Why is number of real characters mod $q$ a multiplicative function?












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Let $R(q)$ be the number of real characters mod $q$. A character $chi mod q$ is called real if $chi(a)inmathbb{R}$ for every $ain mathbb{Z}$, which means $chi(a)in{-1,1}$ for every $ainmathbb{Z}$ with gcd$(a,q)=1$.



I want to show that this $R$ is multiplicative, so $R(ab)=R(a)R(b)$ for $a,binmathbb{Z}$ with gcd$(a,b)=1$. I'm trying to prove this with induced characters, but I'm not getting it completely. Can someone help me to understand it? Thanks!










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  • $begingroup$
    Do you mean number of irreducible real characters? Also, how would induced characters help?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:37












  • $begingroup$
    @A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:43












  • $begingroup$
    There are infinitely many real characters $pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters.
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:45










  • $begingroup$
    @A.Pongrácz I think we only consider irreducible characters then in my course! :)
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:47










  • $begingroup$
    It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:47
















0












$begingroup$


Let $R(q)$ be the number of real characters mod $q$. A character $chi mod q$ is called real if $chi(a)inmathbb{R}$ for every $ain mathbb{Z}$, which means $chi(a)in{-1,1}$ for every $ainmathbb{Z}$ with gcd$(a,q)=1$.



I want to show that this $R$ is multiplicative, so $R(ab)=R(a)R(b)$ for $a,binmathbb{Z}$ with gcd$(a,b)=1$. I'm trying to prove this with induced characters, but I'm not getting it completely. Can someone help me to understand it? Thanks!










share|cite|improve this question









$endgroup$












  • $begingroup$
    Do you mean number of irreducible real characters? Also, how would induced characters help?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:37












  • $begingroup$
    @A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:43












  • $begingroup$
    There are infinitely many real characters $pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters.
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:45










  • $begingroup$
    @A.Pongrácz I think we only consider irreducible characters then in my course! :)
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:47










  • $begingroup$
    It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:47














0












0








0





$begingroup$


Let $R(q)$ be the number of real characters mod $q$. A character $chi mod q$ is called real if $chi(a)inmathbb{R}$ for every $ain mathbb{Z}$, which means $chi(a)in{-1,1}$ for every $ainmathbb{Z}$ with gcd$(a,q)=1$.



I want to show that this $R$ is multiplicative, so $R(ab)=R(a)R(b)$ for $a,binmathbb{Z}$ with gcd$(a,b)=1$. I'm trying to prove this with induced characters, but I'm not getting it completely. Can someone help me to understand it? Thanks!










share|cite|improve this question









$endgroup$




Let $R(q)$ be the number of real characters mod $q$. A character $chi mod q$ is called real if $chi(a)inmathbb{R}$ for every $ain mathbb{Z}$, which means $chi(a)in{-1,1}$ for every $ainmathbb{Z}$ with gcd$(a,q)=1$.



I want to show that this $R$ is multiplicative, so $R(ab)=R(a)R(b)$ for $a,binmathbb{Z}$ with gcd$(a,b)=1$. I'm trying to prove this with induced characters, but I'm not getting it completely. Can someone help me to understand it? Thanks!







analytic-number-theory characters






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asked Dec 30 '18 at 20:28









jbuser430jbuser430

365110




365110












  • $begingroup$
    Do you mean number of irreducible real characters? Also, how would induced characters help?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:37












  • $begingroup$
    @A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:43












  • $begingroup$
    There are infinitely many real characters $pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters.
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:45










  • $begingroup$
    @A.Pongrácz I think we only consider irreducible characters then in my course! :)
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:47










  • $begingroup$
    It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:47


















  • $begingroup$
    Do you mean number of irreducible real characters? Also, how would induced characters help?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:37












  • $begingroup$
    @A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:43












  • $begingroup$
    There are infinitely many real characters $pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters.
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:45










  • $begingroup$
    @A.Pongrácz I think we only consider irreducible characters then in my course! :)
    $endgroup$
    – jbuser430
    Dec 30 '18 at 20:47










  • $begingroup$
    It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$?
    $endgroup$
    – A. Pongrácz
    Dec 30 '18 at 20:47
















$begingroup$
Do you mean number of irreducible real characters? Also, how would induced characters help?
$endgroup$
– A. Pongrácz
Dec 30 '18 at 20:37






$begingroup$
Do you mean number of irreducible real characters? Also, how would induced characters help?
$endgroup$
– A. Pongrácz
Dec 30 '18 at 20:37














$begingroup$
@A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help
$endgroup$
– jbuser430
Dec 30 '18 at 20:43






$begingroup$
@A.Pongrácz not specifically irreducible.. Also, I thought, characters mod $ab$ might be induced by characters mod $a$ and mod $b$? Not sure if this could help
$endgroup$
– jbuser430
Dec 30 '18 at 20:43














$begingroup$
There are infinitely many real characters $pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters.
$endgroup$
– A. Pongrácz
Dec 30 '18 at 20:45




$begingroup$
There are infinitely many real characters $pmod q$ for every $q$. E.g., any positive integer multiple of the trivial character is a real character. (Check out the definition of a character again.) I believe you want to talk about irreducible characters.
$endgroup$
– A. Pongrácz
Dec 30 '18 at 20:45












$begingroup$
@A.Pongrácz I think we only consider irreducible characters then in my course! :)
$endgroup$
– jbuser430
Dec 30 '18 at 20:47




$begingroup$
@A.Pongrácz I think we only consider irreducible characters then in my course! :)
$endgroup$
– jbuser430
Dec 30 '18 at 20:47












$begingroup$
It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$?
$endgroup$
– A. Pongrácz
Dec 30 '18 at 20:47




$begingroup$
It seems to me that you are also using the word "induced character" in a loose sense. Mind that it has a meaning. How would you "induce" a(n irreducible!) character of $ab$ from a(n irreducible!) character of $a$ and one of $b$?
$endgroup$
– A. Pongrácz
Dec 30 '18 at 20:47










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