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Is $langle x^2+1,yrangle$ maximal or prime in $BbbR[x,y]$ or $BbbC[x,y]$

Maximal ideals and Prime ideals.To check whether $langle 3+irangle$ is a maximal ideal or not in the ring of Gaussian integers $Bbb Z[i]$Determine all maximal and prime ideals of the polynomial ring $Bbb C[x]$Find all prime and maximal ideals of ring $mathbbZ[x,y]/langle 6, (x-2)^2, y^6rangle$.When a prime ideal is maximal differential ideal in a UFD?The ideals $langle y-x-1rangle$ and $langle x-2,y-3rangle$ in $mathbb C[x,y]$ are primeShow $ langle 5,7 rangle $ and $ langle 6,9 rangle $ are principal ideals in $Bbb Z$Determine maximal ideal in $(mathbbZ[x], langle f(x)rangle)$Show that $Bbb R[x]/langle x^2+1rangle$ is a field.Show that $P[x]+langle xrangle $ is a prime ideal of $R[x]$.

4

2

$begingroup$

Let $R$ be a ring and $R[x,y]$ be the ring of all polynomials in variables $x$ and $y$ with coefficients in $R$. Then I need to check whether the ideal $langle x^2+1,yrangle$ is
maximal, prime in $R[x,y]$?

My understanding is that $langle x^2+1,yrangle$ is not maximal, as it is a proper subset of the ideals $langle x^2+1rangle$ and $langle yrangle$.

However I think it is a prime ideal if $R=BbbR$, and not prime if $R=BbbC$. Am I right? Then how do I show it?.

abstract-algebra ring-theory maximal-and-prime-ideals

share|cite|improve this question

edited Apr 18 '18 at 4:12

Abishanka Saha

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think it is prime for $mathbbR$ and not for $mathbbC$? In addition, think about whether $y in langle x^2 + 1 rangle$. Do you have theorems relating maximal/prime ideals to field/integral domain quotients? If so, those are the best approach for this question.
  $endgroup$
  – B. Mehta
  Apr 18 '18 at 4:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $x^2+1$ is irreducible in the first case, but reducible in the latter case.
  $endgroup$
  – Abishanka Saha
  Apr 18 '18 at 4:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's a good observation - in the case of $mathbbC$, there's an easy way to show that the ideal is not prime now, using the reduction you mention.
  $endgroup$
  – B. Mehta
  Apr 18 '18 at 4:16
  

  
  
  
  

add a comment | 

4

2

$begingroup$

Let $R$ be a ring and $R[x,y]$ be the ring of all polynomials in variables $x$ and $y$ with coefficients in $R$. Then I need to check whether the ideal $langle x^2+1,yrangle$ is
maximal, prime in $R[x,y]$?

My understanding is that $langle x^2+1,yrangle$ is not maximal, as it is a proper subset of the ideals $langle x^2+1rangle$ and $langle yrangle$.

However I think it is a prime ideal if $R=BbbR$, and not prime if $R=BbbC$. Am I right? Then how do I show it?.

abstract-algebra ring-theory maximal-and-prime-ideals

share|cite|improve this question

edited Apr 18 '18 at 4:12

Abishanka Saha

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you think it is prime for $mathbbR$ and not for $mathbbC$? In addition, think about whether $y in langle x^2 + 1 rangle$. Do you have theorems relating maximal/prime ideals to field/integral domain quotients? If so, those are the best approach for this question.
  $endgroup$
  – B. Mehta
  Apr 18 '18 at 4:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $x^2+1$ is irreducible in the first case, but reducible in the latter case.
  $endgroup$
  – Abishanka Saha
  Apr 18 '18 at 4:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's a good observation - in the case of $mathbbC$, there's an easy way to show that the ideal is not prime now, using the reduction you mention.
  $endgroup$
  – B. Mehta
  Apr 18 '18 at 4:16
  

  
  
  
  

add a comment | 

$begingroup$

Let $R$ be a ring and $R[x,y]$ be the ring of all polynomials in variables $x$ and $y$ with coefficients in $R$. Then I need to check whether the ideal $langle x^2+1,yrangle$ is
maximal, prime in $R[x,y]$?

My understanding is that $langle x^2+1,yrangle$ is not maximal, as it is a proper subset of the ideals $langle x^2+1rangle$ and $langle yrangle$.

However I think it is a prime ideal if $R=BbbR$, and not prime if $R=BbbC$. Am I right? Then how do I show it?.

abstract-algebra ring-theory maximal-and-prime-ideals

share|cite|improve this question

edited Apr 18 '18 at 4:12

Abishanka Saha

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

$endgroup$

share|cite|improve this question

edited Apr 18 '18 at 4:12

Abishanka Saha

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

share|cite|improve this question

edited Apr 18 '18 at 4:12

Abishanka Saha

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

asked Apr 18 '18 at 4:09

Abishanka SahaAbishanka Saha

7,90511022

2 Answers
2

active

oldest

votes

1

$begingroup$

$langle x^2+1, y rangle$ is not a subset of $langle y rangle$ or of $langle x^2+1 rangle$. It's a superset of those.

In $BbbC[x,y]$ it is a proper subset of $langle x+i, y rangle$ and hence not maximal. It's also not prime since factoring out $langle y rangle$ maps it to the ideal $langle x^2+1 rangle$ in $BbbC[x]$, which is not maximal and therefore not prime since $BbbC[x]$ is a PID. (And therefore $BbbC[x,y]/langle x^2+1,y rangle cong BbbC[x]/langle x^2+1 rangle$ which is not an integral domain.)

In $BbbR[x,y]$ it is both maximal and prime, since $BbbR[x,y]/langle x^2+1,y rangle$ is isomorphic to a field, namely $BbbC$.

share|cite|improve this answer

edited Mar 22 at 23:45

Andrews

1,2812423

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

$endgroup$

add a comment | 

2

$begingroup$

The ideal $(x^2+1, y)$ is maximal in $mathbbR[x, y]$, since $mathbbR[x, y] / (x^2+1, y) simeq mathbbR[x] / (x^2 +1)$ by the third isomorphism theorem, and $mathbbR[x] / (x^2 +1)$ is $mathbbC.$

On the other hand, it is not a maximal ideal in $mathbbC[x, y]$ (it is not even prime, in fact). This is because $mathbbC[x, y] / (x^2+1, y) simeq mathbbC[x] / (x^2+1) simeq mathbbC[x] / (x - i) times mathbbC[x] / (x + i) simeq mathbbC^2$.

share|cite|improve this answer

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214

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add a comment | 

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1

$begingroup$

$langle x^2+1, y rangle$ is not a subset of $langle y rangle$ or of $langle x^2+1 rangle$. It's a superset of those.

In $BbbC[x,y]$ it is a proper subset of $langle x+i, y rangle$ and hence not maximal. It's also not prime since factoring out $langle y rangle$ maps it to the ideal $langle x^2+1 rangle$ in $BbbC[x]$, which is not maximal and therefore not prime since $BbbC[x]$ is a PID. (And therefore $BbbC[x,y]/langle x^2+1,y rangle cong BbbC[x]/langle x^2+1 rangle$ which is not an integral domain.)

In $BbbR[x,y]$ it is both maximal and prime, since $BbbR[x,y]/langle x^2+1,y rangle$ is isomorphic to a field, namely $BbbC$.

share|cite|improve this answer

edited Mar 22 at 23:45

Andrews

1,2812423

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

$endgroup$

add a comment | 

1

$begingroup$

$langle x^2+1, y rangle$ is not a subset of $langle y rangle$ or of $langle x^2+1 rangle$. It's a superset of those.

In $BbbC[x,y]$ it is a proper subset of $langle x+i, y rangle$ and hence not maximal. It's also not prime since factoring out $langle y rangle$ maps it to the ideal $langle x^2+1 rangle$ in $BbbC[x]$, which is not maximal and therefore not prime since $BbbC[x]$ is a PID. (And therefore $BbbC[x,y]/langle x^2+1,y rangle cong BbbC[x]/langle x^2+1 rangle$ which is not an integral domain.)

In $BbbR[x,y]$ it is both maximal and prime, since $BbbR[x,y]/langle x^2+1,y rangle$ is isomorphic to a field, namely $BbbC$.

share|cite|improve this answer

edited Mar 22 at 23:45

Andrews

1,2812423

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

$endgroup$

add a comment | 

$begingroup$

$langle x^2+1, y rangle$ is not a subset of $langle y rangle$ or of $langle x^2+1 rangle$. It's a superset of those.

In $BbbC[x,y]$ it is a proper subset of $langle x+i, y rangle$ and hence not maximal. It's also not prime since factoring out $langle y rangle$ maps it to the ideal $langle x^2+1 rangle$ in $BbbC[x]$, which is not maximal and therefore not prime since $BbbC[x]$ is a PID. (And therefore $BbbC[x,y]/langle x^2+1,y rangle cong BbbC[x]/langle x^2+1 rangle$ which is not an integral domain.)

In $BbbR[x,y]$ it is both maximal and prime, since $BbbR[x,y]/langle x^2+1,y rangle$ is isomorphic to a field, namely $BbbC$.

share|cite|improve this answer

edited Mar 22 at 23:45

Andrews

1,2812423

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

$endgroup$

share|cite|improve this answer

edited Mar 22 at 23:45

Andrews

1,2812423

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

edited Mar 22 at 23:45

Andrews

1,2812423

edited Mar 22 at 23:45

Andrews

1,2812423

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

answered Apr 18 '18 at 4:23

C MonsourC Monsour

6,3391326

2

$begingroup$

The ideal $(x^2+1, y)$ is maximal in $mathbbR[x, y]$, since $mathbbR[x, y] / (x^2+1, y) simeq mathbbR[x] / (x^2 +1)$ by the third isomorphism theorem, and $mathbbR[x] / (x^2 +1)$ is $mathbbC.$

On the other hand, it is not a maximal ideal in $mathbbC[x, y]$ (it is not even prime, in fact). This is because $mathbbC[x, y] / (x^2+1, y) simeq mathbbC[x] / (x^2+1) simeq mathbbC[x] / (x - i) times mathbbC[x] / (x + i) simeq mathbbC^2$.

share|cite|improve this answer

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214

$endgroup$

add a comment | 

2

$begingroup$

The ideal $(x^2+1, y)$ is maximal in $mathbbR[x, y]$, since $mathbbR[x, y] / (x^2+1, y) simeq mathbbR[x] / (x^2 +1)$ by the third isomorphism theorem, and $mathbbR[x] / (x^2 +1)$ is $mathbbC.$

On the other hand, it is not a maximal ideal in $mathbbC[x, y]$ (it is not even prime, in fact). This is because $mathbbC[x, y] / (x^2+1, y) simeq mathbbC[x] / (x^2+1) simeq mathbbC[x] / (x - i) times mathbbC[x] / (x + i) simeq mathbbC^2$.

share|cite|improve this answer

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214

$endgroup$

add a comment | 

$begingroup$

The ideal $(x^2+1, y)$ is maximal in $mathbbR[x, y]$, since $mathbbR[x, y] / (x^2+1, y) simeq mathbbR[x] / (x^2 +1)$ by the third isomorphism theorem, and $mathbbR[x] / (x^2 +1)$ is $mathbbC.$

On the other hand, it is not a maximal ideal in $mathbbC[x, y]$ (it is not even prime, in fact). This is because $mathbbC[x, y] / (x^2+1, y) simeq mathbbC[x] / (x^2+1) simeq mathbbC[x] / (x - i) times mathbbC[x] / (x + i) simeq mathbbC^2$.

share|cite|improve this answer

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214

$endgroup$

share|cite|improve this answer

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214

answered Mar 22 at 15:02

Luca MorstabiliniLuca Morstabilini

214