Proposition 1

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

Algebra, chapter IV, § 3. Rational fractions

— Let r, s be two rational fractions.

(i) If deg r ≠ deg s, we have

r + s \neq 0 \quad \text{and} \quad \deg(r + s) = \sup(\deg r, \deg s)

If deg r = deg s, we have deg(r + s) ≤ deg r.

(ii) We have deg(rs) = deg r + deg s.

We can limit ourselves to the case where r and s are non-zero.

Let us write r = $\frac{u}{v}$ , s = $\frac{w}{z}$ , where u, v, w, z are non-zero polynomials. We have rs = $\frac{uw}{vz}$ , and so

\deg(rs) = \deg(uw) - \deg(vz) = \deg u - \deg v + \deg w - \deg z = = \deg r + \deg s.

On the other hand, we have r + s = $\frac{uz + vw}{vz}$ . Suppose that deg r ≠ deg s, in other words deg u + deg z ≠ deg w + deg v. Then uz + vw ≠ 0, and

\begin{align*} \deg(r + s) &= \deg(uz + vw) - \deg(vz) \\ &= \sup(\deg(uz), \deg(vw)) - \deg(vz) \\ &= \sup(\deg(uz) - \deg(vz), \deg(wv) - \deg(vz)) \\ &= \sup(\deg r, \deg s). \end{align*}

Suppose that deg r = deg s, that is deg u + deg z = deg w + deg v. If r + s ≠ 0, then we have

\begin{align*} \deg(r + s) &= \deg(uz + vw) - \deg(vz) \\ &\leq \deg(uz) - \deg(vz) = \deg r \end{align*}

\* The mapping r ↦ −deg r is thus a discrete valuation on the field K((X_i)_{i ∈ I}). \*

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

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