How do the definitions of "irreducible" and "prime" elements differ?
The most widespread terminology would be “irreducible” for that case. However there's no international math police, so everybody is entitled the right of naming concepts as they like; surely, using “prime” for “irreducible” doesn't do a good service to students. $\endgroup$ –
What are examples of irreducible but not prime elements?
The irreducible elements are $\pm$ primes and polynomials irreducible in $\mathbb{Q}[x]$ with $\pm 1$ as a constant term. The ring $\mathbb{Q} + x\mathbb{R}[x]$ works because $\mathbb{Q}$ is a subfield of $\mathbb{R}$. $\endgroup$
What are irreducible factors? - Mathematics Stack Exchange
Irreducible factors are like prime numbers, but for polynomials. You can't write them as a product of lower order polynomials.
Is zero irreducible in a ring? (esp. in a domain or field)
2024年8月27日 · For example, see the definition of associate and irreducible in the paper below, where it turns out that, for all $3$ notions of "irreducible" (based on $3$ notions of "associate"), one has $\,0\,$ is irreducible iff the ring is a domain (see the paragraph following Definition $2.4).$ Factorization in Commutative Rings with Zero-divisors.
abstract algebra - $1-i$ is irreducible in $\mathbb {Z} [i ...
This together with (1) implies $1-i$ can only be factored as a unit times a non-unit (norm is 2) and thus $1-i$ is irreducible by definition. Often to show that elements are irreducible or units we use a norm function.
Is an irreducible polynomial the minimal polynomial of its roots?
Therefore we know that any "candidate" for the minimal polynomial is a multiple of the minimal polynomial and if our candidate is irreducible, then it is the minimal polynomial and we're done. Note that this works over any field, not just $\mathbb{Q}$.
What is an irreducible matrix? - Mathematics Stack Exchange
Therefore, it cannot be broken into smaller strongly connected components and that's why it is called irreducible. E.g. $\pmatrix{0&1\\ 1&0}$ is irreducible while $\pmatrix{1&1\\ 0&1}$ is reducible (node 2 is reachable from node 1, but node 1 is not reachable from node 2).
abstract algebra - difference between irreducible element and ...
2017年10月27日 · This shows you can factor this polynomial into a product of two non-units, so it is $\textbf{not}$ irreducible. This definition just means, $3$ is still irreducible even though you can factor it $(-3)\cdot (-1)$. Since $-1$ is a unit, it doesn't count as a factorization which would make it fail to be irreducible.
Irreducibles are prime in a UFD - Mathematics Stack Exchange
Any irreducible element of a factorial ring $D$ is a prime element of $D$. Proof. Let $p$ be an arbitrary irreducible element of $ D$. Thus $ p$ is a non-unit. If ...
About irreducible morphisms - Mathematics Stack Exchange
Proving irreducible representation splits into direct sum of two irreducible representations when restricted to normal subgroup 2 Why are morphisms between non-isomorphic indecomposable modules always radical?