is finite. Inevitably that means that there must exist two polynominal that generate identical polynomial functions, i.e. there exist two different polynomials f and g such that f(x) = g(x), for all x in the finite field. Then the polynomial equation f(x) - g(x) + 1 = 0 has no root in that field.

[@rv706]

Since you're talking about fields, PLEASE don't denote by *Z*_p anything different from the *p-adics* !!