Exercise 12.2. Let F be a field and n Є N. (1) For every subset S of F", show that I(S) := {f(x1, . ,xn) Є F[x1,...,xn] | f (a₁,, an) = 0, V(a1, ..., an) E S} is an ideal of the polynomial ring F[x1, ..., xn], called the vanishing ideal of S. (2) Given two subsets S and T of F", prove or disprove that I(SUT) = I(S)NI(T) · and I(SNT) = I(S) + I(T).
Exercise 12.2. Let F be a field and n Є N. (1) For every subset S of F", show that I(S) := {f(x1, . ,xn) Є F[x1,...,xn] | f (a₁,, an) = 0, V(a1, ..., an) E S} is an ideal of the polynomial ring F[x1, ..., xn], called the vanishing ideal of S. (2) Given two subsets S and T of F", prove or disprove that I(SUT) = I(S)NI(T) · and I(SNT) = I(S) + I(T).
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.2: Divisibility And Greatest Common Divisor
Problem 34E
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![Exercise 12.2. Let F be a field and n Є N.
(1) For every subset S of F", show that
I(S) := {f(x1,
.
,xn) Є F[x1,...,xn] | f (a₁,, an) = 0, V(a1,
..., an) E S}
is an ideal of the polynomial ring F[x1, ..., xn], called the vanishing ideal of S.
(2) Given two subsets S and T of F", prove or disprove that
I(SUT) = I(S)NI(T)
·
and
I(SNT) = I(S) + I(T).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff2e948f6-fd6f-485f-942e-c931230f8579%2Ffa271370-22ab-47d4-90c9-a7594e478e73%2Fcgt1f8c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Exercise 12.2. Let F be a field and n Є N.
(1) For every subset S of F", show that
I(S) := {f(x1,
.
,xn) Є F[x1,...,xn] | f (a₁,, an) = 0, V(a1,
..., an) E S}
is an ideal of the polynomial ring F[x1, ..., xn], called the vanishing ideal of S.
(2) Given two subsets S and T of F", prove or disprove that
I(SUT) = I(S)NI(T)
·
and
I(SNT) = I(S) + I(T).
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