Is the 4 option is right of this problem?
-4
$begingroup$
Let R be the ring C[x]/(x^2+1). Pick the correct statements from below:
1 dimc R =3
2.R has exactly two prime ideals
3.R is UFD
4.(X) is a maximal ideal of R
ring-theory
asked Jan 20 at 8:14
user582791user582791
12
$endgroup$
add a comment |
-4
$begingroup$
Let R be the ring C[x]/(x^2+1). Pick the correct statements from below:
1 dimc R =3
2.R has exactly two prime ideals
3.R is UFD
4.(X) is a maximal ideal of R
ring-theory
asked Jan 20 at 8:14
user582791user582791
12
$endgroup$
$begingroup$
Please rewrite the title to be informative, and clarify the problem itself. Otherwise we can't help you and future readers will never find these answers by searching.
$endgroup$
– David G. Stork
Jan 20 at 8:18
$begingroup$
Since given that R=C[X]/(X^2+1) ,then by using Chinese remainder theorem it will be as R=CC (since X^2+1 reduces in (X+i).(X-i) ) now by first option dimension of CC(C) =2 and C*C has exactly two prime ideal like as 1*0. & 0*1 type
$endgroup$
– user582791
Jan 20 at 9:30
$begingroup$
Since R is not an integral domain because when we take two non zero elements like As. a= (1,0) & b=(0,1) then a.b=(1,0)(0,1) =(0,0) so it is not an integral domain. But I am still confusing how I check that (X) is a maximal ideal or not of R
$endgroup$
– user582791
Jan 20 at 9:34
add a comment |
-4
-4
-4
$begingroup$
Let R be the ring C[x]/(x^2+1). Pick the correct statements from below:
1 dimc R =3
2.R has exactly two prime ideals
3.R is UFD
4.(X) is a maximal ideal of R
ring-theory
asked Jan 20 at 8:14
user582791user582791
12
$endgroup$
Let R be the ring C[x]/(x^2+1). Pick the correct statements from below:
1 dimc R =3
2.R has exactly two prime ideals
3.R is UFD
4.(X) is a maximal ideal of R
ring-theory
ring-theory
asked Jan 20 at 8:14
user582791user582791
12
asked Jan 20 at 8:14
user582791user582791
12
asked Jan 20 at 8:14
user582791user582791
12
asked Jan 20 at 8:14
user582791user582791
12
asked Jan 20 at 8:14
user582791user582791
12
12
$begingroup$
Please rewrite the title to be informative, and clarify the problem itself. Otherwise we can't help you and future readers will never find these answers by searching.
$endgroup$
– David G. Stork
Jan 20 at 8:18
$begingroup$
Since given that R=C[X]/(X^2+1) ,then by using Chinese remainder theorem it will be as R=CC (since X^2+1 reduces in (X+i).(X-i) ) now by first option dimension of CC(C) =2 and C*C has exactly two prime ideal like as 1*0. & 0*1 type
$endgroup$
– user582791
Jan 20 at 9:30
$begingroup$
Since R is not an integral domain because when we take two non zero elements like As. a= (1,0) & b=(0,1) then a.b=(1,0)(0,1) =(0,0) so it is not an integral domain. But I am still confusing how I check that (X) is a maximal ideal or not of R
$endgroup$
– user582791
Jan 20 at 9:34
add a comment |
$begingroup$
Please rewrite the title to be informative, and clarify the problem itself. Otherwise we can't help you and future readers will never find these answers by searching.
$endgroup$
– David G. Stork
Jan 20 at 8:18
$begingroup$
Since given that R=C[X]/(X^2+1) ,then by using Chinese remainder theorem it will be as R=CC (since X^2+1 reduces in (X+i).(X-i) ) now by first option dimension of CC(C) =2 and C*C has exactly two prime ideal like as 1*0. & 0*1 type
$endgroup$
– user582791
Jan 20 at 9:30
$begingroup$
Since R is not an integral domain because when we take two non zero elements like As. a= (1,0) & b=(0,1) then a.b=(1,0)(0,1) =(0,0) so it is not an integral domain. But I am still confusing how I check that (X) is a maximal ideal or not of R
$endgroup$
– user582791
Jan 20 at 9:34
$begingroup$
Please rewrite the title to be informative, and clarify the problem itself. Otherwise we can't help you and future readers will never find these answers by searching.
$endgroup$
– David G. Stork
Jan 20 at 8:18
$begingroup$
Please rewrite the title to be informative, and clarify the problem itself. Otherwise we can't help you and future readers will never find these answers by searching.
$endgroup$
– David G. Stork
Jan 20 at 8:18
$begingroup$
Since given that R=C[X]/(X^2+1) ,then by using Chinese remainder theorem it will be as R=CC (since X^2+1 reduces in (X+i).(X-i) ) now by first option dimension of CC(C) =2 and C*C has exactly two prime ideal like as 1*0. & 0*1 type
$endgroup$
– user582791
Jan 20 at 9:30
$begingroup$
Since given that R=C[X]/(X^2+1) ,then by using Chinese remainder theorem it will be as R=CC (since X^2+1 reduces in (X+i).(X-i) ) now by first option dimension of CC(C) =2 and C*C has exactly two prime ideal like as 1*0. & 0*1 type
$endgroup$
– user582791
Jan 20 at 9:30
$begingroup$
Since R is not an integral domain because when we take two non zero elements like As. a= (1,0) & b=(0,1) then a.b=(1,0)(0,1) =(0,0) so it is not an integral domain. But I am still confusing how I check that (X) is a maximal ideal or not of R
$endgroup$
– user582791
Jan 20 at 9:34
$begingroup$
Since R is not an integral domain because when we take two non zero elements like As. a= (1,0) & b=(0,1) then a.b=(1,0)(0,1) =(0,0) so it is not an integral domain. But I am still confusing how I check that (X) is a maximal ideal or not of R
$endgroup$
– user582791
Jan 20 at 9:34
add a comment |
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$begingroup$
Please rewrite the title to be informative, and clarify the problem itself. Otherwise we can't help you and future readers will never find these answers by searching.
$endgroup$
– David G. Stork
Jan 20 at 8:18
$begingroup$
Since given that R=C[X]/(X^2+1) ,then by using Chinese remainder theorem it will be as R=CC (since X^2+1 reduces in (X+i).(X-i) ) now by first option dimension of CC(C) =2 and C*C has exactly two prime ideal like as 1*0. & 0*1 type
$endgroup$
– user582791
Jan 20 at 9:30
$begingroup$
Since R is not an integral domain because when we take two non zero elements like As. a= (1,0) & b=(0,1) then a.b=(1,0)(0,1) =(0,0) so it is not an integral domain. But I am still confusing how I check that (X) is a maximal ideal or not of R
$endgroup$
– user582791
Jan 20 at 9:34