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Evaluation at a point has kernel (x_1-a_1, ..., x_n-a_n)

Statement

Let $k$ be a field and let $a = (a_{1}, \dots, a_{n}) \in k^{n}$ . The evaluation map ${ev}_{a} : k [x_{1}, \dots, x_{n}] \to k, f \mapsto f (a),$ has kernel $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ . In particular, this ideal is maximal.

Facts & Assumptions

Given: A field $k$ , a point $a = (a_{1}, \dots, a_{n}) \in k^{n}$ , and the evaluation map at $a$ .

[L1]

Polynomial evaluation at a point is well defined (Evaluation and roots of a polynomial in a commutative target ring).

[L2]

Division by a monic polynomial produces quotient and remainder (Division by a monic polynomial over a commutative ring).

[L3]

A quotient ring is a field exactly when the ideal is maximal ( $R / M$ is a field if and only if $M$ is a maximal ideal).

Proof

technique · induction on the number of variables

1.1L1L2base

Base case $n = 1$ : divide $f (x_{1})$ by the monic polynomial $x_{1} - a_{1}$ . By [L2], $f (x_{1}) = (x_{1} - a_{1}) q (x_{1}) + r$ with $r \in k$ . Evaluating at $x_{1} = a_{1}$ gives $r = f (a_{1})$ by [L1], so $f - f (a_{1}) \in (x_{1} - a_{1})$ . Therefore the kernel of ${ev}_{a_{1}}$ is $(x_{1} - a_{1})$ .

1.2ih

Inductive hypothesis: assume the statement for $n - 1$ variables.

2.1L1L2step 1.2

Write $f$ as a polynomial in $x_{n}$ with coefficients in $k [x_{1}, \dots, x_{n - 1}]$ and divide by the monic polynomial $x_{n} - a_{n}$ : $f = (x_{n} - a_{n}) q + r,$ where $r \in k [x_{1}, \dots, x_{n - 1}]$ . Evaluating at $a$ gives $f (a) = r (a_{1}, \dots, a_{n - 1})$ . By the induction hypothesis, $r - r (a_{1}, \dots, a_{n - 1}) \in (x_{1} - a_{1}, \dots, x_{n - 1} - a_{n - 1})$ , and hence $f - f (a) \in (x_{1} - a_{1}, \dots, x_{n} - a_{n}) .$

3.1L3step 1.1step 2.1discharge-induction∎

Step 2.1 shows that every polynomial differs from its value at $a$ by an element of $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ , so the quotient by that ideal is isomorphic to $k$ . By [L3] the ideal is maximal, and its kernel description is the one established above.

Depends on

Used by

Over an algebraically closed field, every maximal ideal is an evaluation ideal Corollary
A nonradical ideal need not enlarge every tangent space Counterexample
The equation must define the intended scheme Counterexample
A node has two distinct tangent directions Example
A projective cone with a smooth conic base Example
An irreducible curve can have arbitrarily large tangent dimension Example
Different singularities can share a tangent cone Example
Over R, not every maximal ideal is an evaluation ideal Example
The cusp retains a doubled tangent line Example
The product of two parabolas: a block Jacobian and the direct-sum formula Example
The tangent space of the parabola at a general point and the local parameter Example
Three axes are not three coplanar lines Example
Classical affine points are maximal ideals Lemma
Finite local length exactly when no common local branch Lemma
Lengths of truncated plane local rings Lemma

Dependency tree · two levels

13 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, A Primer of Commutative Algebra, v4.03, Corollary 13.9 (standard reference, not scraped)