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Find a maximal ideal in $F_7[x]/(x^2+5)$

Finding an idempotent that satisfies certain conditions in a matrix ring.Find all prime and maximal ideals of $mathbb C[x,y]$ that contain $I=langle x^2 + 1, y + 3rangle$Preimage of maximal ideal is maximalUnique maximal ideal in the ring of fractionNon-invertible elements form an idealShow that the ideal generated by $x^2-2$ is maximalExample of a non-trivial ideal of ring of all continuous functions defined on all $mathbbR$Find not irreducible polynomialTo prove that a polynomial is irreducible in $mathbbF_p[x]$.Maximal Ideal in a polynomial ring

0

1

$begingroup$

$F_p=Z/(pZ)$

I'm stuck at this problem. I tried to find the irreducible factorisation of $x^2+5$, but that didn't help too much. Any ideas? and also could anyone explain the general philosophy with regard to solving this kind of problem?

abstract-algebra ideals

share|cite|improve this question

edited Mar 18 at 0:16

David

69.6k668130

asked Mar 18 at 0:13

davidhdavidh

3178

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: over $F_7$ we have $x^2+5=x^2-9$.
  $endgroup$
  – David
  Mar 18 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @David Okay. So my factorisation turns out to be $(x+3)(x-3)$. But this is where I seem to have stuck upon. Edit: sorry I made a silly mistake.
  $endgroup$
  – davidh
  Mar 18 at 0:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @David Should I invoke the Chinese remainder theorem here?
  $endgroup$
  – davidh
  Mar 18 at 0:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  since your ring is actually an algebra over $mathbbF_7$ with dimesion $2$. It actually suffices to find a nonzero zero divisor. Maybe that helps (but I would also make sure you understand why what I just said holds)
  $endgroup$
  – Enkidu
  Mar 18 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Enkidu I'm sorry I didn't quite catch that. Could you explain why is this? I suppose x+3 and x-3 are the nonzero zero divisor we are looking for? I don't see why (x+3) should be maximal.
  $endgroup$
  – davidh
  Mar 18 at 17:05
  

  
  
  
  

add a comment | 

0

1

$begingroup$

$F_p=Z/(pZ)$

I'm stuck at this problem. I tried to find the irreducible factorisation of $x^2+5$, but that didn't help too much. Any ideas? and also could anyone explain the general philosophy with regard to solving this kind of problem?

abstract-algebra ideals

share|cite|improve this question

edited Mar 18 at 0:16

David

69.6k668130

asked Mar 18 at 0:13

davidhdavidh

3178

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Hint: over $F_7$ we have $x^2+5=x^2-9$.
  $endgroup$
  – David
  Mar 18 at 0:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @David Okay. So my factorisation turns out to be $(x+3)(x-3)$. But this is where I seem to have stuck upon. Edit: sorry I made a silly mistake.
  $endgroup$
  – davidh
  Mar 18 at 0:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @David Should I invoke the Chinese remainder theorem here?
  $endgroup$
  – davidh
  Mar 18 at 0:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  since your ring is actually an algebra over $mathbbF_7$ with dimesion $2$. It actually suffices to find a nonzero zero divisor. Maybe that helps (but I would also make sure you understand why what I just said holds)
  $endgroup$
  – Enkidu
  Mar 18 at 8:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Enkidu I'm sorry I didn't quite catch that. Could you explain why is this? I suppose x+3 and x-3 are the nonzero zero divisor we are looking for? I don't see why (x+3) should be maximal.
  $endgroup$
  – davidh
  Mar 18 at 17:05
  

  
  
  
  

add a comment | 

$begingroup$

$F_p=Z/(pZ)$

I'm stuck at this problem. I tried to find the irreducible factorisation of $x^2+5$, but that didn't help too much. Any ideas? and also could anyone explain the general philosophy with regard to solving this kind of problem?

abstract-algebra ideals

share|cite|improve this question

edited Mar 18 at 0:16

David

69.6k668130

asked Mar 18 at 0:13

davidhdavidh

3178

$endgroup$

share|cite|improve this question

edited Mar 18 at 0:16

David

69.6k668130

asked Mar 18 at 0:13

davidhdavidh

3178

share|cite|improve this question

edited Mar 18 at 0:16

David

69.6k668130

asked Mar 18 at 0:13

davidhdavidh

3178

edited Mar 18 at 0:16

David

69.6k668130

edited Mar 18 at 0:16

David

69.6k668130

asked Mar 18 at 0:13

davidhdavidh

3178

asked Mar 18 at 0:13

davidhdavidh

3178

1 Answer
1

active

oldest

votes

0

$begingroup$

sorry, but the further explanation in the comments would actually give the solution, so I just write it here.]

We know for that $mathbbF_7[x]/(x^2+5)$ some things:

1.$(x^2+5)=(x^2-9)=(x-3)(x+3)$

2.$mathbbF_7[x]/(x^2+5)cong mathbbF_7 oplus [x]mathbbF_7$ as vectorspaces.

3.The above isomorphism makes $mathbbF_7[x]/(x^2+5)$ into a $mathbbF_7$ algebra.

Now we get that any ideal is a sub-$mathbbF_7$-vectorspace of $mathbbF_7[x]/(x^2+5)$ not containing $1$ in particular it cannot contain any element of $mathbbF_7 hookrightarrow mathbbF_7[x]/(x^2+5)$. Now for dimensionreasons this means that our maximal ideal has maximal dimension one, which means that any nontrivial ideal is already maximal, so it suffices to find a nontrivial nonunit element and take the ideal generated by it. So a nontrivial zero-devisor has to do the trick of which we clearly already have 2 obvious possible choices: $(x-3)$ and $(x+3)$.

And we are done, actually no proper algebra involved, only linear algebra.

share|cite|improve this answer

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419

$endgroup$

add a comment | 

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0

$begingroup$

sorry, but the further explanation in the comments would actually give the solution, so I just write it here.]

We know for that $mathbbF_7[x]/(x^2+5)$ some things:

1.$(x^2+5)=(x^2-9)=(x-3)(x+3)$

2.$mathbbF_7[x]/(x^2+5)cong mathbbF_7 oplus [x]mathbbF_7$ as vectorspaces.

3.The above isomorphism makes $mathbbF_7[x]/(x^2+5)$ into a $mathbbF_7$ algebra.

Now we get that any ideal is a sub-$mathbbF_7$-vectorspace of $mathbbF_7[x]/(x^2+5)$ not containing $1$ in particular it cannot contain any element of $mathbbF_7 hookrightarrow mathbbF_7[x]/(x^2+5)$. Now for dimensionreasons this means that our maximal ideal has maximal dimension one, which means that any nontrivial ideal is already maximal, so it suffices to find a nontrivial nonunit element and take the ideal generated by it. So a nontrivial zero-devisor has to do the trick of which we clearly already have 2 obvious possible choices: $(x-3)$ and $(x+3)$.

And we are done, actually no proper algebra involved, only linear algebra.

share|cite|improve this answer

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419

$endgroup$

add a comment | 

0

$begingroup$

sorry, but the further explanation in the comments would actually give the solution, so I just write it here.]

We know for that $mathbbF_7[x]/(x^2+5)$ some things:

1.$(x^2+5)=(x^2-9)=(x-3)(x+3)$

2.$mathbbF_7[x]/(x^2+5)cong mathbbF_7 oplus [x]mathbbF_7$ as vectorspaces.

3.The above isomorphism makes $mathbbF_7[x]/(x^2+5)$ into a $mathbbF_7$ algebra.

Now we get that any ideal is a sub-$mathbbF_7$-vectorspace of $mathbbF_7[x]/(x^2+5)$ not containing $1$ in particular it cannot contain any element of $mathbbF_7 hookrightarrow mathbbF_7[x]/(x^2+5)$. Now for dimensionreasons this means that our maximal ideal has maximal dimension one, which means that any nontrivial ideal is already maximal, so it suffices to find a nontrivial nonunit element and take the ideal generated by it. So a nontrivial zero-devisor has to do the trick of which we clearly already have 2 obvious possible choices: $(x-3)$ and $(x+3)$.

And we are done, actually no proper algebra involved, only linear algebra.

share|cite|improve this answer

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419

$endgroup$

add a comment | 

$begingroup$

sorry, but the further explanation in the comments would actually give the solution, so I just write it here.]

We know for that $mathbbF_7[x]/(x^2+5)$ some things:

1.$(x^2+5)=(x^2-9)=(x-3)(x+3)$

2.$mathbbF_7[x]/(x^2+5)cong mathbbF_7 oplus [x]mathbbF_7$ as vectorspaces.

3.The above isomorphism makes $mathbbF_7[x]/(x^2+5)$ into a $mathbbF_7$ algebra.

Now we get that any ideal is a sub-$mathbbF_7$-vectorspace of $mathbbF_7[x]/(x^2+5)$ not containing $1$ in particular it cannot contain any element of $mathbbF_7 hookrightarrow mathbbF_7[x]/(x^2+5)$. Now for dimensionreasons this means that our maximal ideal has maximal dimension one, which means that any nontrivial ideal is already maximal, so it suffices to find a nontrivial nonunit element and take the ideal generated by it. So a nontrivial zero-devisor has to do the trick of which we clearly already have 2 obvious possible choices: $(x-3)$ and $(x+3)$.

And we are done, actually no proper algebra involved, only linear algebra.

share|cite|improve this answer

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419

$endgroup$

share|cite|improve this answer

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419

answered Mar 19 at 14:00

EnkiduEnkidu

1,44419