4. The following polynomial is irreducible in Q[x]: m(x) = x² - 2x²-1 € Q[x] Let's consider the ideal I = (m(x)), and the quotient ring F = = Q[x]/I. (a) Explain why F is a field. (b) Consider the element a = x + I of F. Show that a is a root of m(x). In other words, show that is zero in the ring F. a4-2a2-1=0
4. The following polynomial is irreducible in Q[x]: m(x) = x² - 2x²-1 € Q[x] Let's consider the ideal I = (m(x)), and the quotient ring F = = Q[x]/I. (a) Explain why F is a field. (b) Consider the element a = x + I of F. Show that a is a root of m(x). In other words, show that is zero in the ring F. a4-2a2-1=0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![4. The following polynomial is irreducible in Q[x]:
m(x) = x² — 2x² – 1 € Q[x]
Let's consider the ideal I = (m(x)), and the quotient ring F = Q[x]/I.
(a) Explain why F is a field.
(b) Consider the element a = x + I of F. Show that a is a root of m(x). In other words, show that
is zero in the ring F.
a42a2-10
Remark: Question 4 shows that we can construct a field extension QC F where the polynomial m(x) has a
root: we define F = Q[x]/[m(x)). The same construction works in much greater generality.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf4d5614-e5fa-4399-aabc-c345eeef0588%2F3d9f16a2-6f2e-4283-a7d4-a83c9a233cf5%2F3z9hrzi_processed.png&w=3840&q=75)
Transcribed Image Text:4. The following polynomial is irreducible in Q[x]:
m(x) = x² — 2x² – 1 € Q[x]
Let's consider the ideal I = (m(x)), and the quotient ring F = Q[x]/I.
(a) Explain why F is a field.
(b) Consider the element a = x + I of F. Show that a is a root of m(x). In other words, show that
is zero in the ring F.
a42a2-10
Remark: Question 4 shows that we can construct a field extension QC F where the polynomial m(x) has a
root: we define F = Q[x]/[m(x)). The same construction works in much greater generality.
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