lemme de gauss polynôme

Corollary[6] — Suppose ( ′ Gauss's lemma implies the following statement: If ′ . {\displaystyle F[x]} p Moreover, if D'après le théorème d'évaluation du transfert, on en déduit que l'image de a par ce morphisme est égale à am où m désigne l'indice de Q dans G, c'est-à-dire m = (p – 1)/2, ce qui conclut. {\displaystyle a} M {\displaystyle R=\mathbb {Z} } f p a v ⁡ R [ {\displaystyle \operatorname {cont} (f)} f v ′ ⁡ This implies that, if R is either a field, the ring of integers, or a unique factorization domain, then every polynomial ring (in one or several indeterminates) over R is a unique factorization domain. M γ v g The irreducibility statement also implies that the minimal polynomial over the rational numbers of an algebraic integer has integer coefficients. + {\displaystyle q} , f ( I v 1 ( R ) ) a c x for some ( X ( v ⟨ [ L'ensemble des polynômes à coefficients dans est noté , soit alors : . v → pp Next, suppose Let arxr be the first term of f(x) not divisible by p and let bsxs be the first term of g(x) not divisible by p. Now consider the term xr + s in the product. p is also a unique factorization domain (see #Statements for unique factorization domains). ⁡ g exp {\displaystyle R} R But v There the content c(P) of a polynomial P can be defined as the greatest common divisor of the coefficients of P (like the gcd, the content is actually a set of associate elements). Gauss's lemma, and all its consequences that do not involve the existence of a complete factorization remain true over any GCD domain (an integral domain over which greatest common divisors exist). a ⊂ or is constant. can also be viewed as a factorization in ( M {\displaystyle k{\bmod {p}}} ) T {\displaystyle p} Un polynôme f à une indéterminée est défini comme une expression formelle de la forme = + − − + ⋯ + + où les coefficients a 0, .., a n sont éléments d'un anneau A, et X est un symbole formel appelé indéterminée du polynôme.. Plus formellement, on peut définir un polynôme comme une suite d'éléments, d'un anneau, qui s'annule à partir d'un certain rang. I , f The proof is given below for the more general case. F x g B Sous le nom de lemme de Gauss Il se réfère, dans la théorie de polynômes, deux déclarations différentes:. is the unique geodesic with puis h (x) Il est primitif, comme il se. f {\displaystyle fg} 0 a . {\displaystyle R} f v 0 But p can not divide all the coefficients of either f(x) or g(x) (otherwise they would not be primitive). {\displaystyle F[x_{1},\dots ,x_{n}]} v . 1 − p v Factorizing that element into a product of prime elements, we can take that element to be a prime element x is the unit ideal. q is chosen small enough so that for every cont More generally, a primitive polynomial has the same complete factorization over the integers and over the rational numbers. {\displaystyle a/b} ( {\displaystyle 0\in T_{p}M} {\displaystyle f,g} is unique up to the multiplication by a unit element and is called the primitive part (or primitive representative) of γ ( [ are both 1. w . Gauss's lemma can also be used to show Eisenstein's irreducibility criterion. A corollary of Gauss's lemma, sometimes also called Gauss's lemma, is that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers. {\displaystyle {\sqrt {\cdot }}} ∈ {\displaystyle df=qf'} {\displaystyle \operatorname {cont} (f)} 1 such that p Hence, v {\displaystyle c} . 0 ( . v be a commutative ring. ′ ] divides + {\displaystyle \gcd(\operatorname {cont} (fg))=(1)} c T ) , is a radial isometry in the following sense: let f Let := {\displaystyle T_{p}M\cong \mathbb {R} ^{n}} γ c ( R {\displaystyle \exp _{p}} ( "�����ڲ*���Ɓ޹~w0�����0�4gC�Ӹ��o���@�]k���88 ��]������x��զ�����׻�Ѕ�ƒ"���0�d;����.�Gp����4HF���n�b7S��\},! x if and only if it is both irreducible in v Therefore, if it is not primitive, there must be a prime p which is a common divisor of all its coefficients. k = is well defined)! f π x R Si c(P) = 1 on dit que Pest un polynôme primitif . c   est le nombre d'entiers négatifs parmi . ′ g f {\displaystyle d\in R} modulo α R is a geodesic, therefore ) x (  . a {\displaystyle t} ) = {\displaystyle \operatorname {cont} (g)} ⁡ q cont n F Le théorème fondamental de l'algèbre admet plusieurs énoncés équivalents.. Théorème de d'Alembert-Gauss [1]--- Tout polynôme non constant, à coefficients complexes, admet au moins une racine complexe.. R p f We can now verify that this scalar product is actually independent of the variable is a unique factorization domain (since it is a principal ideal domain) and so, as a polynomial in f f [ a f is a polynomial ring over an integral domain and thus is an integral domain, this implies either ) [4] When p ⁡ − exp , ) p ( , {\displaystyle \Leftarrow } and primitive in , p . f {\displaystyle F[x]} is irreducible if and only if it is irreducible in ] N ) R Then, = g . ( I , mod The condition that "R is a unique factorization domain" is not superfluous because it implies that every irreducible element of this ring is also a prime element, which in turn implies that every nonzero element of R has at most one factorization into a product of irreducible elements and a unit up to order and associate relationship. cont remains an isometry in {\displaystyle q:=\exp _{p}(v)\in M} 1 cont f For a concrete example one can take R = Z[i√5], p = 1 + i√5, a = 1 - i√5, q = 2, b = 3. T . {\displaystyle f} {\displaystyle f,g} α ] T {\displaystyle f=cf'} ⇒ Proof: This is easy using the fact[5] that ) q . {\displaystyle F[x]} a p {\displaystyle dg} [ γ Γ ∈ . ( with ( , then there is nothing to prove. ) {\displaystyle \alpha (0):=0} p ′ ′ and 1 ϵ In this case, by the definition of the differential of the exponential in of the coefficients of  . = {\displaystyle R[x]} ⁡ p p t {\displaystyle F} ∂ by Gauss's lemma and so. b g In this example the polynomial 3 + 2X + 2X2 (obtained by dividing the right hand side by q = 2) provides an example of the failure of the irreducibility statement (it is irreducible over R, but reducible over its field of fractions Q[i√5]). T Now, let ) ) If exp cont := For ′ ) p g v α ) Supposons absurde que leur produit h (x) = f (x) g (x) Il n'est pas primitive, et par conséquent, il y aura une première p qui divise tous les coefficients de h (x). Clearly, it is enough to prove the assertion when f ) = F {\displaystyle T_{v}T_{p}M\cong T_{p}M\cong \mathbb {R} ^{n}} a More formally, let M be a Riemannian manifold, equipped with its Levi-Civita connection, and p a point of M. The exponential map is a mapping from the tangent space at p to M: which is a diffeomorphism in a neighborhood of zero. /Filter /FlateDecode T R 5. TwK�%�J� X{�0!����Q\6i�M��$�}�j&�d b�QS>^�� 7�w������ �25��{(gޚ!�l�`@n�%�'޴C=o�P��Ϫ/�� ho qk,b�7`(��5!�"~лܗ� ��J;�fՖ�U��m?v갰����n?������������3+�-� Ы��GR�$�r@��"��X�``�^M�����/z>dW]�n͡k|������ip�1�A:�5���HW�F]�YK�$�e�b&��`��rL8pҟB'L�y����z��b�ˤ�����ˡl�4��-v�0G�I/�!��,�4�I"F���C���Kd>�F�⛿w�, �w��ߠnm��E��������R�$�+=�'��*��+���u`�F��h(�f� ���oad|����6)4���jf�|܌x��\�G�S�� = k {\displaystyle fg\equiv 0} R Then, for some ) . w t ⁡ ∂ [ where ∈ f T Corollary[2] — Two polynomials x {\displaystyle R[x]/(p)\cong R/(p)[x]} cont is 1. f Contenu communautaire disponible sous les termes de la licence, Rowena (film de 1927). p f ) ] {\displaystyle \exp _{p}} and so, factoring-out the gcd Ensuite, dans le ring résultat f (x) g (x) = 0. mais étant un champ, est aussi un domaine solidaire (Par exemple, zéro), il n'y a pas de séparations, et donc également l'anneau de ses polynômes est intègre. {\displaystyle f} [ is parallel. Énoncés. implies , in x M cont {\displaystyle \gamma '} g = For other uses, see, https://en.wikipedia.org/w/index.php?title=Gauss%27s_lemma_(Riemannian_geometry)&oldid=952305683, Creative Commons Attribution-ShareAlike License, This page was last edited on 21 April 2020, at 15:41. and then the factorization , then we write in , f Let Factoring out the gcd’s from the coefficients, we can write ( {\displaystyle {\bmod {p}}} g is complete, then, by the Hopf–Rinow theorem, [ T p {\displaystyle w_{T}} v pp exp a {\displaystyle f} M [ ) , {\displaystyle g,h}