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Proof that the algebraic integers form a subring of $mathbbC$
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In some notes on abstract algebra I found a number theory problem in which I was given to prove that the algebraic integers, $overlinemathbbZ$ form a subring of $mathbbC$ using the tensor product. The solution that I've found kinda generalizes some notions of field extensions. But DK if it's right.
Here's my work:
I took two arbitrary elements, $lambda$ and $lambda '$ in $overlinemathbbZ$ and then, since $mathbbZ[lambda]$ and $mathbbZ[lambda ']$ are $mathbbZ$ algebras, their tensor product, $mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda ']$ is also a $mathbbZ$ algebra (the general result is for any ring $R$ and any algebras $A$ and $B$ over it).
Since $lambda in overlinemathbbZ$, we could choose a monic, irreductible polynominal, $P_lambda$ such that $lambda$ is a root of this polynominal. If we were to choose another polynominal of smaller degree that's not monic in $mathbbZ[T]$ for which $lambda$ is not a root, its product with another polynominal won't be monic anymore. Since $P_lambda$ is monic, by the division algorithm, $forall f in mathbbZ[T]$, $exists$ unique $q;g in mathbbZ[T]$ such that $f= qP_lambda+g$. Take the evaluation homomorphism $psi_lambda:mathbbZ[T] longmapsto mathbbC$ and hence, by the first ISO theorem, $$fracmathbbZ[T]ker(psi_lambda) cong Im( psi_lambda)=mathbbZ[lambda]$$ and by the previous result, we get that $ker(psi_lambda)=<P_lambda>$. The same is true for $lambda '$.
Hence, $$mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong fracmathbbZ[T]<P_lambda> otimes_mathbbZ fracmathbbZ[T]<P_lambda '> $$.
Now we will take a bilinear map $varphi :mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: (a;b) longmapsto ab$ which is surjective. By the universal property of the tensor prodcut, we get a $mathbbZ$-module homomorphism $widetildevarphi:mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: a otimes_mathbbZ b longmapsto ab$ which is also a ring homomorphism but $varphi$ and $mu : mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] : (a;b) longmapsto a otimes_mathbbZ b$ are surjective and so is $widetildevarphi$, since $ varphi= widetildevarphi circ mu$.
Since for any monic polynominal $f in mathbbZ[T],, mathbbZ[T]/<f> cong mathbbZ^deg(f)$, we have: $$ mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong mathbbZ^n otimes_mathbbZ mathbbZ^m cong mathbbZ^mn$$ where $n:=deg(P_lambda)$; $m:=deg( P_lambda') $ and hence, $exists$ $k(a) in mathbbN^*$ such that for every $a in mathbbZ[lambda;lambda '],$ $$fracmathbbZ[T] <P_a> cong mathbbZ[a] cong mathbbZ^k(a)$$ and $a$ could take values like $lambda + lambda '$ or $lambda lambda'$ for which there is a monic polynominal $P_a$ and $a$ is a root of it.
Therefore, $<overlinemathbbZ;+;times>$ is a ring.
ring-theory algebraic-number-theory tensor-products integers
asked Jul 16 at 18:28
Mario 04
6013
 |Â
show 2 more comments
up vote
1
down vote
favorite
In some notes on abstract algebra I found a number theory problem in which I was given to prove that the algebraic integers, $overlinemathbbZ$ form a subring of $mathbbC$ using the tensor product. The solution that I've found kinda generalizes some notions of field extensions. But DK if it's right.
Here's my work:
I took two arbitrary elements, $lambda$ and $lambda '$ in $overlinemathbbZ$ and then, since $mathbbZ[lambda]$ and $mathbbZ[lambda ']$ are $mathbbZ$ algebras, their tensor product, $mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda ']$ is also a $mathbbZ$ algebra (the general result is for any ring $R$ and any algebras $A$ and $B$ over it).
Since $lambda in overlinemathbbZ$, we could choose a monic, irreductible polynominal, $P_lambda$ such that $lambda$ is a root of this polynominal. If we were to choose another polynominal of smaller degree that's not monic in $mathbbZ[T]$ for which $lambda$ is not a root, its product with another polynominal won't be monic anymore. Since $P_lambda$ is monic, by the division algorithm, $forall f in mathbbZ[T]$, $exists$ unique $q;g in mathbbZ[T]$ such that $f= qP_lambda+g$. Take the evaluation homomorphism $psi_lambda:mathbbZ[T] longmapsto mathbbC$ and hence, by the first ISO theorem, $$fracmathbbZ[T]ker(psi_lambda) cong Im( psi_lambda)=mathbbZ[lambda]$$ and by the previous result, we get that $ker(psi_lambda)=<P_lambda>$. The same is true for $lambda '$.
Hence, $$mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong fracmathbbZ[T]<P_lambda> otimes_mathbbZ fracmathbbZ[T]<P_lambda '> $$.
Now we will take a bilinear map $varphi :mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: (a;b) longmapsto ab$ which is surjective. By the universal property of the tensor prodcut, we get a $mathbbZ$-module homomorphism $widetildevarphi:mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: a otimes_mathbbZ b longmapsto ab$ which is also a ring homomorphism but $varphi$ and $mu : mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] : (a;b) longmapsto a otimes_mathbbZ b$ are surjective and so is $widetildevarphi$, since $ varphi= widetildevarphi circ mu$.
Since for any monic polynominal $f in mathbbZ[T],, mathbbZ[T]/<f> cong mathbbZ^deg(f)$, we have: $$ mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong mathbbZ^n otimes_mathbbZ mathbbZ^m cong mathbbZ^mn$$ where $n:=deg(P_lambda)$; $m:=deg( P_lambda') $ and hence, $exists$ $k(a) in mathbbN^*$ such that for every $a in mathbbZ[lambda;lambda '],$ $$fracmathbbZ[T] <P_a> cong mathbbZ[a] cong mathbbZ^k(a)$$ and $a$ could take values like $lambda + lambda '$ or $lambda lambda'$ for which there is a monic polynominal $P_a$ and $a$ is a root of it.
Therefore, $<overlinemathbbZ;+;times>$ is a ring.
ring-theory algebraic-number-theory tensor-products integers
asked Jul 16 at 18:28
Mario 04
6013
1
It seems ok to me
– usr0192
Jul 16 at 18:44
Thanks. I was quite sure before asking but I wanted to see if I made any silly mistake.
– Mario 04
Jul 16 at 18:47
why are $varphi$ and $mu$ surjective ?
– mercio
Jul 16 at 23:26
Because of the definition of $mathbbZ[lambda][lambda ']$ and because $mu$ is the function that transforms elements into their tensor product and every element has that form.
– Mario 04
Jul 17 at 8:16
That should be $mathbbZ[lambda;lambda ']$ instead of $mathbbZ[lambda][lambda ']$.
– Mario 04
Jul 17 at 9:44
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In some notes on abstract algebra I found a number theory problem in which I was given to prove that the algebraic integers, $overlinemathbbZ$ form a subring of $mathbbC$ using the tensor product. The solution that I've found kinda generalizes some notions of field extensions. But DK if it's right.
Here's my work:
I took two arbitrary elements, $lambda$ and $lambda '$ in $overlinemathbbZ$ and then, since $mathbbZ[lambda]$ and $mathbbZ[lambda ']$ are $mathbbZ$ algebras, their tensor product, $mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda ']$ is also a $mathbbZ$ algebra (the general result is for any ring $R$ and any algebras $A$ and $B$ over it).
Since $lambda in overlinemathbbZ$, we could choose a monic, irreductible polynominal, $P_lambda$ such that $lambda$ is a root of this polynominal. If we were to choose another polynominal of smaller degree that's not monic in $mathbbZ[T]$ for which $lambda$ is not a root, its product with another polynominal won't be monic anymore. Since $P_lambda$ is monic, by the division algorithm, $forall f in mathbbZ[T]$, $exists$ unique $q;g in mathbbZ[T]$ such that $f= qP_lambda+g$. Take the evaluation homomorphism $psi_lambda:mathbbZ[T] longmapsto mathbbC$ and hence, by the first ISO theorem, $$fracmathbbZ[T]ker(psi_lambda) cong Im( psi_lambda)=mathbbZ[lambda]$$ and by the previous result, we get that $ker(psi_lambda)=<P_lambda>$. The same is true for $lambda '$.
Hence, $$mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong fracmathbbZ[T]<P_lambda> otimes_mathbbZ fracmathbbZ[T]<P_lambda '> $$.
Now we will take a bilinear map $varphi :mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: (a;b) longmapsto ab$ which is surjective. By the universal property of the tensor prodcut, we get a $mathbbZ$-module homomorphism $widetildevarphi:mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: a otimes_mathbbZ b longmapsto ab$ which is also a ring homomorphism but $varphi$ and $mu : mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] : (a;b) longmapsto a otimes_mathbbZ b$ are surjective and so is $widetildevarphi$, since $ varphi= widetildevarphi circ mu$.
Since for any monic polynominal $f in mathbbZ[T],, mathbbZ[T]/<f> cong mathbbZ^deg(f)$, we have: $$ mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong mathbbZ^n otimes_mathbbZ mathbbZ^m cong mathbbZ^mn$$ where $n:=deg(P_lambda)$; $m:=deg( P_lambda') $ and hence, $exists$ $k(a) in mathbbN^*$ such that for every $a in mathbbZ[lambda;lambda '],$ $$fracmathbbZ[T] <P_a> cong mathbbZ[a] cong mathbbZ^k(a)$$ and $a$ could take values like $lambda + lambda '$ or $lambda lambda'$ for which there is a monic polynominal $P_a$ and $a$ is a root of it.
Therefore, $<overlinemathbbZ;+;times>$ is a ring.
ring-theory algebraic-number-theory tensor-products integers
asked Jul 16 at 18:28
Mario 04
6013
In some notes on abstract algebra I found a number theory problem in which I was given to prove that the algebraic integers, $overlinemathbbZ$ form a subring of $mathbbC$ using the tensor product. The solution that I've found kinda generalizes some notions of field extensions. But DK if it's right.
Here's my work:
I took two arbitrary elements, $lambda$ and $lambda '$ in $overlinemathbbZ$ and then, since $mathbbZ[lambda]$ and $mathbbZ[lambda ']$ are $mathbbZ$ algebras, their tensor product, $mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda ']$ is also a $mathbbZ$ algebra (the general result is for any ring $R$ and any algebras $A$ and $B$ over it).
Since $lambda in overlinemathbbZ$, we could choose a monic, irreductible polynominal, $P_lambda$ such that $lambda$ is a root of this polynominal. If we were to choose another polynominal of smaller degree that's not monic in $mathbbZ[T]$ for which $lambda$ is not a root, its product with another polynominal won't be monic anymore. Since $P_lambda$ is monic, by the division algorithm, $forall f in mathbbZ[T]$, $exists$ unique $q;g in mathbbZ[T]$ such that $f= qP_lambda+g$. Take the evaluation homomorphism $psi_lambda:mathbbZ[T] longmapsto mathbbC$ and hence, by the first ISO theorem, $$fracmathbbZ[T]ker(psi_lambda) cong Im( psi_lambda)=mathbbZ[lambda]$$ and by the previous result, we get that $ker(psi_lambda)=<P_lambda>$. The same is true for $lambda '$.
Hence, $$mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong fracmathbbZ[T]<P_lambda> otimes_mathbbZ fracmathbbZ[T]<P_lambda '> $$.
Now we will take a bilinear map $varphi :mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: (a;b) longmapsto ab$ which is surjective. By the universal property of the tensor prodcut, we get a $mathbbZ$-module homomorphism $widetildevarphi:mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] longmapsto mathbbZ[lambda;lambda ']: a otimes_mathbbZ b longmapsto ab$ which is also a ring homomorphism but $varphi$ and $mu : mathbbZ[lambda] times mathbbZ[lambda '] longmapsto mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] : (a;b) longmapsto a otimes_mathbbZ b$ are surjective and so is $widetildevarphi$, since $ varphi= widetildevarphi circ mu$.
Since for any monic polynominal $f in mathbbZ[T],, mathbbZ[T]/<f> cong mathbbZ^deg(f)$, we have: $$ mathbbZ[lambda] otimes_mathbbZ mathbbZ[lambda '] cong mathbbZ^n otimes_mathbbZ mathbbZ^m cong mathbbZ^mn$$ where $n:=deg(P_lambda)$; $m:=deg( P_lambda') $ and hence, $exists$ $k(a) in mathbbN^*$ such that for every $a in mathbbZ[lambda;lambda '],$ $$fracmathbbZ[T] <P_a> cong mathbbZ[a] cong mathbbZ^k(a)$$ and $a$ could take values like $lambda + lambda '$ or $lambda lambda'$ for which there is a monic polynominal $P_a$ and $a$ is a root of it.
Therefore, $<overlinemathbbZ;+;times>$ is a ring.
ring-theory algebraic-number-theory tensor-products integers
asked Jul 16 at 18:28
Mario 04
6013
asked Jul 16 at 18:28
Mario 04
6013
asked Jul 16 at 18:28
Mario 04
6013
asked Jul 16 at 18:28
Mario 04
6013
6013
1
It seems ok to me
– usr0192
Jul 16 at 18:44
Thanks. I was quite sure before asking but I wanted to see if I made any silly mistake.
– Mario 04
Jul 16 at 18:47
why are $varphi$ and $mu$ surjective ?
– mercio
Jul 16 at 23:26
Because of the definition of $mathbbZ[lambda][lambda ']$ and because $mu$ is the function that transforms elements into their tensor product and every element has that form.
– Mario 04
Jul 17 at 8:16
That should be $mathbbZ[lambda;lambda ']$ instead of $mathbbZ[lambda][lambda ']$.
– Mario 04
Jul 17 at 9:44
 |Â
show 2 more comments
1
It seems ok to me
– usr0192
Jul 16 at 18:44
Thanks. I was quite sure before asking but I wanted to see if I made any silly mistake.
– Mario 04
Jul 16 at 18:47
why are $varphi$ and $mu$ surjective ?
– mercio
Jul 16 at 23:26
Because of the definition of $mathbbZ[lambda][lambda ']$ and because $mu$ is the function that transforms elements into their tensor product and every element has that form.
– Mario 04
Jul 17 at 8:16
That should be $mathbbZ[lambda;lambda ']$ instead of $mathbbZ[lambda][lambda ']$.
– Mario 04
Jul 17 at 9:44
1
1
It seems ok to me
– usr0192
Jul 16 at 18:44
It seems ok to me
– usr0192
Jul 16 at 18:44
Thanks. I was quite sure before asking but I wanted to see if I made any silly mistake.
– Mario 04
Jul 16 at 18:47
Thanks. I was quite sure before asking but I wanted to see if I made any silly mistake.
– Mario 04
Jul 16 at 18:47
why are $varphi$ and $mu$ surjective ?
– mercio
Jul 16 at 23:26
why are $varphi$ and $mu$ surjective ?
– mercio
Jul 16 at 23:26
Because of the definition of $mathbbZ[lambda][lambda ']$ and because $mu$ is the function that transforms elements into their tensor product and every element has that form.
– Mario 04
Jul 17 at 8:16
Because of the definition of $mathbbZ[lambda][lambda ']$ and because $mu$ is the function that transforms elements into their tensor product and every element has that form.
– Mario 04
Jul 17 at 8:16
That should be $mathbbZ[lambda;lambda ']$ instead of $mathbbZ[lambda][lambda ']$.
– Mario 04
Jul 17 at 9:44
That should be $mathbbZ[lambda;lambda ']$ instead of $mathbbZ[lambda][lambda ']$.
– Mario 04
Jul 17 at 9:44
 |Â
show 2 more comments
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1
It seems ok to me
– usr0192
Jul 16 at 18:44
Thanks. I was quite sure before asking but I wanted to see if I made any silly mistake.
– Mario 04
Jul 16 at 18:47
why are $varphi$ and $mu$ surjective ?
– mercio
Jul 16 at 23:26
Because of the definition of $mathbbZ[lambda][lambda ']$ and because $mu$ is the function that transforms elements into their tensor product and every element has that form.
– Mario 04
Jul 17 at 8:16
That should be $mathbbZ[lambda;lambda ']$ instead of $mathbbZ[lambda][lambda ']$.
– Mario 04
Jul 17 at 9:44