Invertible elements of $mathbb{Z}_3[x] / (x^4+x^3-1)^3$
$begingroup$
Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.
I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$
I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):
$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)
$|R|$ = $mathrm{3}^{12}$
$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)
Then:
$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$
But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?
ring-theory galois-theory ideals finite-fields irreducible-polynomials
edited Jan 17 at 20:15
user289143
1,069313
asked Jan 17 at 19:00
MattP9MattP9
304
$endgroup$
add a comment |
$begingroup$
Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.
I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$
I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):
$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)
$|R|$ = $mathrm{3}^{12}$
$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)
Then:
$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$
But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?
ring-theory galois-theory ideals finite-fields irreducible-polynomials
edited Jan 17 at 20:15
user289143
1,069313
asked Jan 17 at 19:00
MattP9MattP9
304
$endgroup$
$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17
$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20
add a comment |
$begingroup$
Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.
I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$
I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):
$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)
$|R|$ = $mathrm{3}^{12}$
$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)
Then:
$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$
But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?
ring-theory galois-theory ideals finite-fields irreducible-polynomials
edited Jan 17 at 20:15
user289143
1,069313
asked Jan 17 at 19:00
MattP9MattP9
304
$endgroup$
Let $F=mathbb{Z}/3mathbb{Z}$, $h(x)=x^4+x^3-1$, $R = F[x]/(h(x)^3)$.
I know $R$ has $4$ ideals and $1$ maximal ideal. Let $M$ be the maximal ideal $(h(x))/(h(x)^3)$
I need to find the number of invertible elements of $R$ and in order to do so I need the number of elements in $M$. I think this number is $3^8$ since we must have (I'm not sure about that):
$|R| = |R/M||M|$ ($R/M$ is the quotient additive group)
$|R|$ = $mathrm{3}^{12}$
$|R/M| = |F[x]/(h(x))| = 3^4$ (is that so?)
Then:
$mathrm{3}^{12} = 3^4 cdot |M|$ so $|M| = 3^8$
But according to the solution of the exercice, $|M| = 3^9$. Where did I go wrong?
ring-theory galois-theory ideals finite-fields irreducible-polynomials
ring-theory galois-theory ideals finite-fields irreducible-polynomials
edited Jan 17 at 20:15
user289143
1,069313
asked Jan 17 at 19:00
MattP9MattP9
304
edited Jan 17 at 20:15
user289143
1,069313
asked Jan 17 at 19:00
MattP9MattP9
304
edited Jan 17 at 20:15
user289143
1,069313
edited Jan 17 at 20:15
user289143
1,069313
edited Jan 17 at 20:15
user289143
1,069313
1,069313
asked Jan 17 at 19:00
MattP9MattP9
304
asked Jan 17 at 19:00
MattP9MattP9
304
asked Jan 17 at 19:00
MattP9MattP9
304
304
$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17
$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20
add a comment |
$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17
$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20
$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17
$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17
$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20
$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20
add a comment |
1 Answer
1
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Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
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add a comment |
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1 Answer
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$begingroup$
Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
$endgroup$
Your procedure is entirely valid; you can verify that $|R/M|=3^4$ by simply noting that $R/M$ is a vector space over $F$ with basis ${1,x,x^2,x^3}$. In fact, whatever argument you use to prove that $|R|=3^{12}$ undoubtedly also proves that $|R/M|=3^4$. Indeed it follows that $|M|=3^8$, and the solution given to you is wrong.
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
answered Jan 17 at 19:30
ServaesServaes
30.6k342101
30.6k342101
add a comment |
add a comment |
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$begingroup$
Another argument would be that $M simeq F[x] / langle h^2 rangle$ (the right-to-left map would be induced by multiplication by $h$) and the right hand side has size $3^8$ agreeing with your calculation.
$endgroup$
– Daniel Schepler
Jan 17 at 19:17
$begingroup$
I suspect the solution on the paper is wrong but, most importantly, is my procedure valid?
$endgroup$
– MattP9
Jan 17 at 19:20