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A finitely localized polynomial ring in positive dimension is not a field

Statement

Let $k$ be a field, let $r > 0$ , and let $s \in k [t_{1}, \dots, t_{r}]$ be nonzero. Then the localization $k [t_{1}, \dots, t_{r}] [\frac{1}{s}]$ is not a field.

Facts & Assumptions

Given: A field $k$ , an integer $r > 0$ , and a nonzero polynomial $s \in k [t_{1}, \dots, t_{r}]$ .

[L1]

A $k$ -algebra map out of a polynomial ring is determined by the images of the indeterminates (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[L2]

Maps inverting the denominators extend uniquely to a localization (Universal property of localisation: maps that invert $S$ factor uniquely through $S^{- 1} R$ ).

[L3]

The polynomial ring $k [u]$ is a domain inside its fraction field; localizing it at powers of a nonzero polynomial is injective, and equality of fractions permits cross-multiplication (For a field $F$ , $F (t) = Frac (F [t])$ is its rational function field; in particular $R (t) = Frac (R [t])$ , Equality, vanishing, and the kernel of the localisation map).

Proof

technique · direct

1.1L1givenchoosealgebra

Choose an integer $N > 1$ larger than every exponent occurring in $s$ . Iterating [L1] defines $φ : k [t_{1}, \dots, t_{r}] \to k [u]$ by $t_{i} \mapsto u^{N^{i - 1}}$ . Distinct exponent tuples with entries below $N$ give distinct sums $\sum_{i} e_{i} N^{i - 1}$ : reduction modulo $N$ identifies the first entry, and subtraction and division by $N$ repeats the argument. Thus distinct monomials of $s$ retain distinct degrees, and $g : = φ (s) \neq 0$ . Also $φ (t_{1}) = u$ .

2.1L2L3step 1.1algebra

In $k [u] [1 / g]$ the image of $s$ is the unit $g$ , so [L2] gives a unital homomorphism $Φ : k [t_{1}, \dots, t_{r}] [1 / s] \to k [u] [1 / g]$ . It is surjective: $h (u) / g^{m}$ is the image of $h (t_{1}) / s^{m}$ . The target is nonzero by [L3]. If the source were a field, every nonzero target element would be the image of a nonzero element and its inverse would be the image of the source inverse. Hence the target would be a field.

3.1L3step 1.1step 2.1algebra

The target is not a field. If $g$ is constant, it is nonzero, so the target is $k [u]$ ; the polynomial $u$ has no inverse since a nonzero product $u h$ has positive degree. If $g$ is nonconstant, then $g + 1$ is a nonzero nonconstant polynomial and remains nonzero after localization. An inverse $h / g^{m}$ would imply $g^{m} = h (g + 1)$ in $k [u]$ by [L3]. But $g^{m} - (- 1)^{m}$ is divisible by $g + 1$ (expand $(g + 1 - 1)^{m}$ ), so this equality would make $g + 1$ divide the nonzero constant $(- 1)^{m}$ , impossible by degrees. This also covers $m = 0$ .

4.1step 2.1step 3.1∎

By step 2.1 a field source would give a field target, which step 3.1 rules out. Therefore $k [t_{1}, \dots, t_{r}] [1 / s]$ is not a field.

Depends on

Dependency tree · 2 levels

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, Lemma 13.6 and Proposition 13.7