2 X is not a principal ideal:Proof verification: ⟨2
Proof verification: ⟨2
2015年12月4日—Iwanttoprovethat⟨2,x⟩isaprime,butnotprincipal,idealofZ[X].⟨2,x⟩isprime:Supposef(x)=fnxn+…+f1x+f0andg(x)=gmxm+…。其他文章還包含有:「⟨2,x⟩isanon」、「(2」、「Provethat(2」、「IsI=⟨5」、「Provingthatanidealisnotaprincipalideal」、「ExampleofanidealwhichisnotprincipalintheringZ[x]...」、「Is$mathbbZ}[x]$aprincipalidealdomain?」、「1CHAPTER3」
查看更多 離開網站⟨2,x⟩ is a non
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Therefore, we conclude that (2,X) can never be a principal ideal of Z[X]. Share.
(2
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Suppose that (2,x)=(p), this implies that 2=pq this implies that p and q are constant and p divides 2. Since (2,x)≠Z[X], p=2.
Prove that (2
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Proof. Assume by contradiction that ( 2 , x ) (2,x) (2,x) is a principal ideal. Then ( 2 , x ) = ( f ( x ) ) (2,x) = (f(x)) (2,x)=(f(x)), where ...
Is I=⟨5
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1 Answer 1 ... Z[X]/⟨5,X⟩≅Z5 is a field so ⟨5,X⟩ is a prime ideal (it is even maximal). On the other hand, note that Z[X]/⟨2,X2⟩≅Z2[X]/(X2) ...
Proving that an ideal is not a principal ideal
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1 Answer 1 ... You correctly state that A(X) should be a common factor of 2 and X. However the two facts you deem “obvious” aren't. Not difficult, ...
Example of an ideal which is not principal in the ring Z[x] ...
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Try the following one: I=⟨2,x⟩≤Z[x] . Try first to characterize the elements of the ideal (take a good peek at the free coefficient of ...
Is $mathbbZ}[x]$ a principal ideal domain?
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No. Consider the ideal (2,x). In general, ...
1 CHAPTER 3
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Solution. Suppose that the ideal I =< 2,X > in Z[X] is principal. Then there exists f(X) ∈ Z[X] such that I =< f(X) >. As 2 ∈ I there exists.