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 ...
In general, when does a ring have a division algorithm?
The example given is a polynomial ring, but it never once mentions the relationship between a ring and a polynomial ring. It specifically says "requirements on a ring" $\endgroup$ – Stella Biderman
abstract algebra - Characterizing units in polynomial rings ...
But there is a trick to make it work at least without the axiom of choice, by reducing (with standard methods) to the case of a finitely generated and hence countable commutative non-trivial ring, which always has a maximal ideal without using the (full) axiom of choice. $\endgroup$
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.
Laurent Polynomials Ring - Mathematics Stack Exchange
Wikipedia says: "The Laurent polynomial ring $R[X, X^{−1}]$ is isomorphic to the group ring of the group $\mathbb{Z}$ of integers over $R$".
How can i find a inverse of a polynomial in a quotient ring?
2019年2月4日 · I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I. In previous question, we have calculated the inverse of $\widehat {X} $ …
linear algebra - Prove that a Polynomial ring is a vector space ...
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Modulo polynomial in ring theory - Mathematics Stack Exchange
2015年2月27日 · $\begingroup$ Yes it still works. The ideal $(x^4-16)$ is a submodule, so you can quotient out by it. The equivalence relation corresponding to $(x^2-16)$ viewed as an ideal or a submodule are the same, and as sets the quotient module/quotient rings are exactly the same.
Does $A$ a UFD imply that $A[T]$ is also a UFD?
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How to represent polynomial rings in multiple variables.
2016年7月23日 · You define it recursively. $$ R[x_1,x_2, \dots, x_{n+1}] = \left\{ \sum_{j=0}^k a_j x_{n+1}^j \mid a_j \in R[x_1, x_2, \dots, x_n], k \in \Bbb{Z}_{\geq 0} \right\} $$