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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=(a1,,an)kn. The evaluation map

eva:k[x1,,xn]k,ff(a),

has kernel (x1a1,,xnan). In particular, this ideal is maximal.

Facts & Assumptions

Given: A field k, a point a=(a1,,an)kn, 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
1.1

Base case n=1: divide f(x1) by the monic polynomial x1a1. By [L2], f(x1)=(x1a1)q(x1)+r with rk. Evaluating at x1=a1 gives r=f(a1) by [L1], so ff(a1)(x1a1). Therefore the kernel of eva1 is (x1a1).

L1L2base
1.2

Inductive hypothesis: assume the statement for n1 variables.

ih
2.1

Write f as a polynomial in xn with coefficients in k[x1,,xn1] and divide by the monic polynomial xnan: f=(xnan)q+r, where rk[x1,,xn1]. Evaluating at a gives f(a)=r(a1,,an1). By the induction hypothesis, rr(a1,,an1)(x1a1,,xn1an1), and hence ff(a)(x1a1,,xnan).

L1L2step 1.2
3.1

Step 2.1 shows that every polynomial differs from its value at a by an element of (x1a1,,xnan), 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.

L3step 1.1step 2.1discharge-induction

Depends on

Used by

Dependency tree · two levels

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Sources