Let R be a domain and let Q=Frac(R). (i) If r ∈R is nonzero and A is an R-module for which r A=

step 1 prove that a general polynomial in real space is a vector space. So a general polynomial would b

Let R be a domain and let Q=Frac(R). (i) If r ∈R is nonzero...

Key Concepts

Torsion Modules and Annihilation

A module A is called a torsion module with respect to an element r of R if r annihilates every element of A, meaning r · a = 0 for all a in A. This property leads to the absence of nonzero R-homomorphisms from or into certain R-modules when r becomes invertible in an extension such as Q, thereby having important implications for derived functors like Ext and Tor.

Vector Spaces over a Field

A vector space over a field Q is a module with an additional scalar multiplication by elements of Q, which is inherently free and has well-understood structural properties. When R-modules are extended to Q-vector spaces, they inherit properties that simplify the analysis of homological problems, such as calculating Ext and Tor, because many complications arising in R are resolved in the field Q where nonzero elements are invertible.

Flat Modules

A module is flat if the tensor product with it preserves the exactness of sequences. In the context of localization, the field of fractions Q is flat over R, meaning that when tensoring with Q, any torsion phenomena arising from annihilation by a nonzero element are effectively ‘killed’ in the derived functors. This flatness property is key to showing that certain Tor groups vanish, which has consequences for the computation of Ext groups as well.

Localization of a Domain

In commutative algebra, given an integral domain R, its field of fractions Q is obtained by localizing R at the set of nonzero elements. This process inverts every nonzero element of R, thereby creating Q where every nonzero element becomes a unit. The localization is a fundamental tool to extend properties from R to Q and to study the behavior of modules when passing from the ring to the field of fractions.

Derived Functors Ext and Tor

Ext and Tor are derived functors in homological algebra that measure, respectively, the failure of Hom to be exact and the failure of the tensor product to be exact. They are constructed through projective or injective resolutions of modules and are instrumental in computing and understanding deeper module properties. The vanishing of these functors in certain cases indicates a form of ‘homological triviality’ that reflects underlying structural simplicity.

Transcript

00:01 Prove that a general polynomial in real space is a vector space.

00:14 So a general polynomial would be of the form a sub 0 plus a sub 1x plus a sub 2 x squared plus dot dot plus a sub n x to the n power.

00:40 All right, a1.

00:47 And i'm trying to put a few things away here, just a minute.

00:55 A1.

00:55 Closure under addition.

00:58 So if i take one polynomial and i add another polynomial to it, b2x squared, then i get, i'm grouping them together, the coefficients together.

01:57 And so we see that all of the coefficients, a -n plus b -n, for example, are real coefficients.

02:11 And so, yes, it is closed under addition.

02:15 A -2, is this closed? under scalar multiplication.

02:19 So if i take a real number k times the general polynomial, then that's going to give me, now i'll have to scroll down a little bit, k times a sub 0 plus k times a sub 1, x plus k times a sub 2, x squared, and every term is still a real number.

03:07 A 3, commutativity of addition.

03:22 Oopsies, and i'm gonna add a second polynomial to it.

03:56 So that's going to give me this.

04:06 So we've already done this before.

04:32 Okay, but now we can just reverse the order of everything because these are just standard numbers.

04:41 We're just applying the commutative property for real numbers.

05:15 And now we can separate the two nine polynomials apart again b0 plus b1 x plus b sub 2 x squared plus a okay and we're done a 4 this is a long one i'm trying to get there we go a 4 now we need three polynomials we already know that a 1 plus b1 is is a1 plus b1.

06:22 I mean, we take these two polynomials and add them together, like this.

06:31 Then we get a1 plus b1.

06:34 A2 plus b2, x squared plus a3 plus a3.

06:49 Oh my goodness, i didn't do cubes before.

06:53 To erase that.

06:56 Dot, dot, that, plus.

07:12 Now we're going to have a third polynomial.

07:15 C1 plus c2x plus what in the world.

07:27 Okay, i see what i'm doing wrong.

07:29 Let's try this again and start with a sub -zero.

07:33 A -0 plus b -0 plus a -1 plus a -1 plus a -1 plus a -2, plus b2 x squared plus dot dot dot plus a n plus b n plus now we need c 0 c1 x plus c2 x squared plus dot dot dot plus c n x x okay now we're doing good.

08:25 So now a -0 plus b -0 plus c -0 plus a1 plus c -1 plus c1 plus a2 plus a2 plus b2 plus c2 x squared plus dot dot dot plus a n plus b n plus c n x n okay now we can rearrange them into whatever order we want i'm going to write b0 plus c 0 plus a plus zero plus zero plus zero plus um b1 plus c 1 plus c 1 plus a1x plus c2 plus b2 plus a2 plus a2 x squared plus dot dot plus c n plus b.

09:49 Wait, i didn't want c first, did i? i wanted b and then c.

10:00 B, n plus c, n plus a n.

10:05 X, n.

10:07 And back here i wanted b and then c.

10:11 So b2 plus c2.

10:19 Okay, now i can start splitting it apart.

10:25 B0 plus b1 x plus b2x squared plus dot dot dot plus bnx.

10:46 Plus c0 plus a 0 plus c1 plus c1 x and we keep going and we do it again we don't change this b polynomial c0 plus c1 x c n x x c n x and plus n x and plus a 0 plus a nx to n.

12:15 And so we've successfully reversed or switched the order, and the associative property shows that we can add in any order.

12:25 So a4 is satisfied.

12:30 A5, existence of a zero vector.

12:37 So a0 plus a1x plus.

12:48 A 2x squared plus a 2 x squared plus a 2 x what in the world n x n plus 0 the 0 vector that's going to be a 0 plus 0 plus a 1 x1 plus a 2 1 plus a 2 plus a 2 plus a 2 plus and x n and we notice that a 0 plus 0 is just a 0 so yes that one is satisfied a 6 existence of additive inverses so we need to add the opposite of a 0 minus a 1x minus a 2 x squared minus dot, dot, dot, minus.

14:25 Okay, so that's going to give us, well, notice that a0 minus a0 cancel out.

14:39 A1x minus a1x cancel out.

14:42 That cancels out, and the n terms cancel out, and that does give us zero, which is the zero vector.

14:50 So that's good.

14:52 A7.

14:57 Unit property, one time.

15:00 Okay, that's just going to equal 1a0 plus 1a1x plus 1a2x squared plus dot dot dot plus 1a1n x and 1 times anything is itself.

15:41 1 times a0 is a0.

15:44 1 times a1 is a1.

15:48 1 times a2 is a2 plus 1 times a0 is an.

15:59 And so we've verified this one...