PolynomialsOct7
... Problem A. Show that if P and D have coefficients in , and if D is monic, then P can be written as P = QD + S where S has smaller degree than D and both Q and S have coefficients in . Conclude that if P has coefficients in and an integer n is a root of P, then P(x) = (x-n) Q(x) where Q also has coef ...
... Problem A. Show that if P and D have coefficients in , and if D is monic, then P can be written as P = QD + S where S has smaller degree than D and both Q and S have coefficients in . Conclude that if P has coefficients in and an integer n is a root of P, then P(x) = (x-n) Q(x) where Q also has coef ...
We`ve Got to Operate Name
... 8. A set has the closure property under a particular operation if the result of the operation is always an element in the set. If a set has the closure property under a particular operation, then we say that the set is “closed under the operation.” It is much easier to understand a property by looki ...
... 8. A set has the closure property under a particular operation if the result of the operation is always an element in the set. If a set has the closure property under a particular operation, then we say that the set is “closed under the operation.” It is much easier to understand a property by looki ...
TRANSCENDENCE BASES AND N
... Reversing the roles of {xi} and {yj} in the lemma, you see that any two finite transcendence bases have the same cardinality. The lemma also implies that if one transcendence base is finite then so is any other. PROOF OF LEMMA: By the hypothesis on {yj}, x1 satisfies some non-trivial polynomial P(yj ...
... Reversing the roles of {xi} and {yj} in the lemma, you see that any two finite transcendence bases have the same cardinality. The lemma also implies that if one transcendence base is finite then so is any other. PROOF OF LEMMA: By the hypothesis on {yj}, x1 satisfies some non-trivial polynomial P(yj ...
