Factoring operator and fundamental theorem of algebra

• tim_lou
In summary, the conversation discusses the factorization of differential operators and their relation to the ring of polynomials in one variable over the complex numbers. The speaker asks about the algebraic structure that allows for this factorization and the axioms that a random operator must satisfy in order for it to be factorizable like a polynomial. The other person explains that the set of differential operators is essentially the same as the polynomial ring and that proving this involves showing that the operators commute with each other and satisfy certain identities. They also mention that if the operator satisfies a specific equation, the ring of polynomials in that operator acting on solutions of the equation is a quotient of the polynomial ring.

tim_lou

I haven't taken any abstract algebra course so I do not know if this is the right section to post this question in.

Anyway, I am learning differential equation right now and my prof. recently showed factorization of differential operator.

For example, let D be the differential operator, he then said that any thing in the form:
$$(D^n+A_{n-1}D^{n-1}+...+A_1D+A_0I)f$$
where I is the identity can be factorized into:
$$(D-z_1I)(D-z_2I)...(D-z_nI)f$$
where multiplication means composition of operator

I am curious to know what kind of algebraic structure has this property... and if I have a random operator, let's say F, what axioms does F have to satisfy in order for
$$F^n+A_{n-1}F^{n-1}+...+A_1F+A_0I$$
to be factorizable like a polynomial?

more specifically, If I define an operator $$\Delta x(n)=x(n+1)-x(n)$$
how can I show (what axioms does Delta have to satisfy) that polynomials in $$\Delta$$ can be factorized?
(I'm reading a book on differences equation... so I would like to know)

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Observe that the set of all such differential operators is nothing more than the ring of polynomials in one variable over C. You can factor polynomials, therefore you can factor such differential operators.

commutativity is useful, which is why he stuck to linear combinations of the D operator. those all commute with each other.

then they behave exactly like polynomials in X. almost.

they may also satisfy relations that polynomials do niot, but that means they behave even better.

thx for the quick response.

Hurkyl said:
Observe that the set of all such differential operators is nothing more than the ring of polynomials in one variable over C. You can factor polynomials, therefore you can factor such differential operators.

but what do I need to do to show that they are the ring of polynomials?

if I want to prove something is a group, i show a set of elements and a binary operator. then I prove associativity, identity and the inverse axioms... what about proving something is the "ring of polynomials"?

for instance, to prove that factoring works for differential operators, what do I have to show?

Do I have to show D commute with D and Identity and stuffs like:
$$DI=ID$$
$$D(D+I)=D^2+D$$
$$I(D+I)=ID+II=D+I$$

... or something similar?

the polynomial ring k[X] has the nice property that given any algebra R over k, and any element D of that algebra, there is a unique k algebra map from k[X] to R sending X to D.

Thus if D satisfies no relations, the map is an isomorphism. on the iother hand if you view D as an operator satisfyijng some equation like D^2+1 = 0, then the ring of poloynomials in D acting on the solutiions of that equation is a quotient of k[X] by the ideal generated by X^2+1.

1. What is a factoring operator?

A factoring operator is a mathematical symbol or function used to simplify or break down a mathematical expression into smaller factors. It is often represented by the symbol "x" or "⋅" and is used to indicate multiplication.

2. How is the factoring operator used in algebra?

In algebra, the factoring operator is used to simplify algebraic expressions by identifying common factors and grouping them together. This helps to solve equations, find roots, and factor polynomials.

3. What is the fundamental theorem of algebra?

The fundamental theorem of algebra states that every non-constant polynomial equation with complex coefficients has at least one complex root. In other words, every polynomial equation of degree n has exactly n complex roots.

4. How is the fundamental theorem of algebra related to the factoring operator?

The fundamental theorem of algebra can be proven using the factoring operator. By factoring a polynomial equation, we can determine its roots and show that it has n complex roots, which aligns with the fundamental theorem of algebra.

5. Why is the factoring operator and fundamental theorem of algebra important?

The factoring operator and fundamental theorem of algebra are important concepts in mathematics because they provide a foundation for solving polynomial equations and understanding complex numbers. They are also used in many applications, such as engineering, physics, and economics.