Sur-anneau (traduction)

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Connaissez vous la traduction anglaise d'un "sur-anneau" (en algebre)? Merci de vos réponses.
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[invite286face9]

Connaissez vous la traduction anglaise d'un "sur-anneau" (en algebre)? Merci de vos réponses.

je crois avoir trouvé

http://mathworld.wolfram.com/ExtensionRing.html

[GuYem]

Oui, ça me parait naturel : A est un suranneau de B si B est un sous-anneau de A.

[invite286face9]

Oui, ça me parait naturel : A est un suranneau de B si B est un sous-anneau de A.

D'accord mon texte ne passe pas correctement. Mais ça donne une idée:

Construction of the ring R[X] of polynomials: Let (R;+,⋅) be a unit ring. We build the set S of all sequences (a₀,a₁,...) with a_{v}∈R, and comprising only a finite number of a_{v}≠0. S is then the set of all maps f:N→R such that f(N)∩R\{0} is a finite set. We define on S:
.Equality: (a₀,a₁,...)=(b₀,b₁,...):⇔a₀=b₀ ,a₁=b₁,...
.Addition: (a₀,a₁,...)+(b₀,b₁,...):=(a₀+b ₀,a₁+b₁,...)
.Multiplication: (a₀,a₁,...)⋅(b₀,b₁,...):=(a₀b₀ ,...,a₀b_{i}+
a₁b_{i-1}+...+a_{i}b₀,...)
.Neutral element for addition: (0,0,...)
.Unit element: (1,0,0,...)
.Opposite element: (-a₀,-a₁,...) of (a₀,a₁,...)
It follows that (R;+,⋅) is an unit ring.
Furthermore, ({(a,0,0,...)|a∈R};+,⋅) is a subring of S isomorphic to R (that we can identified with R). Then S can be taken as an extension ring of R. S possesses a remarkable element: X:=(0,1,0,0,...). We have X²:=(0,0,1,0,0,...), X³:=X⋅X²:=(0,0,0,1,0,0,...) etc, then (a₀,a₁,a₂,...)=(a₀,0,0,...)+(a ₁,0,0,...)⋅X+(a₂,0,0,...)⋅X²+⋯ , which is a finite sum because v such that a_{v}≠0 constitutes a finite set.
The identification mentioned above makes possible to write:

(a₀,a₁,...)=∑_{v=0}^{r}a_{v}X^ {v}

S is by definition the ring of polynomials in X on R; it is denoted R[X]: It is the smallest ring containing R and X. X is called "indeterminate" because ∑_{v=0}^{r}a_{v}X^{v}=0 belongs to R implies always whatever v, a_{v}=0∈R.

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