Show that the ordinary polynomial ring R[X] is a right R[x ; δ]-module with R acting as right mu

Show that the ordinary polynomial ring R[X] is a right R[x ;...

Key Concepts

Module Theory

Module theory is a branch of algebra that generalizes the concept of vector spaces by allowing the scalars to come from a ring rather than a field. In this context, a module is an additive abelian group together with an action by a ring that satisfies certain compatibility conditions such as distributivity and associativity. When a ring acts on a module from the right, the module is referred to as a right module, and the action must satisfy (m · a) · b = m · (ab) for all module elements m and ring elements a and b.

Ore Extensions

An Ore extension is a way to construct noncommutative polynomial rings by introducing a twist in the multiplication process. Specifically, in an Ore extension R[x; ?], the multiplication of the indeterminate x with an element r of R is modified to x * r = r * x + ?(r), where ? is a derivation on R. This construction generalizes the ordinary polynomial ring and is fundamental in studying rings with noncommutative features due to the differential behavior introduced by ?.

Derivations

A derivation is a function ? from a ring R to itself that satisfies the Leibniz rule: ?(rs) = ?(r)s + r?(s) for all r, s in R, and is additive. This property is crucial in defining the noncommutative multiplication in Ore extensions because it controls how the indeterminate interacts with the elements of R. The derivation provides the 'twist' or 'skew' which differentiates these structures from typical polynomial rings.

Right Module Structure over a Skew Polynomial Ring

Establishing a right module structure for the ordinary polynomial ring R[X] over an Ore extension R[x; ?] involves specifying how the elements of R[x; ?] act on R[X]. Here, the action is defined so that the elements of R act by right multiplication and the indeterminate x acts in a way that incorporates both a shift and an application of the derivation ?. Verifying this structure means showing that the specified actions respect the module axioms, particularly the associative property, ensuring that the module action is consistent with the noncommutative multiplication of R[x; ?].

Transcript

00:01 In this problem given two polynomial functions b of x and the q of x and given a function between the two functions b and q defined as the constant multiplied by each other 2 multiplied by a1 b1 which is the coefficients of x plus a2 b2 which is the coefficients of x squared we want to show that this function is an inner product between the two functions in b2 we can do this by proving the following properties the first one the inner product or the function between b and the q is the same as the function between q and b a constant to multiply it by b and q equals the function between a constant multiply by b and q then the function between between b and the q plus m where m is a polynomial function in the same way as b and q equals the function between b and q plus the function between b and m self is always greater than equal 0 and equal 0 only when the coefficients of the function equals 0 let's start by the first one starting by the left hand side b and the q it's defined here and we have a property of the product between two real numbers it's commutative then we can switch each product here then it equals b node a node plus 2 b1 a1 plus b2 a2 this means we can start by q the coefficients of q then multiply by the coefficients of b or we can switch between b and the q here it's q the function between q and b then the first the first property is proven let's move to the second one constant multiplied by the function between b and the q equals we multiply c by each term here then it's c a node b node plus 2 c a 1 b1 plus a 2 b2 by gathering c and a like this, i forget seeing.

03:29 By gathering c and a like this, we can notice that we can multiply the function by c first, then substitute in the function here as cb and the q.

03:48 Then it's a function between cb and q.

03:53 Then the second property is proven.

03:55 Let's move to the third one.

03:59 It's b and the q...

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