statement theorem
Polynomial ring over a UFD is a UFD
Statement
Let be a unique factorisation domain. Then the polynomial ring is a unique factorisation domain.
Remarks
The proof uses Gauss's lemma and unique factorisation in , where is the field of fractions of . The irreducible elements of are the irreducible elements of and the primitive polynomials of positive degree that are irreducible in . By induction, is a unique factorisation domain. In particular, and are. These rings are not principal ideal domains: in and in are not principal.
Explanations
L1 human
Integer polynomials factor uniquely into irreducible pieces, like integers do. The pieces are prime numbers and irreducible primitive polynomials. For example, .
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What this needs
To read this node, read these 21 nodes in this order. The foundations come first, and the list ends with this node.
- no evidence level
- Group concept
- Ring definition
- Commutative ring definition
- Polynomial ring definition
- Unit of a ring definition
- Zero divisor definition
- Divisibility in a commutative ring definition
- Field definition
- Integral domain definition
- Associate elements definition
- Prime element definition
- Field of fractions definition
- Irreducible element definition
- Greatest common divisor in an integral domain definition
- Unique factorisation domain definition
- Irreducible polynomial definition
- Content of a polynomial definition
- Primitive polynomial definition
- Product of primitive polynomials is primitive statement
- Gauss's lemma statement
- Polynomial ring over a UFD is a UFD statement
What depends on this
4 nodes use this node, directly or through other nodes.
- Existence of splitting fields 01441KDGBY
- Galois field 04SGPNT8R9
- Existence of finite fields of every prime power order 0T8M3YC5RG
- Every polynomial has a root in some extension field 0VG02M1FEZ