A polynomial that is reducible under every finite field?
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Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.
Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.
I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.
Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!
abstract-algebra ring-theory field-theory finite-fields extension-field
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add a comment |
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Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.
Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.
I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.
Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!
abstract-algebra ring-theory field-theory finite-fields extension-field
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4
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You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
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– Ethan Alwaise
Dec 9 '16 at 7:50
4
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@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
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– Alex Wertheim
Dec 9 '16 at 7:57
1
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Ah right, good point.
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– Ethan Alwaise
Dec 9 '16 at 19:23
add a comment |
$begingroup$
Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.
Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.
I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.
Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!
abstract-algebra ring-theory field-theory finite-fields extension-field
$endgroup$
Q: Prove that for any finite field $mathbb{F_q}$, the ring $mathbb{F_q}[x]/(x^9+x^5+x^3+x+1)$ cannot be a field.
Upon first glance I am really not sure where to start. Intuitively, it seems that I should be finding a way to show that $x^9+x^5+x^3+x+1$ is reducible over every finite field.
I have that:
$x^9+x^5+x^3+x+1 = (x^2-x+1)(x^7+x^6-x^4+x^2+2x+1)$ but I don't see where to go from here.
Am I even on the right track? Is there some well-known theorem I've blanked on that would help me out? Thanks in advance!
abstract-algebra ring-theory field-theory finite-fields extension-field
abstract-algebra ring-theory field-theory finite-fields extension-field
asked Dec 9 '16 at 7:45
Jane DoeJane Doe
19112
19112
4
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You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
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– Ethan Alwaise
Dec 9 '16 at 7:50
4
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@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
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– Alex Wertheim
Dec 9 '16 at 7:57
1
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Ah right, good point.
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– Ethan Alwaise
Dec 9 '16 at 19:23
add a comment |
4
$begingroup$
You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 7:50
4
$begingroup$
@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
$endgroup$
– Alex Wertheim
Dec 9 '16 at 7:57
1
$begingroup$
Ah right, good point.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 19:23
4
4
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You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 7:50
$begingroup$
You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 7:50
4
4
$begingroup$
@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
$endgroup$
– Alex Wertheim
Dec 9 '16 at 7:57
$begingroup$
@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
$endgroup$
– Alex Wertheim
Dec 9 '16 at 7:57
1
1
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Ah right, good point.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 19:23
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Ah right, good point.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 19:23
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1 Answer
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From the comments above.
If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.
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From the comments above.
If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.
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add a comment |
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From the comments above.
If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.
$endgroup$
add a comment |
$begingroup$
From the comments above.
If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.
$endgroup$
From the comments above.
If $f(x) = x^9 + x^5 + x^3 + x + 1 in mathbb{Z}[x]$, then you have shown that $f = gh$, where $g(x) = x^2 - x + 1 in mathbb{Z}[x]$ and $h(x) = x^7 + x^6 - x^4 + x^2 + 2x + 1 in mathbb{Z}[x]$. Let $p$ be a prime and $bar{f}$, $bar{g}$ and $bar{h}$ denote the images of $f$, $g$ and $h$, respectively, under the canonical map $mathbb{Z} to mathbb{F}_p$. Then, $bar{f} = bar{g} bar{h}$ and $bar{g}$ and $bar{h}$ are nonconstant polynomials in $mathbb{F}_p[x]$. Hence, $bar{f}$ is irreducible over every finite field, as was needed to be shown.
answered Nov 30 '18 at 10:44
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Brahadeesh
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You just factored it over $mathbb{Z}$. You are done. If it's reducible over $mathbb{Z}$ then it's reducible over $mathbb{F}_p$ for every prime $p$.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 7:50
4
$begingroup$
@EthanAlwaise: well, one must be a little careful: $3X^{2}+X$ is reducible as $X(3X+1)$ over $mathbb{Z}$, but is irreducible over $mathbb{F}_{3}$. But since both of the factors in the product above are monic, there are no issues of this kind.
$endgroup$
– Alex Wertheim
Dec 9 '16 at 7:57
1
$begingroup$
Ah right, good point.
$endgroup$
– Ethan Alwaise
Dec 9 '16 at 19:23