Does there exist a $1-1$ ring homomorphism from $M_d(mathbb{F})$ to $M_n(mathbb{F})$ for $d <n$?












4












$begingroup$


Let $mathbb{F}$ be a field and $d,n$ be positive integers with $d <n.$ Then does there exist a injective ring homomorphism from $M_d(mathbb{F})$ to $M_n(mathbb{F})$ ? (BTW A ring map sends $1$ to $1$)



I failed to produce a $1-1$ ring map. This appears in the process of solving a field theory exercise from Dummit and Foote. For your ref. it is Problem number $19.(b)$ in sec. $13$ (Second Edition). Any help will be appreciated. Thanks.



Edited Later:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition) is stated below.




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 $leq n.$







linear-algebra abstract-algebra matrices ring-theory field-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Could you say what the Dummit and Foote exercise is?
    $endgroup$
    – Slade
    Dec 29 '18 at 8:01






  • 1




    $begingroup$
    And such a map does exist when $dmid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form.
    $endgroup$
    – Slade
    Dec 29 '18 at 8:04










  • $begingroup$
    Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:15
















4












$begingroup$


Let $mathbb{F}$ be a field and $d,n$ be positive integers with $d <n.$ Then does there exist a injective ring homomorphism from $M_d(mathbb{F})$ to $M_n(mathbb{F})$ ? (BTW A ring map sends $1$ to $1$)



I failed to produce a $1-1$ ring map. This appears in the process of solving a field theory exercise from Dummit and Foote. For your ref. it is Problem number $19.(b)$ in sec. $13$ (Second Edition). Any help will be appreciated. Thanks.



Edited Later:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition) is stated below.




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 $leq n.$







linear-algebra abstract-algebra matrices ring-theory field-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Could you say what the Dummit and Foote exercise is?
    $endgroup$
    – Slade
    Dec 29 '18 at 8:01






  • 1




    $begingroup$
    And such a map does exist when $dmid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form.
    $endgroup$
    – Slade
    Dec 29 '18 at 8:04










  • $begingroup$
    Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:15














4












4








4


1



$begingroup$


Let $mathbb{F}$ be a field and $d,n$ be positive integers with $d <n.$ Then does there exist a injective ring homomorphism from $M_d(mathbb{F})$ to $M_n(mathbb{F})$ ? (BTW A ring map sends $1$ to $1$)



I failed to produce a $1-1$ ring map. This appears in the process of solving a field theory exercise from Dummit and Foote. For your ref. it is Problem number $19.(b)$ in sec. $13$ (Second Edition). Any help will be appreciated. Thanks.



Edited Later:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition) is stated below.




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 $leq n.$







linear-algebra abstract-algebra matrices ring-theory field-theory







share|cite|improve this question











$endgroup$




Let $mathbb{F}$ be a field and $d,n$ be positive integers with $d <n.$ Then does there exist a injective ring homomorphism from $M_d(mathbb{F})$ to $M_n(mathbb{F})$ ? (BTW A ring map sends $1$ to $1$)



I failed to produce a $1-1$ ring map. This appears in the process of solving a field theory exercise from Dummit and Foote. For your ref. it is Problem number $19.(b)$ in sec. $13$ (Second Edition). Any help will be appreciated. Thanks.



Edited Later:Prob. $19.(b).,$ Section $13.2,$ from Abstract Algebra by Dummit and Foote(Second Edition) is stated below.




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 $leq n.$







linear-algebra abstract-algebra matrices ring-theory field-theory




linear-algebra abstract-algebra matrices ring-theory field-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 9:22







user371231

















asked Dec 29 '18 at 7:58









user371231user371231

417511




417511












  • $begingroup$
    Could you say what the Dummit and Foote exercise is?
    $endgroup$
    – Slade
    Dec 29 '18 at 8:01






  • 1




    $begingroup$
    And such a map does exist when $dmid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form.
    $endgroup$
    – Slade
    Dec 29 '18 at 8:04










  • $begingroup$
    Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:15


















  • $begingroup$
    Could you say what the Dummit and Foote exercise is?
    $endgroup$
    – Slade
    Dec 29 '18 at 8:01






  • 1




    $begingroup$
    And such a map does exist when $dmid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form.
    $endgroup$
    – Slade
    Dec 29 '18 at 8:04










  • $begingroup$
    Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:15
















$begingroup$
Could you say what the Dummit and Foote exercise is?
$endgroup$
– Slade
Dec 29 '18 at 8:01




$begingroup$
Could you say what the Dummit and Foote exercise is?
$endgroup$
– Slade
Dec 29 '18 at 8:01




1




1




$begingroup$
And such a map does exist when $dmid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form.
$endgroup$
– Slade
Dec 29 '18 at 8:04




$begingroup$
And such a map does exist when $dmid n$: send $A$ to $n/d$ copies of $A$ in block diagonal form.
$endgroup$
– Slade
Dec 29 '18 at 8:04












$begingroup$
Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote.
$endgroup$
– user371231
Dec 29 '18 at 9:15




$begingroup$
Yes, I stated the Problem from the abstract Algebra book by Dummit and Foote.
$endgroup$
– user371231
Dec 29 '18 at 9:15










1 Answer
1






active

oldest

votes


















7












$begingroup$

As Slade notes in a comment, this is true when $dmid n$, and the homomorphism from $M_d(mathbb F)$ to $M_n(mathbb F)$ can be defined by sending a $dtimes d$ matrix $A$ to a block diagonal matrix with $n/d$ copies of $A$.



However, it is not true for general $d$ and $n$. We can see this, for example, with $mathbb F=mathbb C$, $d=2$, $n=3$.
If we have such an injection, let $A$ be the image of
$$begin{pmatrix}0&1\0&0end{pmatrix}$$
and $B$ be the image of
$$begin{pmatrix}0&0\1&0end{pmatrix}.$$
Then $A^2=0$, $B^2=0$, and $(A+B)^2=I$.
We can see (e.g. using the Jordan form) that the only $3times 3$ matrix $A$ satisfying $Ane0$ but $A^2=0$ is, up to similarity,
$$begin{pmatrix}0&1&0\0&0&0\0&0&0end{pmatrix},$$
so $A$ must have rank $1$. Similarly $B$ also has rank $1$, so their sum $A+B$ has rank at most $2$. But this contradicts $(A+B)^2=I$, which would require $A+B$ to be invertible and so have full rank.



This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $dnotmid n$, although I expect that it is so.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:20










  • $begingroup$
    If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:22








  • 1




    $begingroup$
    Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
    $endgroup$
    – Carmeister
    Dec 29 '18 at 9:32












  • $begingroup$
    Ok then I'll ask a separate question, thanks for the nice counter example.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:38














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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









7












$begingroup$

As Slade notes in a comment, this is true when $dmid n$, and the homomorphism from $M_d(mathbb F)$ to $M_n(mathbb F)$ can be defined by sending a $dtimes d$ matrix $A$ to a block diagonal matrix with $n/d$ copies of $A$.



However, it is not true for general $d$ and $n$. We can see this, for example, with $mathbb F=mathbb C$, $d=2$, $n=3$.
If we have such an injection, let $A$ be the image of
$$begin{pmatrix}0&1\0&0end{pmatrix}$$
and $B$ be the image of
$$begin{pmatrix}0&0\1&0end{pmatrix}.$$
Then $A^2=0$, $B^2=0$, and $(A+B)^2=I$.
We can see (e.g. using the Jordan form) that the only $3times 3$ matrix $A$ satisfying $Ane0$ but $A^2=0$ is, up to similarity,
$$begin{pmatrix}0&1&0\0&0&0\0&0&0end{pmatrix},$$
so $A$ must have rank $1$. Similarly $B$ also has rank $1$, so their sum $A+B$ has rank at most $2$. But this contradicts $(A+B)^2=I$, which would require $A+B$ to be invertible and so have full rank.



This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $dnotmid n$, although I expect that it is so.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:20










  • $begingroup$
    If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:22








  • 1




    $begingroup$
    Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
    $endgroup$
    – Carmeister
    Dec 29 '18 at 9:32












  • $begingroup$
    Ok then I'll ask a separate question, thanks for the nice counter example.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:38


















7












$begingroup$

As Slade notes in a comment, this is true when $dmid n$, and the homomorphism from $M_d(mathbb F)$ to $M_n(mathbb F)$ can be defined by sending a $dtimes d$ matrix $A$ to a block diagonal matrix with $n/d$ copies of $A$.



However, it is not true for general $d$ and $n$. We can see this, for example, with $mathbb F=mathbb C$, $d=2$, $n=3$.
If we have such an injection, let $A$ be the image of
$$begin{pmatrix}0&1\0&0end{pmatrix}$$
and $B$ be the image of
$$begin{pmatrix}0&0\1&0end{pmatrix}.$$
Then $A^2=0$, $B^2=0$, and $(A+B)^2=I$.
We can see (e.g. using the Jordan form) that the only $3times 3$ matrix $A$ satisfying $Ane0$ but $A^2=0$ is, up to similarity,
$$begin{pmatrix}0&1&0\0&0&0\0&0&0end{pmatrix},$$
so $A$ must have rank $1$. Similarly $B$ also has rank $1$, so their sum $A+B$ has rank at most $2$. But this contradicts $(A+B)^2=I$, which would require $A+B$ to be invertible and so have full rank.



This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $dnotmid n$, although I expect that it is so.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:20










  • $begingroup$
    If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:22








  • 1




    $begingroup$
    Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
    $endgroup$
    – Carmeister
    Dec 29 '18 at 9:32












  • $begingroup$
    Ok then I'll ask a separate question, thanks for the nice counter example.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:38
















7












7








7





$begingroup$

As Slade notes in a comment, this is true when $dmid n$, and the homomorphism from $M_d(mathbb F)$ to $M_n(mathbb F)$ can be defined by sending a $dtimes d$ matrix $A$ to a block diagonal matrix with $n/d$ copies of $A$.



However, it is not true for general $d$ and $n$. We can see this, for example, with $mathbb F=mathbb C$, $d=2$, $n=3$.
If we have such an injection, let $A$ be the image of
$$begin{pmatrix}0&1\0&0end{pmatrix}$$
and $B$ be the image of
$$begin{pmatrix}0&0\1&0end{pmatrix}.$$
Then $A^2=0$, $B^2=0$, and $(A+B)^2=I$.
We can see (e.g. using the Jordan form) that the only $3times 3$ matrix $A$ satisfying $Ane0$ but $A^2=0$ is, up to similarity,
$$begin{pmatrix}0&1&0\0&0&0\0&0&0end{pmatrix},$$
so $A$ must have rank $1$. Similarly $B$ also has rank $1$, so their sum $A+B$ has rank at most $2$. But this contradicts $(A+B)^2=I$, which would require $A+B$ to be invertible and so have full rank.



This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $dnotmid n$, although I expect that it is so.






share|cite|improve this answer









$endgroup$



As Slade notes in a comment, this is true when $dmid n$, and the homomorphism from $M_d(mathbb F)$ to $M_n(mathbb F)$ can be defined by sending a $dtimes d$ matrix $A$ to a block diagonal matrix with $n/d$ copies of $A$.



However, it is not true for general $d$ and $n$. We can see this, for example, with $mathbb F=mathbb C$, $d=2$, $n=3$.
If we have such an injection, let $A$ be the image of
$$begin{pmatrix}0&1\0&0end{pmatrix}$$
and $B$ be the image of
$$begin{pmatrix}0&0\1&0end{pmatrix}.$$
Then $A^2=0$, $B^2=0$, and $(A+B)^2=I$.
We can see (e.g. using the Jordan form) that the only $3times 3$ matrix $A$ satisfying $Ane0$ but $A^2=0$ is, up to similarity,
$$begin{pmatrix}0&1&0\0&0&0\0&0&0end{pmatrix},$$
so $A$ must have rank $1$. Similarly $B$ also has rank $1$, so their sum $A+B$ has rank at most $2$. But this contradicts $(A+B)^2=I$, which would require $A+B$ to be invertible and so have full rank.



This proof uses a bit of linear algebra and doesn't generalize to show that the statement is false for all $dnotmid n$, although I expect that it is so.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 8:57









CarmeisterCarmeister

2,8692924




2,8692924












  • $begingroup$
    Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:20










  • $begingroup$
    If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:22








  • 1




    $begingroup$
    Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
    $endgroup$
    – Carmeister
    Dec 29 '18 at 9:32












  • $begingroup$
    Ok then I'll ask a separate question, thanks for the nice counter example.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:38




















  • $begingroup$
    Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:20










  • $begingroup$
    If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:22








  • 1




    $begingroup$
    Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
    $endgroup$
    – Carmeister
    Dec 29 '18 at 9:32












  • $begingroup$
    Ok then I'll ask a separate question, thanks for the nice counter example.
    $endgroup$
    – user371231
    Dec 29 '18 at 9:38


















$begingroup$
Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
$endgroup$
– user371231
Dec 29 '18 at 9:20




$begingroup$
Nice answer, many thanks. And I guess the Dummit Foote problem can be solved without using this.
$endgroup$
– user371231
Dec 29 '18 at 9:20












$begingroup$
If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
$endgroup$
– user371231
Dec 29 '18 at 9:22






$begingroup$
If extension of $F$ means that intermediate subfield of $F$ and $K$ then this answers the question.
$endgroup$
– user371231
Dec 29 '18 at 9:22






1




1




$begingroup$
Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
$endgroup$
– Carmeister
Dec 29 '18 at 9:32






$begingroup$
Yes, I'm guessing the problem comes in with the last statement, "...contains an isomorphic copy of every extension of $F$ of degree $le n$" specifically the $<n$ part. I believe that may be an error in the exercise, and that actually it's only true for the $=n$ case, but I'm not sure - it's worth asking a separate question about that.
$endgroup$
– Carmeister
Dec 29 '18 at 9:32














$begingroup$
Ok then I'll ask a separate question, thanks for the nice counter example.
$endgroup$
– user371231
Dec 29 '18 at 9:38






$begingroup$
Ok then I'll ask a separate question, thanks for the nice counter example.
$endgroup$
– user371231
Dec 29 '18 at 9:38




















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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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Residenzschloss Arolsen

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Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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