Mod $p$ of irreducible representation is irreducible for large $p$?
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Let $G$ be a finite group, and a homomophism $f: G rightarrow GL_n(mathbb Z)$. Assume $f otimes mathbb Q: G rightarrow GL_n(mathbb Z) hookrightarrow GL_n(mathbb Q)$ is an irreducible $mathbb Q$-representation of $G$, must $f otimes mathbb F_p: G rightarrow GL_n(mathbb Z) rightarrow GL_n(mathbb F_p)$ is an irreducible $mathbb F_p$-representation of $G$ for large enough $p$?
If $f otimes mathbb Q$ is absolute irreducible, one can show this by character theory.
group-theory representation-theory
edited Nov 18 at 0:06
Bernard
116k637108
asked Nov 18 at 0:00
zzy
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up vote
3
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Let $G$ be a finite group, and a homomophism $f: G rightarrow GL_n(mathbb Z)$. Assume $f otimes mathbb Q: G rightarrow GL_n(mathbb Z) hookrightarrow GL_n(mathbb Q)$ is an irreducible $mathbb Q$-representation of $G$, must $f otimes mathbb F_p: G rightarrow GL_n(mathbb Z) rightarrow GL_n(mathbb F_p)$ is an irreducible $mathbb F_p$-representation of $G$ for large enough $p$?
If $f otimes mathbb Q$ is absolute irreducible, one can show this by character theory.
group-theory representation-theory
edited Nov 18 at 0:06
Bernard
116k637108
asked Nov 18 at 0:00
zzy
2,2231319
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $G$ be a finite group, and a homomophism $f: G rightarrow GL_n(mathbb Z)$. Assume $f otimes mathbb Q: G rightarrow GL_n(mathbb Z) hookrightarrow GL_n(mathbb Q)$ is an irreducible $mathbb Q$-representation of $G$, must $f otimes mathbb F_p: G rightarrow GL_n(mathbb Z) rightarrow GL_n(mathbb F_p)$ is an irreducible $mathbb F_p$-representation of $G$ for large enough $p$?
If $f otimes mathbb Q$ is absolute irreducible, one can show this by character theory.
group-theory representation-theory
edited Nov 18 at 0:06
Bernard
116k637108
asked Nov 18 at 0:00
zzy
2,2231319
Let $G$ be a finite group, and a homomophism $f: G rightarrow GL_n(mathbb Z)$. Assume $f otimes mathbb Q: G rightarrow GL_n(mathbb Z) hookrightarrow GL_n(mathbb Q)$ is an irreducible $mathbb Q$-representation of $G$, must $f otimes mathbb F_p: G rightarrow GL_n(mathbb Z) rightarrow GL_n(mathbb F_p)$ is an irreducible $mathbb F_p$-representation of $G$ for large enough $p$?
If $f otimes mathbb Q$ is absolute irreducible, one can show this by character theory.
group-theory representation-theory
group-theory representation-theory
edited Nov 18 at 0:06
Bernard
116k637108
asked Nov 18 at 0:00
zzy
2,2231319
edited Nov 18 at 0:06
Bernard
116k637108
asked Nov 18 at 0:00
zzy
2,2231319
edited Nov 18 at 0:06
Bernard
116k637108
edited Nov 18 at 0:06
Bernard
116k637108
edited Nov 18 at 0:06
Bernard
116k637108
116k637108
asked Nov 18 at 0:00
zzy
2,2231319
asked Nov 18 at 0:00
zzy
2,2231319
asked Nov 18 at 0:00
zzy
2,2231319
2,2231319
add a comment |
add a comment |
1 Answer
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No. For instance, let $G$ be cyclic of order $4$ and let $f:Gto GL_2(mathbb{Z})$ send a generator to $begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$. Then for any field $K$, $fotimes K$ is reducible iff this matrix has an eigenvalue in $K$, i.e. iff $-1$ has a square root in $K$. So, $fotimesmathbb{Q}$ is irreducible, but $fotimesmathbb{F}_p$ is reducible for any prime $p$ that is $1$ mod $4$, and there are infinitely many such $p$.
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
No. For instance, let $G$ be cyclic of order $4$ and let $f:Gto GL_2(mathbb{Z})$ send a generator to $begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$. Then for any field $K$, $fotimes K$ is reducible iff this matrix has an eigenvalue in $K$, i.e. iff $-1$ has a square root in $K$. So, $fotimesmathbb{Q}$ is irreducible, but $fotimesmathbb{F}_p$ is reducible for any prime $p$ that is $1$ mod $4$, and there are infinitely many such $p$.
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
add a comment |
up vote
3
down vote
accepted
No. For instance, let $G$ be cyclic of order $4$ and let $f:Gto GL_2(mathbb{Z})$ send a generator to $begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$. Then for any field $K$, $fotimes K$ is reducible iff this matrix has an eigenvalue in $K$, i.e. iff $-1$ has a square root in $K$. So, $fotimesmathbb{Q}$ is irreducible, but $fotimesmathbb{F}_p$ is reducible for any prime $p$ that is $1$ mod $4$, and there are infinitely many such $p$.
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
No. For instance, let $G$ be cyclic of order $4$ and let $f:Gto GL_2(mathbb{Z})$ send a generator to $begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$. Then for any field $K$, $fotimes K$ is reducible iff this matrix has an eigenvalue in $K$, i.e. iff $-1$ has a square root in $K$. So, $fotimesmathbb{Q}$ is irreducible, but $fotimesmathbb{F}_p$ is reducible for any prime $p$ that is $1$ mod $4$, and there are infinitely many such $p$.
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
No. For instance, let $G$ be cyclic of order $4$ and let $f:Gto GL_2(mathbb{Z})$ send a generator to $begin{pmatrix} 0 & -1 \ 1 & 0 end{pmatrix}$. Then for any field $K$, $fotimes K$ is reducible iff this matrix has an eigenvalue in $K$, i.e. iff $-1$ has a square root in $K$. So, $fotimesmathbb{Q}$ is irreducible, but $fotimesmathbb{F}_p$ is reducible for any prime $p$ that is $1$ mod $4$, and there are infinitely many such $p$.
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
answered Nov 18 at 1:23
Eric Wofsey
176k12202326
176k12202326
add a comment |
add a comment |
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