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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
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A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field
Statement
Let be a field and let have degree or . Then is irreducible over if and only if has no root in .
Facts & Assumptions
Given: A field and a polynomial of degree or .
An element is a root of exactly when divides (Factor theorem over a commutative ring).
Degrees add in a product of nonzero polynomials over a field (Over an integral domain, degrees add under multiplication of nonzero polynomials).
The units of are exactly the nonzero constants (The units of over an integral domain are exactly the constant polynomials whose values are units of ).
A nonzero nonunit is irreducible exactly when every factorization has a unit factor (Irreducible and prime elements of an integral domain).
Proof
If has a root , then [L1] gives ; [L2] and make both factors nonunits by [L3], so [L4] shows that is reducible.
Conversely, if with both factors nonunits, [L2] and [L3] give positive degrees summing to or , so one factor has degree ; writing it as with , it has root , and that root is a root of . Thus reducibility implies a root, proving the biconditional.
Depends on
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 25 results over 11 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.
Sources
- Neil Donaldson, Math 120B Notes, Theorem 23.8 (standard reference, not scraped)