Question regarding algebraic closure of$ mathbb{F_2}$












0












$begingroup$



Let $mathbb{F_2} $ be the finite field of order $2$. Then
which of the following statements are true?



$1.$ $mathbb{F_2} [x]$ has only finitely many irreducible
elements.



$2.$ $mathbb{F_2} [x]$ has exactly one irreducible polynomial
of degree $2.$



$3.$$mathbb{F_2} [x]/<x^2+1>$
is a finite dimensional
vector space over .



$4.$ Any irreducible polynomial in $mathbb{F_2} [x]$ of
degree $5$ has distinct roots in any algebraic
closure of $mathbb{F_2}$




My attempt:




  • option 1) will True take $f(x) = x^2+x+1$


  • option $2)$ will false because number irreducible polynomail of degree $2 = frac{p^2-p}{p-1} = 2$


  • option $3)$ will true dimension will be $4$


  • option $4$ i don't have any hints to tackle this option



any hints/solution will be appreciated



thanks u






abstract-algebra finite-fields







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How does your argument prove that 1) is true?
    $endgroup$
    – Arthur
    Dec 11 '18 at 15:50










  • $begingroup$
    @Arthur there is one irreducible elment that is $ x^2 +x+ 1$ in $mathbb{Z_2}$
    $endgroup$
    – jasmine
    Dec 11 '18 at 15:50








  • 6




    $begingroup$
    Q: True or false: there are only finitely many natural numbers. A: True, $1$ is a natural number. Q: How does that prove that there are only finitely many natural numbers? A: There is one natural number, $1$. Q: What about $2$, and $3$, and $4$, and...
    $endgroup$
    – Alex Kruckman
    Dec 11 '18 at 15:53






  • 2




    $begingroup$
    Can you show that $mathbb F_{2^n}$ is a simple extension of $mathbb F_{2}$ for each $n$? Then, the minimal polynomial of the generator must be irreducible and of order $n$, so we have, for each natural number, an irreducible polynomial over $mathbb F_2$ of that degree. Hence, we get infinitely many irreducible polynomials over $mathbb F_2$.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:09






  • 1




    $begingroup$
    @jasmine You are welcome. Also, the answer below takes care of everything else.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:22
















0












$begingroup$



Let $mathbb{F_2} $ be the finite field of order $2$. Then
which of the following statements are true?



$1.$ $mathbb{F_2} [x]$ has only finitely many irreducible
elements.



$2.$ $mathbb{F_2} [x]$ has exactly one irreducible polynomial
of degree $2.$



$3.$$mathbb{F_2} [x]/<x^2+1>$
is a finite dimensional
vector space over .



$4.$ Any irreducible polynomial in $mathbb{F_2} [x]$ of
degree $5$ has distinct roots in any algebraic
closure of $mathbb{F_2}$




My attempt:




  • option 1) will True take $f(x) = x^2+x+1$


  • option $2)$ will false because number irreducible polynomail of degree $2 = frac{p^2-p}{p-1} = 2$


  • option $3)$ will true dimension will be $4$


  • option $4$ i don't have any hints to tackle this option



any hints/solution will be appreciated



thanks u






abstract-algebra finite-fields







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    How does your argument prove that 1) is true?
    $endgroup$
    – Arthur
    Dec 11 '18 at 15:50










  • $begingroup$
    @Arthur there is one irreducible elment that is $ x^2 +x+ 1$ in $mathbb{Z_2}$
    $endgroup$
    – jasmine
    Dec 11 '18 at 15:50








  • 6




    $begingroup$
    Q: True or false: there are only finitely many natural numbers. A: True, $1$ is a natural number. Q: How does that prove that there are only finitely many natural numbers? A: There is one natural number, $1$. Q: What about $2$, and $3$, and $4$, and...
    $endgroup$
    – Alex Kruckman
    Dec 11 '18 at 15:53






  • 2




    $begingroup$
    Can you show that $mathbb F_{2^n}$ is a simple extension of $mathbb F_{2}$ for each $n$? Then, the minimal polynomial of the generator must be irreducible and of order $n$, so we have, for each natural number, an irreducible polynomial over $mathbb F_2$ of that degree. Hence, we get infinitely many irreducible polynomials over $mathbb F_2$.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:09






  • 1




    $begingroup$
    @jasmine You are welcome. Also, the answer below takes care of everything else.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:22














0












0








0





$begingroup$



Let $mathbb{F_2} $ be the finite field of order $2$. Then
which of the following statements are true?



$1.$ $mathbb{F_2} [x]$ has only finitely many irreducible
elements.



$2.$ $mathbb{F_2} [x]$ has exactly one irreducible polynomial
of degree $2.$



$3.$$mathbb{F_2} [x]/<x^2+1>$
is a finite dimensional
vector space over .



$4.$ Any irreducible polynomial in $mathbb{F_2} [x]$ of
degree $5$ has distinct roots in any algebraic
closure of $mathbb{F_2}$




My attempt:




  • option 1) will True take $f(x) = x^2+x+1$


  • option $2)$ will false because number irreducible polynomail of degree $2 = frac{p^2-p}{p-1} = 2$


  • option $3)$ will true dimension will be $4$


  • option $4$ i don't have any hints to tackle this option



any hints/solution will be appreciated



thanks u






abstract-algebra finite-fields







share|cite|improve this question











$endgroup$





Let $mathbb{F_2} $ be the finite field of order $2$. Then
which of the following statements are true?



$1.$ $mathbb{F_2} [x]$ has only finitely many irreducible
elements.



$2.$ $mathbb{F_2} [x]$ has exactly one irreducible polynomial
of degree $2.$



$3.$$mathbb{F_2} [x]/<x^2+1>$
is a finite dimensional
vector space over .



$4.$ Any irreducible polynomial in $mathbb{F_2} [x]$ of
degree $5$ has distinct roots in any algebraic
closure of $mathbb{F_2}$




My attempt:




  • option 1) will True take $f(x) = x^2+x+1$


  • option $2)$ will false because number irreducible polynomail of degree $2 = frac{p^2-p}{p-1} = 2$


  • option $3)$ will true dimension will be $4$


  • option $4$ i don't have any hints to tackle this option



any hints/solution will be appreciated



thanks u






abstract-algebra finite-fields




abstract-algebra finite-fields






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 6:17







jasmine

















asked Dec 11 '18 at 15:48









jasminejasmine

1,704417




1,704417








  • 2




    $begingroup$
    How does your argument prove that 1) is true?
    $endgroup$
    – Arthur
    Dec 11 '18 at 15:50










  • $begingroup$
    @Arthur there is one irreducible elment that is $ x^2 +x+ 1$ in $mathbb{Z_2}$
    $endgroup$
    – jasmine
    Dec 11 '18 at 15:50








  • 6




    $begingroup$
    Q: True or false: there are only finitely many natural numbers. A: True, $1$ is a natural number. Q: How does that prove that there are only finitely many natural numbers? A: There is one natural number, $1$. Q: What about $2$, and $3$, and $4$, and...
    $endgroup$
    – Alex Kruckman
    Dec 11 '18 at 15:53






  • 2




    $begingroup$
    Can you show that $mathbb F_{2^n}$ is a simple extension of $mathbb F_{2}$ for each $n$? Then, the minimal polynomial of the generator must be irreducible and of order $n$, so we have, for each natural number, an irreducible polynomial over $mathbb F_2$ of that degree. Hence, we get infinitely many irreducible polynomials over $mathbb F_2$.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:09






  • 1




    $begingroup$
    @jasmine You are welcome. Also, the answer below takes care of everything else.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:22














  • 2




    $begingroup$
    How does your argument prove that 1) is true?
    $endgroup$
    – Arthur
    Dec 11 '18 at 15:50










  • $begingroup$
    @Arthur there is one irreducible elment that is $ x^2 +x+ 1$ in $mathbb{Z_2}$
    $endgroup$
    – jasmine
    Dec 11 '18 at 15:50








  • 6




    $begingroup$
    Q: True or false: there are only finitely many natural numbers. A: True, $1$ is a natural number. Q: How does that prove that there are only finitely many natural numbers? A: There is one natural number, $1$. Q: What about $2$, and $3$, and $4$, and...
    $endgroup$
    – Alex Kruckman
    Dec 11 '18 at 15:53






  • 2




    $begingroup$
    Can you show that $mathbb F_{2^n}$ is a simple extension of $mathbb F_{2}$ for each $n$? Then, the minimal polynomial of the generator must be irreducible and of order $n$, so we have, for each natural number, an irreducible polynomial over $mathbb F_2$ of that degree. Hence, we get infinitely many irreducible polynomials over $mathbb F_2$.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:09






  • 1




    $begingroup$
    @jasmine You are welcome. Also, the answer below takes care of everything else.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 11 '18 at 16:22








2




2




$begingroup$
How does your argument prove that 1) is true?
$endgroup$
– Arthur
Dec 11 '18 at 15:50




$begingroup$
How does your argument prove that 1) is true?
$endgroup$
– Arthur
Dec 11 '18 at 15:50












$begingroup$
@Arthur there is one irreducible elment that is $ x^2 +x+ 1$ in $mathbb{Z_2}$
$endgroup$
– jasmine
Dec 11 '18 at 15:50






$begingroup$
@Arthur there is one irreducible elment that is $ x^2 +x+ 1$ in $mathbb{Z_2}$
$endgroup$
– jasmine
Dec 11 '18 at 15:50






6




6




$begingroup$
Q: True or false: there are only finitely many natural numbers. A: True, $1$ is a natural number. Q: How does that prove that there are only finitely many natural numbers? A: There is one natural number, $1$. Q: What about $2$, and $3$, and $4$, and...
$endgroup$
– Alex Kruckman
Dec 11 '18 at 15:53




$begingroup$
Q: True or false: there are only finitely many natural numbers. A: True, $1$ is a natural number. Q: How does that prove that there are only finitely many natural numbers? A: There is one natural number, $1$. Q: What about $2$, and $3$, and $4$, and...
$endgroup$
– Alex Kruckman
Dec 11 '18 at 15:53




2




2




$begingroup$
Can you show that $mathbb F_{2^n}$ is a simple extension of $mathbb F_{2}$ for each $n$? Then, the minimal polynomial of the generator must be irreducible and of order $n$, so we have, for each natural number, an irreducible polynomial over $mathbb F_2$ of that degree. Hence, we get infinitely many irreducible polynomials over $mathbb F_2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 11 '18 at 16:09




$begingroup$
Can you show that $mathbb F_{2^n}$ is a simple extension of $mathbb F_{2}$ for each $n$? Then, the minimal polynomial of the generator must be irreducible and of order $n$, so we have, for each natural number, an irreducible polynomial over $mathbb F_2$ of that degree. Hence, we get infinitely many irreducible polynomials over $mathbb F_2$.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 11 '18 at 16:09




1




1




$begingroup$
@jasmine You are welcome. Also, the answer below takes care of everything else.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 11 '18 at 16:22




$begingroup$
@jasmine You are welcome. Also, the answer below takes care of everything else.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 11 '18 at 16:22










1 Answer
1






active

oldest

votes


















2












$begingroup$

$mathbb{F}_2[x]$ has an irreducible element of degree $n$ for every $n$, so certainly infinitely many.



But it only has one of degree $2$: there are only $4$ candidates to check and only $x^2 +x +1$ is irreducible.



$mathbb{F}_2[x]/(x^2+1)$ is essentially the set of linear polynomials (we identify $x^2$ with $1$), which is a vector space of dimension $2$ over $mathbb{F}_2$.



For 4. yes, see this question and answer






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks u @Henno sir can u give any hints of option $4$
    $endgroup$
    – jasmine
    Dec 11 '18 at 16:15













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

$mathbb{F}_2[x]$ has an irreducible element of degree $n$ for every $n$, so certainly infinitely many.



But it only has one of degree $2$: there are only $4$ candidates to check and only $x^2 +x +1$ is irreducible.



$mathbb{F}_2[x]/(x^2+1)$ is essentially the set of linear polynomials (we identify $x^2$ with $1$), which is a vector space of dimension $2$ over $mathbb{F}_2$.



For 4. yes, see this question and answer






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks u @Henno sir can u give any hints of option $4$
    $endgroup$
    – jasmine
    Dec 11 '18 at 16:15


















2












$begingroup$

$mathbb{F}_2[x]$ has an irreducible element of degree $n$ for every $n$, so certainly infinitely many.



But it only has one of degree $2$: there are only $4$ candidates to check and only $x^2 +x +1$ is irreducible.



$mathbb{F}_2[x]/(x^2+1)$ is essentially the set of linear polynomials (we identify $x^2$ with $1$), which is a vector space of dimension $2$ over $mathbb{F}_2$.



For 4. yes, see this question and answer






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks u @Henno sir can u give any hints of option $4$
    $endgroup$
    – jasmine
    Dec 11 '18 at 16:15
















2












2








2





$begingroup$

$mathbb{F}_2[x]$ has an irreducible element of degree $n$ for every $n$, so certainly infinitely many.



But it only has one of degree $2$: there are only $4$ candidates to check and only $x^2 +x +1$ is irreducible.



$mathbb{F}_2[x]/(x^2+1)$ is essentially the set of linear polynomials (we identify $x^2$ with $1$), which is a vector space of dimension $2$ over $mathbb{F}_2$.



For 4. yes, see this question and answer






share|cite|improve this answer











$endgroup$



$mathbb{F}_2[x]$ has an irreducible element of degree $n$ for every $n$, so certainly infinitely many.



But it only has one of degree $2$: there are only $4$ candidates to check and only $x^2 +x +1$ is irreducible.



$mathbb{F}_2[x]/(x^2+1)$ is essentially the set of linear polynomials (we identify $x^2$ with $1$), which is a vector space of dimension $2$ over $mathbb{F}_2$.



For 4. yes, see this question and answer







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 16:18

























answered Dec 11 '18 at 16:10









Henno BrandsmaHenno Brandsma

108k347114




108k347114












  • $begingroup$
    thanks u @Henno sir can u give any hints of option $4$
    $endgroup$
    – jasmine
    Dec 11 '18 at 16:15




















  • $begingroup$
    thanks u @Henno sir can u give any hints of option $4$
    $endgroup$
    – jasmine
    Dec 11 '18 at 16:15


















$begingroup$
thanks u @Henno sir can u give any hints of option $4$
$endgroup$
– jasmine
Dec 11 '18 at 16:15






$begingroup$
thanks u @Henno sir can u give any hints of option $4$
$endgroup$
– jasmine
Dec 11 '18 at 16:15




















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