statement theorem
Product of primitive polynomials is primitive
Statement
Let be a unique factorisation domain and let be primitive polynomials. Then is primitive.
Remarks
The proof: suppose an irreducible divides all the coefficients of . Since is prime, is an integral domain. The images of and in are nonzero, but their product is zero, which is a contradiction. Many texts call this statement Gauss's lemma. The node Gauss's lemma states the form about irreducibility. Over : and are primitive, and so is .
Explanations
L1 human
Suppose that no prime divides all the coefficients of , and no prime divides all the coefficients of . Then no prime divides all the coefficients of . The proof reduces the coefficients modulo a prime and uses that the result is an integral domain.
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What this needs
To read this node, read these 15 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
- Unit of a ring definition
- Zero divisor definition
- Divisibility in a commutative ring definition
- Integral domain definition
- Associate elements definition
- Prime element definition
- Irreducible element definition
- Greatest common divisor in an integral domain definition
- Unique factorisation domain definition
- Content of a polynomial definition
- Primitive polynomial definition
- Product of primitive polynomials is primitive statement
What depends on this
9 nodes use this node, directly or through other nodes.
- Existence of splitting fields 01441KDGBY
- Eisenstein's criterion 02BPKD7TMD
- Galois field 04SGPNT8R9
- Impossibility of doubling the cube 0CQ357M0QS
- Polynomial ring over a UFD is a UFD 0D6KH8Y59N
- Reduction modulo p test for irreducibility 0KWWEXW0RN
- Gauss's lemma 0QFV5Y1Z5X
- Existence of finite fields of every prime power order 0T8M3YC5RG
- Every polynomial has a root in some extension field 0VG02M1FEZ
Sources
- Web https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials), Statements for unique factorization domains
What links here
Used by
- Gauss's lemma 0QFV5Y1Z5X