$M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $d leq n.$
Source of the following Problem:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition).
Let $K$ be an extension of $F$ of degree $n.$ Prove that $K$ is isomorphic to a subfield of the ring $M_n(F),$ so $M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $d leq n.$
I proved that $K$ is isomorphic to a subfield of the ring $M_n(F)$. I stuck in the next part, though from my previous question this it is clear that whenever $d | n$ it is true. But
Is it true if $d<n$ and $d$ does not divide $n.$
linear-algebra abstract-algebra matrices ring-theory field-theory
edited Dec 29 '18 at 10:55
jmerry
2,894312
asked Dec 29 '18 at 10:40
user371231user371231
756511
add a comment |
Source of the following Problem:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition).
Let $K$ be an extension of $F$ of degree $n.$ Prove that $K$ is isomorphic to a subfield of the ring $M_n(F),$ so $M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $d leq n.$
I proved that $K$ is isomorphic to a subfield of the ring $M_n(F)$. I stuck in the next part, though from my previous question this it is clear that whenever $d | n$ it is true. But
Is it true if $d<n$ and $d$ does not divide $n.$
linear-algebra abstract-algebra matrices ring-theory field-theory
edited Dec 29 '18 at 10:55
jmerry
2,894312
asked Dec 29 '18 at 10:40
user371231user371231
756511
2
Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $dtimes d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $dtimes d$ matrices into upper left corners of $ntimes n$ matrices.
– Jyrki Lahtonen
Dec 29 '18 at 10:47
1
On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $ell$, then, when viewed as a vector space over $F$, it will have dimension $dell$. So $dell=n$ forcing $d$ to be a factor of $n$.
– Jyrki Lahtonen
Dec 29 '18 at 10:52
Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-)
– Jyrki Lahtonen
Dec 29 '18 at 10:54
add a comment |
Source of the following Problem:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition).
Let $K$ be an extension of $F$ of degree $n.$ Prove that $K$ is isomorphic to a subfield of the ring $M_n(F),$ so $M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $d leq n.$
I proved that $K$ is isomorphic to a subfield of the ring $M_n(F)$. I stuck in the next part, though from my previous question this it is clear that whenever $d | n$ it is true. But
Is it true if $d<n$ and $d$ does not divide $n.$
linear-algebra abstract-algebra matrices ring-theory field-theory
edited Dec 29 '18 at 10:55
jmerry
2,894312
asked Dec 29 '18 at 10:40
user371231user371231
756511
Source of the following Problem:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition).
Let $K$ be an extension of $F$ of degree $n.$ Prove that $K$ is isomorphic to a subfield of the ring $M_n(F),$ so $M_n(F)$ contains an isomorphic copy of every extension of $F$ of degree $d leq n.$
I proved that $K$ is isomorphic to a subfield of the ring $M_n(F)$. I stuck in the next part, though from my previous question this it is clear that whenever $d | n$ it is true. But
Is it true if $d<n$ and $d$ does not divide $n.$
linear-algebra abstract-algebra matrices ring-theory field-theory
linear-algebra abstract-algebra matrices ring-theory field-theory
edited Dec 29 '18 at 10:55
jmerry
2,894312
asked Dec 29 '18 at 10:40
user371231user371231
756511
edited Dec 29 '18 at 10:55
jmerry
2,894312
asked Dec 29 '18 at 10:40
user371231user371231
756511
edited Dec 29 '18 at 10:55
jmerry
2,894312
edited Dec 29 '18 at 10:55
jmerry
2,894312
edited Dec 29 '18 at 10:55
jmerry
2,894312
2,894312
asked Dec 29 '18 at 10:40
user371231user371231
756511
asked Dec 29 '18 at 10:40
user371231user371231
756511
asked Dec 29 '18 at 10:40
user371231user371231
756511
756511
2
Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $dtimes d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $dtimes d$ matrices into upper left corners of $ntimes n$ matrices.
– Jyrki Lahtonen
Dec 29 '18 at 10:47
1
On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $ell$, then, when viewed as a vector space over $F$, it will have dimension $dell$. So $dell=n$ forcing $d$ to be a factor of $n$.
– Jyrki Lahtonen
Dec 29 '18 at 10:52
Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-)
– Jyrki Lahtonen
Dec 29 '18 at 10:54
add a comment |
2
Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $dtimes d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $dtimes d$ matrices into upper left corners of $ntimes n$ matrices.
– Jyrki Lahtonen
Dec 29 '18 at 10:47
1
On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $ell$, then, when viewed as a vector space over $F$, it will have dimension $dell$. So $dell=n$ forcing $d$ to be a factor of $n$.
– Jyrki Lahtonen
Dec 29 '18 at 10:52
Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-)
– Jyrki Lahtonen
Dec 29 '18 at 10:54
2
2
Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $dtimes d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $dtimes d$ matrices into upper left corners of $ntimes n$ matrices.
– Jyrki Lahtonen
Dec 29 '18 at 10:47
Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $dtimes d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $dtimes d$ matrices into upper left corners of $ntimes n$ matrices.
– Jyrki Lahtonen
Dec 29 '18 at 10:47
1
1
On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $ell$, then, when viewed as a vector space over $F$, it will have dimension $dell$. So $dell=n$ forcing $d$ to be a factor of $n$.
– Jyrki Lahtonen
Dec 29 '18 at 10:52
On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $ell$, then, when viewed as a vector space over $F$, it will have dimension $dell$. So $dell=n$ forcing $d$ to be a factor of $n$.
– Jyrki Lahtonen
Dec 29 '18 at 10:52
Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-)
– Jyrki Lahtonen
Dec 29 '18 at 10:54
Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-)
– Jyrki Lahtonen
Dec 29 '18 at 10:54
add a comment |
1 Answer
1
active
oldest
votes
That's simply false, at least if one assumes the identity of $K$ maps
to the identity of $M_n(F)$. For an extension $K/F$ of degree $d$, $K$
can only have
an $F$-embedding in $M_n(F)$ if $dmid n$.
Suppose $K$ embeds in $M_n(F)$. Then the space $C$ of column vectors of
height $n$ is a left $M_n(F)$-module, and so by restriction of scalars becomes a
vector space over $K$. If $dim_K C=m$, then $n=dim_K C=md$. Therefore
$dmid n$.
But if one admits non-unital ring homomorphisms, then $M_d(F)$ embeds
in $M_n(F)$ for $dle n$ by
$$Amapstopmatrix{A&0\0&0}$$
so the answer becomes yes if you accept this.
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
add a comment |
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That's simply false, at least if one assumes the identity of $K$ maps
to the identity of $M_n(F)$. For an extension $K/F$ of degree $d$, $K$
can only have
an $F$-embedding in $M_n(F)$ if $dmid n$.
Suppose $K$ embeds in $M_n(F)$. Then the space $C$ of column vectors of
height $n$ is a left $M_n(F)$-module, and so by restriction of scalars becomes a
vector space over $K$. If $dim_K C=m$, then $n=dim_K C=md$. Therefore
$dmid n$.
But if one admits non-unital ring homomorphisms, then $M_d(F)$ embeds
in $M_n(F)$ for $dle n$ by
$$Amapstopmatrix{A&0\0&0}$$
so the answer becomes yes if you accept this.
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
add a comment |
That's simply false, at least if one assumes the identity of $K$ maps
to the identity of $M_n(F)$. For an extension $K/F$ of degree $d$, $K$
can only have
an $F$-embedding in $M_n(F)$ if $dmid n$.
Suppose $K$ embeds in $M_n(F)$. Then the space $C$ of column vectors of
height $n$ is a left $M_n(F)$-module, and so by restriction of scalars becomes a
vector space over $K$. If $dim_K C=m$, then $n=dim_K C=md$. Therefore
$dmid n$.
But if one admits non-unital ring homomorphisms, then $M_d(F)$ embeds
in $M_n(F)$ for $dle n$ by
$$Amapstopmatrix{A&0\0&0}$$
so the answer becomes yes if you accept this.
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
add a comment |
That's simply false, at least if one assumes the identity of $K$ maps
to the identity of $M_n(F)$. For an extension $K/F$ of degree $d$, $K$
can only have
an $F$-embedding in $M_n(F)$ if $dmid n$.
Suppose $K$ embeds in $M_n(F)$. Then the space $C$ of column vectors of
height $n$ is a left $M_n(F)$-module, and so by restriction of scalars becomes a
vector space over $K$. If $dim_K C=m$, then $n=dim_K C=md$. Therefore
$dmid n$.
But if one admits non-unital ring homomorphisms, then $M_d(F)$ embeds
in $M_n(F)$ for $dle n$ by
$$Amapstopmatrix{A&0\0&0}$$
so the answer becomes yes if you accept this.
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
That's simply false, at least if one assumes the identity of $K$ maps
to the identity of $M_n(F)$. For an extension $K/F$ of degree $d$, $K$
can only have
an $F$-embedding in $M_n(F)$ if $dmid n$.
Suppose $K$ embeds in $M_n(F)$. Then the space $C$ of column vectors of
height $n$ is a left $M_n(F)$-module, and so by restriction of scalars becomes a
vector space over $K$. If $dim_K C=m$, then $n=dim_K C=md$. Therefore
$dmid n$.
But if one admits non-unital ring homomorphisms, then $M_d(F)$ embeds
in $M_n(F)$ for $dle n$ by
$$Amapstopmatrix{A&0\0&0}$$
so the answer becomes yes if you accept this.
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
edited Dec 29 '18 at 10:58
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
answered Dec 29 '18 at 10:54
Lord Shark the UnknownLord Shark the Unknown
102k959132
102k959132
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
add a comment |
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
Amen :-) ${}{}$
– Jyrki Lahtonen
Dec 29 '18 at 10:57
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@Lord Shark the Unknown: Can you explain what $C$ is ?
– user371231
Dec 29 '18 at 15:26
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@user371231 The space of $1times n$ column vectors over $F$.
– Lord Shark the Unknown
Dec 29 '18 at 15:28
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
@Lord Shark the Unknown: $n times 1$ right ? So it is clear to me that $C=F^n$ is a $K$ module. Then you assumed that $text{dim}_{K}C=m.$ I guess it will be $n=text{dim}_{F}C$ and considering the tower we get $n=md.$
– user371231
Dec 29 '18 at 15:47
add a comment |
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2
Somebody told me that for Dummit & Foote a ring does not need to have a multiplicative neutral element. Such a heretical view has stopped me from ever reading their tome. But it does save the day here. When your "subring" need not share the neutral element, then you can turn $K$ into a set of $dtimes d$ matrices over $F$. And fill those matrices with a bunch of zeros to turn those $dtimes d$ matrices into upper left corners of $ntimes n$ matrices.
– Jyrki Lahtonen
Dec 29 '18 at 10:47
1
On the other hand, if we sensibly assume that rings have a unit, shared by all the subrings, then this is possible only when $d$ is a factor of $n$. Observe that embedding $K$ into $M_n(F)$ in such a way that $1_K$ becomes $I_n$ turns the vector space $F^n$ also into a vector space over $K$. If a vector space $V$ over $K$ has dimension $ell$, then, when viewed as a vector space over $F$, it will have dimension $dell$. So $dell=n$ forcing $d$ to be a factor of $n$.
– Jyrki Lahtonen
Dec 29 '18 at 10:52
Unless, of course, Dummit & Foote also drop $1x=x$ from the list of vector space axioms. But in that case all linear algebra gets broken, and nobody would buy their book :-)
– Jyrki Lahtonen
Dec 29 '18 at 10:54