Polynomial Rings and Division -- Abstract Algebra 22

[@synaestheziac]

@52:20 Michael I hope you’re okay, it sounds like you forgot to breathe before squeezing out The Last Seven Words

[@robshaw2639]

A good follow-on course for this syllabus would be Galois theory.... Or accelerated group theory, assuming knowledge of the group theory introduced in this course.

[@QuantumHistorian]

Representation theory would be another one - and would link in to the abstract linear algebra course too. Most applications of groups in physics are really about representations. I'd love to have an understandable explanation of Young's Tableaux for once.

[@robshaw2639]

@@QuantumHistorian I intentionally didn't mention representation theory, since i don't think Micheal's background there is very deep -- he would need to do some learning/research on his own before being confident to prepare a syllabus... I've watched the Borcherds grad course (among others), and yes, alot of the material seems to be lacking a compelling motivation why is it the way it is... or a deep enough working of examples to really gain an intuition...

[@jplikesmaths]

Love this series so far, can't wait for Modules and Fields (if you're planning on doing them)

[@iabervon]

"That's not actually the answer I was looking for, because the book I got this problem from wants you to do it in Z_11[x]. But don't worry, Z_11[x] is just like base ten, if you've got an extra finger and don't carry between terms." (cf Tom Lehrer, New Math)

[@Happy_Abe]

@35:45 wouldn’t the case of degree=0 contain a counterexample to the way it’s written here. Let f(x)=0 then f has degree 0, yet it has infinite many zeroes. Any value for x would satisfy f(x)=0 so this result isn’t true for the boring 0 polynomial.

[@schweinmachtbree1013]

Yes, this is why the degree of the zero polynomial is often left undefined or defined to be negative infinity. Alternatively, if you don't like segregating the constant polynomial 0 from the rest of the constant polynomials, then since the number of zeros is obvious for constant polynomials you could state the result just for non-constant polynomials.

[@Happy_Abe]

@@schweinmachtbree1013 thank you! What’s the intuition for some defining it as negative infinity?

[@schweinmachtbree1013]

​@@Happy_Abe So that the rules deg(p+q) ≤ max(deg(p), deg(q)) and deg(pq) ≤ deg(p) + deg(q) hold uniformly (i.e. even if p or q is the zero polynomial)

[@Happy_Abe]

@@schweinmachtbree1013 why do we care about the degree of the product , pq

[@zadsar3406]

The term "polynomial" can be rigorously defined as follows: a polynomial f on a ring R is a function from N0 (the natural numbers including 0) to R which is nonzero for only finitely many values. The interpretation of this definition is that our function gives us the coefficients of the polynomial. Clearly, these coefficients, along with the operations defined as in the video, uniquely determine what a polynomial "is". A nice generalization can be made as follows: a formal series on a ring R is a function f from N to R. Here, we don't require the function to be nonzero for finitely many values, so we get "infinite polynomials", i.e. power series. We can then work with such objects without worrying about convergence.

[@schweinmachtbree1013]

Such functions are called finitely-supported and infinitely-supported respectively, and the product is called the (discrete) convolution. To me "function from *N* _0 to R" is slightly overcomplicated; in my opinion "(zero-indexed) sequence" is conceptually simpler, and we also have notation for sequences so a polynomial over a ring R is a sequence (r_0, r_1, r_2, ...), and the product of polynomials is given by convolution of sequences.

[@synaestheziac]

Is there a generalization of polynomial rings to non-commutative rings, or rings without 1?

[@schweinmachtbree1013]

There is, but it's... very messy. To get the familiar nice properties of polynomials you need the coefficient ring to be both commutative and a domain: - For an example that commutativity is essential, consider polynomials over the quaternions *H* (*H* is a even a (non-commutative) domain): there are infinitely many square roots of -1 in the quaternions, so the polynomial x^2 + 1 has more than 2 zeros. i, j, and k are the obvious ones, but more generally we have (a + bi + cj + dk)^2 = (a^2 - b^2 - c^2 - d^2) + 2ab i + 2ac j + 2ad k - for this to equal -1 we need a=0 (because if a≠0 then b=c=d=0 and then (a + bi + cj + dk)^2 = a^2 is non-negative), and hence we have the equation (bi + cj + dk)^2 = -(b^2 + c^2 + d^2) = -1, so the solutions are parameterized by points on a sphere (since they are of the form (x, y, z) = (b, c, d) where x^2 + y^2 + z^2 = 1). - There was already an example that being a domain is essential in the video, with the coefficient ring being *Z* _12. The same thing happens over *Z* _8: we have 1^2 = (-1)^2 = 1 (mod 8) and 3^2 = (-5)^2 ≡ 1 (mod 8), so the polynomial x^2 - 1 has four zeros over *Z* _8, and in addition to the usual factorisation x^2 - 1 = (x-1)(x+1) we also have the factorization (x-3)(x+3). The reason that ordinary polynomials don't behave nicely over non-commutative domains is because ordinary polynomials are an inherently commutative notion: when we write p(x) = a_n x^n + ... + a_1 x + a_0, we are assuming that the a_k's commute with "x", but over a non-commutative ring this will usually not be the case (when we substitute something for x), and therefore a "non-commutative linear polynomial" should rather be something of the form "axb + c", and a "non-commutative quadratic polynomial" should be something of the form "axbxc + dxe + f", and in general a "degree" n "non-commutative polynomial" would have 1+2+...+(n+1) = (n+1)(n+2)/2 coefficients. An ordinary polynomial is still meaningful over a non-commutative ring R however: an expression of the form a_n x^n + ... + a_1 x + a_0 corresponds to a polynomial in a *central* variable x - that is, x can't be any element of R; rather it must come from the center Z(R). One can justify this by proving that for any z ϵ Z(R) there is an evaluation homomorphism ev_z: R[x] → R - if z is not central then this does not give a homomorphism. This comment is already getting too long so I will describe the situation for rings without identity in another comment

[@synaestheziac]

@@schweinmachtbree1013 wow, thank you! I will need to read through this slowly.

[@schweinmachtbree1013]

@@synaestheziac you're welcome, just let me know if there are any sentences that you don't understand and I'll explain them better/more fully.

[@kono152]

oh man Michael these big breaths you're taking are concerning, i know the video is a year old but i really hope you're feeling better now.