As it turns out, both $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{6}]$ are UFDs, and what is probably the simplest, most common way of proving this requires a full understanding of ideals. Adapting the proof that $\mathbb{Z}$ is a UFD to these domains can …

2019年3月10日 · The ring $\mathbb Z$ is a UFD, in spite of the fact that $12=2\times6=4\times3$. There is no contradiction, since these decompositions don't express $12$ as a product of irreducible integers.

2018年8月16日 · Any $f(x)$ in $\mathbb{Z}[x]$ can be written uniquely as $f(x) = c(f)p(x)$ where $p$ is primitive. As $\mathbb{Z}$ is a UFD then $c(f)$ can be uniquely written as a product of prime (integers). As $\mathbb{Q}[x]$ is a UFD then $p(x)$ can be written as a product of irreducible rational polynomials.

In mathematics, a unique factorization domain (UFD) (also sometimes called a factorial ring following the terminology of Bourbaki) is a ring in which a statement analogous to the fundamental theorem of arithmetic holds.

primes can be rewritten as \Z is a UFD". We are going to explore why this is true and we are going to try to prove that a few other domains are also UFDs. Even before that, we will see some examples of non-UFDs. Let’s introduce an important family of examples. Example 9 (Quadratic Rings). Let D2Z be a number that is not a square. We de ne Z[p ...

The polynomial ring Z[x] is a UFD, because every polynomial can be factored into irreducibles polynomials. However, it is not a PID because the ideal. is not principle. We've shown that (ii) holds for PIDs.

2021年1月28日 · Let $d$ be a positive integer that is not a perfect square. In this note we give necessary and sufficient conditions for $\mathbb {Z} [\sqrt { d}]$ to be a unique factorization domain.

Now we can go through the proof that Z[X] is a ufd, by replacing every polynomial by its primitive form whenever possible. The point is that the primitive polynomials have the same divisibility properties in Q[X] as in Z[X], allowing us to bring unique factorization down from Q[X] to Z[X]. (We emphasize that primitive polynomials are always

that Z[x] is a UFD. In Z[x], 1 is a greatest common divisor of 2 and x, but 1 ∈ 2Z[x]+xZ[x]. Lemma 6.6.4. In a unique factorization domain, every irreducible is prime. Proof. Suppose an irreducible p in the unique factorization R di-vides a product ab. If b is a unit, then p divides a. So we can assume that neither a nor b is a unit.

Z[x] is a UFD. We can also take polynomial rings in several variables, as discussed pre-viously. Corollary 16. Let R be a UFD. Then R[x 1,...,x n] is also a UFD. Proof. Induct on n! n = 0 is R, which follows by hypothesis. But then R[x 1,...,x n] = R[x 1,...,x n−1][x n] by definition, and since R[x 1,...,x n−1] is a UFD by induction ...