proof of Eisenstein criterion
Let be a polynomial satisfying Eisenstein’s Criterion with prime .
Let , , be the expansions of , and respectively.
Let be the natural homomorphism from to . Note that since for and , we have for and
Therefore we have so we must have and for some and some integers .
Clearly and , and therefore since , we must have and . Thus and .
If , then for . In particular, , hence . Similarly if , then .
Since , by equating coefficients we see that .
If and , then and , which implies that . But this contradicts our assumptions on , and therefore we must have or , that is, we must have a trivial factorization. Therefore is irreducible.
|Title||proof of Eisenstein criterion|
|Date of creation||2013-03-22 12:42:11|
|Last modified on||2013-03-22 12:42:11|
|Last modified by||rspuzio (6075)|