Mixing
Proving a polynomial is irreducible over a finite field
$begingroup$
Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.
Proof.
Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$
So by the quadratic formula we have
$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$
However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$
Therefore $m(X)$ is irreducible.
Would this be correct?
abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials
edited Oct 7 '16 at 13:53
user26857
39.4k124183
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
$endgroup$
add a comment |
$begingroup$
Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.
Proof.
Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$
So by the quadratic formula we have
$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$
However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$
Therefore $m(X)$ is irreducible.
Would this be correct?
abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials
edited Oct 7 '16 at 13:53
user26857
39.4k124183
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
$endgroup$
2
$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52
1
$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59
2
$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03
$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12
$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20
add a comment |
$begingroup$
Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.
Proof.
Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$
So by the quadratic formula we have
$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$
However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$
Therefore $m(X)$ is irreducible.
Would this be correct?
abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials
edited Oct 7 '16 at 13:53
user26857
39.4k124183
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
$endgroup$
Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.
Proof.
Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$
So by the quadratic formula we have
$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$
However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$
Therefore $m(X)$ is irreducible.
Would this be correct?
abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials
abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials
edited Oct 7 '16 at 13:53
user26857
39.4k124183
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
edited Oct 7 '16 at 13:53
user26857
39.4k124183
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
edited Oct 7 '16 at 13:53
user26857
39.4k124183
edited Oct 7 '16 at 13:53
user26857
39.4k124183
edited Oct 7 '16 at 13:53
user26857
39.4k124183
39.4k124183
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
asked Oct 7 '16 at 10:43
aoshdosiaoshdosi
111
111
2
$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52
1
$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59
2
$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03
$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12
$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20
add a comment |
2
$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52
1
$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59
2
$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03
$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12
$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20
2
2
$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52
$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52
1
1
$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59
$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59
2
2
$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03
$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03
$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12
$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12
$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20
$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20
add a comment |
1 Answer
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$begingroup$
A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.
Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
1
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
add a comment |
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1 Answer
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A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.
Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
1
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
add a comment |
$begingroup$
A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.
Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
1
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
add a comment |
$begingroup$
A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.
Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
$endgroup$
A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.
Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
answered Oct 7 '16 at 11:09
Jack D'AurizioJack D'Aurizio
291k33284667
291k33284667
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
1
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
add a comment |
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
1
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13
1
1
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16
add a comment |
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2
$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52
1
$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59
2
$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03
$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12
$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20