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Normalization of the t³,t⁴,t⁵ monomial curve

Statement

Let $k$ be any field, let $t$ be an indeterminate and let $A : = k [t^{3}, t^{4}, t^{5}] \subseteq k [t]$ be the $k$ -subalgebra generated by $t^{3}, t^{4}, t^{5}$ . Then the integral closure of $A$ in the rational function field $k (t)$ is exactly $k [t]$ , and $k [t] = A + A t + A t^{2}$ is a finite $A$ -module. The conductor ideal of $A$ in $k [t]$ is $(t^{3}, t^{4}, t^{5})$ and is strictly smaller than $k [t]$ ; it is not the object computed here.

Facts & Assumptions

Given: a field $k$ and the $k$ -subalgebras $A = k [t^{3}, t^{4}, t^{5}] \subseteq k [t] \subseteq k (t)$ of the polynomial ring in one indeterminate and its fraction field.

[L1]

For a field $k$ and elements $s_{1}, \dots, s_{r}$ of a $k$ -algebra, the subalgebra generated by them is written $k [s_{1}, \dots, s_{r}]$ and is the smallest subring containing $k$ and the $s_{i}$ ; $Frac (D)$ is the fraction field of a domain $D$ (Subalgebra generated by a subset, algebras of finite type, and module-finite algebras, Field extensions, generated subrings $F [S]$ , generated subfields $F (S)$ , and simple extensions, The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain).

[L2]

The polynomial ring $k [t]$ over a field is an integral domain with fraction field $k (t)$ (A polynomial ring over an integral domain is an integral domain, Zero divisor, and integral domain: a commutative ring with $1 \neq 0$ and no zero divisors, The field of fractions $Frac (D) = (D ∖ {0})^{- 1} D$ of an integral domain).

[L3]

For every field $F$ and $d \geq 0$ the polynomial ring $F [y_{1}, \dots, y_{d}]$ is an integrally closed domain (Finite-variable polynomial algebras over fields are integrally closed, Integral closure in an extension ring and integrally closed domains).

[L4]

Integrality over a subring means being a root of a monic polynomial over that subring; integrality is transitive along domain inclusions; integral elements form a subring; an integrally closed domain contains every integral element of its fraction field (Integral elements over a commutative ring and algebraic integers, Integral extensions are transitive, Integral elements over a nonzero base ring form a subring, Integral closure in an extension ring and integrally closed domains).

Proof

technique · direct

1.1L1L2given

$t = t^{4} / t^{3}$ is a quotient of two elements of $A$ with $t^{3} \neq 0$ , so $t \in Frac (A)$ by [L1]; hence $k [t] \subseteq Frac (A)$ , and $k (t) = Frac (k [t]) \subseteq Frac (A) \subseteq k (t)$ by [L2] because $A \subseteq k [t]$ . Thus $Frac (A) = k (t)$ .

1.2L4given

The element $t$ is integral over $A$ , being a root of the monic polynomial $T^{3} - t^{3} \in A [T]$ ; and $k [t] = A + A t + A t^{2}$ is a finite $A$ -module, because every polynomial in $t$ is a finite $k$ -linear combination of powers of $t$ and the relation $t^{3} = t^{3} \in A$ reduces the exponents modulo $3$ : $t^{3 m + r} = t^{3 m} t^{r} = (t^{3})^{m} t^{r}$ with $t^{3} \in A$ and $r \in {0, 1, 2}$ . Hence every element of $k [t]$ is integral over $A$ .

2.1L2L3L4step 1.1step 1.2∎

As in the cusp computation: if $z \in k (t) = Frac (A)$ is integral over $A$ , every monic equation for $z$ over $A$ also has coefficients in $k [t]$ , so $z$ is integral over $k [t]$ , and since $k [t]$ is integrally closed in its fraction field $k (t)$ by [L3] and [L2], $z \in k [t]$ . With step 1.2 this shows that the integral closure of $A = k [t^{3}, t^{4}, t^{5}]$ in $k (t)$ is exactly $k [t] = A + A t + A t^{2}$ , a finite $A$ -module. The conductor ideal $(t^{3}, t^{4}, t^{5}) \subseteq A$ is a proper ideal of $A$ contained in $A$ , not in the closure beyond $A$ , and plays no role in the identification of the closure.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Algebraic Geometry, Proposition 8.3 (standard reference, not scraped)
Stacks Project, Lemma 10.161.13 (polynomial N-2) (standard reference, not scraped)