abstract algebra - Importance and Intuition of Polynomial Rings ...
Dec 14, 2020 · We denote this ring by R[x]. The intuition is that we are "adjoining" an "indeterminate" x to the ring R, which means that we are adding some element that has no constraints on it. In some sense, the element x is "free". In practice, this is how we always think about polynomials. Polynomial rings are extremely important.
Is a polynomial ring over a UFD in countably many variables a UFD?
Key Idea $ $ Each successive polynomial ring extension $\,D\subset D[x]\,$ is factorization inert, i.e. the ring extension introduces no new factorizations, i.e. if $\, 0\ne d\ \in D\,$ factors in $\,D[x]\,$ as $\,d = ab\,$ for $\, a,b\in D[x]\,$ then $\,a,b\in D.\,$ From this one easily deduces that the requisite factorization properties extend to the ascending union $\,R[x_1,x_2,\cdots ...
abstract algebra - How to deal with polynomial quotient rings ...
Often you can use the universal properties (of quotients and polynomial algebras) in order to simplify the rings. For example, F2[x] / (x4 + 1) = F2[x] / ((x + 1)4) ≅ F2[y] / (y4) and this cannot be simplified anymore. You just "cut" the polynomials in y …
Ring of polynomials in infinitely many variables
Sep 16, 2020 · 0. I am reading wiki article on polynomial rings and I want to make sure that I understand the bit about generalization allowing infinitely many variables. Let A = Z[xi | 1 ≤ i ≤ ∞] be the ring in infinitely many variables xi. Each polynomial is a linear combination of monomials of finite degree with all but finitely many coefficients ...
abstract algebra - Characterizing units in polynomial rings ...
Let R be a commutative ring with 1. Then f(X) = a0 + a1X + a2X2 + ⋯ + anXn is a unit in R[X] if and only if a0 is a unit in R and a1, a2, …, an are all nilpotent in R. Proof. Suppose f(X) = a0 + a1X + ⋯ + anXn is such that a0 is a unit in R and a1, a2, …, ar are all nilpotent in R. Since R is commutative, we get that a1X, a2X2, ⋯ ...
Modulo polynomial in ring theory - Mathematics Stack Exchange
Feb 27, 2015 · In this way you get a polynomial r(x) of degree less than x4 − 16, so of degree at most 3, and a polynomial q(x) such that. 7x13 − 11x9 + 5x5 − 2x3 + 3 = q(x)(x4 − 16) + r(x) You see that r(x) is congruent to 7x13 − 11x9 + 5x5 − 2x3 + 3 modulo x4 − 16. Note that this will always work as long as the polynomial that defines your ...
Ideals in Polynomial Rings - Mathematics Stack Exchange
1. I = x2, 2x, 4 I = x 2, 2 x, 4 is an ideal of Z[x] Z [x]. Prove that I I is not a principal ideal and find the size of Z[x]/I Z [x] / I. Using the theorem that ideals are principal iff the generator is irreducible, I think the first part is obvious since x2 x 2 is reduced to x (x). I'm slightly thrown by having multiple elements in the ...
abstract algebra - Maximal ideals of polynomial ring
Jun 15, 2015 · Set R = k[x1...xn] R = k [x 1... x n] and suppose R/m ≅ k R / m ≅ k. Then we have a homomorphism R → k R → k given by composing R → R/m R → R / m with this isomorphism, and each xi x i is sent to some element ai a i of k k. This completely determines the kernel of R → k R → k, for instance since R R is a free algebra, and so ...
Why is the polynomial ring of more than one variable not a PID?
Then R[x] is not a PID. Let r ∈ R, r ≠ 0 and r not invertible. We want to prove that (r, x) is not principal. Suppose r = f(x)g(x), for some f and g. Then both f and g are constant. Thus a generator for (r, x) must be a constant, a. Then x = ah(x), so h must have degree 1, but this implies a is invertible, by comparing coefficients.
Prove that the Polynomial Ring $F[t]$ Is an Integral Domain
In the polynomial ring over an integral domain, monomials decompose into monomials. 1 Prove that a polynomial ring over R in infinitely many variables is non-Noetherian.