Bourbaki

Proposition 12

Tag 06CI, permanent. Cite it as [Bourbaki, Tag 06CI].

Algebra, chapter IV, § 4. Formal power series

— Every non-zero element uu of K((X))K((X)) may be written in a unique way as u=Xkvu = X^k v, where k∈Zk \in \mathbf{Z} and vv is a formal power series in XX of order 0.

Let u=w/tu = w/t, where w,tw, t are non-zero elements of K[[X]]K[[X]]. We have w=Xrw1,t=Xst1w = X^r w_1, t = X^s t_1, where r,s∈Nr, s \in \mathbf{N} and w1,t1w_1, t_1 are formal power series of order 0, hence invertible in K[[X]]K[[X]] (IV, p. 30, Prop. 6). Then u=Xr−sw1t1−1u = X^{r-s} w_1 t_1^{-1} and w1t1−1w_1 t_1^{-1} is a formal power series of order 0.

Let us prove the uniqueness. Suppose that u=Xk1v1=Xk2v2u = X^{k_1} v_1 = X^{k_2} v_2 where k1,k2∈Zk_1, k_2 \in \mathbf{Z} and v1,v2v_1, v_2 are formal power series of order 0. Since Xk1−k2=v2v1−1X^{k_1 - k_2} = v_2 v_1^{-1} is a formal power series of order 0, we have k1=k2k_1 = k_2 whence v1=v2v_1 = v_2 and this proves the uniqueness assertion.

We shall say that the elements of K((X))K((X)) are generalized formal power series in XX with coefficients in KK, or simply formal power series when no confusion can arise (the elements of K[[X]]K[[X]] are then called formal power series with positive exponents); if u≠0u \neq 0, the integer kk defined in Prop. 12 is also called the order of uu and is written ω(u)\omega(u), even if it is <0< 0; we also put ω(0)=∞\omega(0) = \infty. It may be verified at once that

ω(u+v)≥inf⁡(ω(u),ω(v))ω(u+v)=inf⁡(ω(u),ω(v))if  ω(u)≠ω(v)ω(uv)=ω(u)+ω(v) \begin{align*} \omega(u + v) &\geq \inf(\omega(u), \omega(v)) \\ \omega(u + v) &= \inf(\omega(u), \omega(v)) \quad \text{if}\ \ \omega(u) \neq \omega(v) \\ \omega(uv) &= \omega(u) + \omega(v) \end{align*}

still hold for generalized formal power series. In particular, if u≠0u \neq 0, then w(u−1)=−w(u)w(u^{-1}) = -w(u). \* In other words (Comm. Alg., VI, § 3, No. 6, p. 392, Def. 3), w is a normalized discrete valuation of the field K((X))K((X)). \*

For each integer n∈Zn \in \mathbf{Z} let pnp_n be the set of all u∈K((X))u \in K((X)) such that ω(u)≥n\omega(u) \geq n. Then (p.), , , is a decreasing sequence of subgroups of the additive group K((X))K((X)), with intersection 0 ; there exists thus a topology on K((X))K((X)), invariant under translation, for which (pn)n∈Z(p_n)_{n \in \mathbf{Z}} is a fundamental system of neighbourhoods of 0 (Gen. Top., III, p. 223). We can easily verify that K((X))K((X)) is a topological field (Gen. Top., III, p. 281) and that K[[X]]K[[X]] is an open and closed subspace of K((X))K((X)).

Let (αn)n∈Z(\alpha_n)_{n \in \mathbf{Z}} be a family of elements of KK, and suppose that there exists an integer N such that an=0a_n = 0 for all n<Nn < N. Then the family' (αnXn)n∈Z(\alpha_n X^n)_{n \in \mathbf{Z}} is summable in K((X))K((X)) (Gen. Top., III, p. 263, Cor.) ; put u=∑n∈ZαnXnu = \sum_{n \in \mathbf{Z}} \alpha_n X^n, then u=0u = 0 if and only if an=0a_n = 0 for all nn; otherwise the order of uu is the least integer k such that αk≠0\alpha_k \neq 0. Finally every element of K((X))K((X)) may be written in a unique fashion in the form ∑n∈ZαnXn\sum_{n \in \mathbf{Z}} \alpha_n X^n, where the sequence (αn)(\alpha_n) satisfies α−n=0\alpha_{-n} = 0 for all sufficiently large n.

Since the ring K[X]K[X] is a subring of K[[X]]K[[X]], every rational fraction u/v∈K(X)u/v \in K(X) (u, v being polynomials in X) may be identified with the (generalized) formal power series uv−1uv^{-1} of K((X))K((X)), which we shall call its expansion at the origin ; the field K(X)K(X) is thus identified with a subfield of K((X))K((X)).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2003, Springer. Not translated.