I am confused in this question...because according to me A does not satisfy the condition of ideal. It means A is not ideal. But given that A is ideal.A is ideal of Euclidean ring R.I attach a solution but i cant understand it... Please check and tell me is it true.. or send me a clear and well explained solution.How we show this.Please help From defini tion of ideal itis clear7.!thatand r ER.thenMy Question isR is euclidears ting-A isits ideal BadeccOntYa and arare in A.that 3 as element19,ERSuch that A-fay iseR}wePlease solve with ezplanationOY give reason.ive let REu-Ringbeandbean ideadshowThertelemed a, E RCinsnchA=a.ol-weKnowidealeveryprincipal ideal.Gn idealEu-RingasuchThat(a)=ASinceenclideanDomaina'= ar4r?ris remainder)=>eilherora'= agY <>A = (a,r)(1)3D)%3D

The question is, If R is Euclidean ring,A is its ideal.Show that ∃ an element a0∈R such that…

Answered: From definition of ideal it is clear 7!…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

I am confused in this question...because according to me A does not satisfy the condition of ideal. It means A is not ideal. But given that A is ideal.

A is ideal of Euclidean ring R.

I attach a solution but i cant understand it... Please check and tell me is it true.. or send me a clear and well explained solution.

How we show this.

Please help

From defini tion of ideal it
is clear
7.!
that
and r ER.
then
My Question is
R is euclidears ting-A is
its ideal BadeccOnt
Ya and ar
are in A.
that 3 as element
19,ER
Such that A-fay iseR}
we
Please solve with ezplanation
OY give reason.
ive

let R
Eu-Ring
be
and
be
an idead
show
Thert
elemed a, E R
Cin
snch
A=
a.
ol-
we
Know
ideal
every
principal ideal.
Gn ideal
Eu-Ring
a
such
That
(a)=A
Since
enclideanDomain
a'= ar4r?
ris remainder)
=>
eilher
or
a'= agY <>
A = (a,r)
(1)
3D)
%3D

Expert Solution

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Ring$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions