Isomoprhism between the tensor product of modules
$begingroup$
Basically what says in the top line is :Let $k$ be a filed, $V$ a vector space over $k$ and $T$ an endomorphism of V as a $k$-module.Consider the structure of a $K[x]$-module in V induced by the endomorphism.
Can anyone help me prove the following result? Ive tried creating homomorphism with the universal property of the tensor product and then proving they are inverse to one another but im not seeming to get any luck.Thanks in advance.
abstract-algebra modules tensor-products
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
$endgroup$
add a comment |
$begingroup$
Basically what says in the top line is :Let $k$ be a filed, $V$ a vector space over $k$ and $T$ an endomorphism of V as a $k$-module.Consider the structure of a $K[x]$-module in V induced by the endomorphism.
Can anyone help me prove the following result? Ive tried creating homomorphism with the universal property of the tensor product and then proving they are inverse to one another but im not seeming to get any luck.Thanks in advance.
abstract-algebra modules tensor-products
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
$endgroup$
1
$begingroup$
A translation, written up in MathJax, would be highly advisable. I would imagine the vast majority of users do not know Spanish. Even if they didn't, if you were to type it up so that it can be copied and pasted into e.g. Google Translate even that would be better than an image. Languages other than English are not strictly prohibited but it will give much less visibility to your question.
$endgroup$
– Matt Samuel
Dec 22 '18 at 12:13
add a comment |
$begingroup$
Basically what says in the top line is :Let $k$ be a filed, $V$ a vector space over $k$ and $T$ an endomorphism of V as a $k$-module.Consider the structure of a $K[x]$-module in V induced by the endomorphism.
Can anyone help me prove the following result? Ive tried creating homomorphism with the universal property of the tensor product and then proving they are inverse to one another but im not seeming to get any luck.Thanks in advance.
abstract-algebra modules tensor-products
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
$endgroup$
Basically what says in the top line is :Let $k$ be a filed, $V$ a vector space over $k$ and $T$ an endomorphism of V as a $k$-module.Consider the structure of a $K[x]$-module in V induced by the endomorphism.
Can anyone help me prove the following result? Ive tried creating homomorphism with the universal property of the tensor product and then proving they are inverse to one another but im not seeming to get any luck.Thanks in advance.
abstract-algebra modules tensor-products
abstract-algebra modules tensor-products
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
asked Dec 22 '18 at 12:04
Pedro Miguel SantosPedro Miguel Santos
498
498
1
$begingroup$
A translation, written up in MathJax, would be highly advisable. I would imagine the vast majority of users do not know Spanish. Even if they didn't, if you were to type it up so that it can be copied and pasted into e.g. Google Translate even that would be better than an image. Languages other than English are not strictly prohibited but it will give much less visibility to your question.
$endgroup$
– Matt Samuel
Dec 22 '18 at 12:13
add a comment |
1
$begingroup$
A translation, written up in MathJax, would be highly advisable. I would imagine the vast majority of users do not know Spanish. Even if they didn't, if you were to type it up so that it can be copied and pasted into e.g. Google Translate even that would be better than an image. Languages other than English are not strictly prohibited but it will give much less visibility to your question.
$endgroup$
– Matt Samuel
Dec 22 '18 at 12:13
1
1
$begingroup$
A translation, written up in MathJax, would be highly advisable. I would imagine the vast majority of users do not know Spanish. Even if they didn't, if you were to type it up so that it can be copied and pasted into e.g. Google Translate even that would be better than an image. Languages other than English are not strictly prohibited but it will give much less visibility to your question.
$endgroup$
– Matt Samuel
Dec 22 '18 at 12:13
$begingroup$
A translation, written up in MathJax, would be highly advisable. I would imagine the vast majority of users do not know Spanish. Even if they didn't, if you were to type it up so that it can be copied and pasted into e.g. Google Translate even that would be better than an image. Languages other than English are not strictly prohibited but it will give much less visibility to your question.
$endgroup$
– Matt Samuel
Dec 22 '18 at 12:13
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let me simply follow what you tried and see where it doesn't work, if it's ever the case.
A first thing to ask yourself is "what is the structure of $k[x]$-module on the right ?". But $k[x]$ is a $k[x]$-module so this is essentially basic commutative algebra : this structure is simply $lambdacdot (aotimes b) =(lambda a)otimes b$. Let me call this module $N$.
The second thing to note is that the quotient is a quotient as $k[x]$-modules, so the $langle rangle$ denotes the generated sub-$k[x]$-module. Let me call $M$ the quotient module on the right, and $L$ the submodule by which we quotient. In short, $M=N/L$
With these things in mind, let's proceed. By induction, we prove that $x^notimes v = 1otimes T^n v$ in the module on the right.
Indeed $xotimes v - 1otimes Tv$ is in the kernel of the quotient map by definition; and then if it's true for $n$ we have $x^{n+1}otimes v = xcdot (x^notimes v) = xcdot (1otimes T^n v) = xotimes T^nv= 1otimes T^{n+1}v$.
Now from this it follows by linearity that for any $Pin k[x]$, $Potimes v = 1otimes P(T)(v)$.
This proves that $h: Vmapsto M, vmapsto 1otimes v$ is $k[x]$-linear.
Let's now define the converse. Define $f: k[x]times Vto V$, $(P,v)mapsto P(T)(v)$. It's easy to check that this is $k$-bilinear so factors through $k[x]otimes_k V = N$. Let's call $g$ the map $Nto V$ thay we get.
But now on $L$, $g(xotimes v -1otimes Tv) = g(xotimes v) - g(1otimes Tv) = Tv- Tv = 0$. So $g$ factors through $N/L= M$.
What this means is that $k: Potimes vmapsto P(f)(v)$ is well-defined on $M$. Now we check that $k$ and $h$ are inverse to eachother, but this is easy as we already proved that $Potimes v = 1otimes P(T)(v)$ in $M$.
So we proved that $Msimeq V$ as $k[x]$-modules : what does this have to do with your problem ?
Well for any ring $R$ and any $R$-module $K$, $Rotimes_R Ksimeq K$ as $R$-modules via $xotimes mmapsto xm$. So $Vsimeq k[x]otimes_{k[x]}V$ as $k[x]$-modules.
answered Dec 22 '18 at 13:29
MaxMax
15k11143
$endgroup$
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 17:17
$begingroup$
It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
$endgroup$
– Max
Dec 30 '18 at 18:51
|
show 1 more comment
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
active
oldest
votes
$begingroup$
Let me simply follow what you tried and see where it doesn't work, if it's ever the case.
A first thing to ask yourself is "what is the structure of $k[x]$-module on the right ?". But $k[x]$ is a $k[x]$-module so this is essentially basic commutative algebra : this structure is simply $lambdacdot (aotimes b) =(lambda a)otimes b$. Let me call this module $N$.
The second thing to note is that the quotient is a quotient as $k[x]$-modules, so the $langle rangle$ denotes the generated sub-$k[x]$-module. Let me call $M$ the quotient module on the right, and $L$ the submodule by which we quotient. In short, $M=N/L$
With these things in mind, let's proceed. By induction, we prove that $x^notimes v = 1otimes T^n v$ in the module on the right.
Indeed $xotimes v - 1otimes Tv$ is in the kernel of the quotient map by definition; and then if it's true for $n$ we have $x^{n+1}otimes v = xcdot (x^notimes v) = xcdot (1otimes T^n v) = xotimes T^nv= 1otimes T^{n+1}v$.
Now from this it follows by linearity that for any $Pin k[x]$, $Potimes v = 1otimes P(T)(v)$.
This proves that $h: Vmapsto M, vmapsto 1otimes v$ is $k[x]$-linear.
Let's now define the converse. Define $f: k[x]times Vto V$, $(P,v)mapsto P(T)(v)$. It's easy to check that this is $k$-bilinear so factors through $k[x]otimes_k V = N$. Let's call $g$ the map $Nto V$ thay we get.
But now on $L$, $g(xotimes v -1otimes Tv) = g(xotimes v) - g(1otimes Tv) = Tv- Tv = 0$. So $g$ factors through $N/L= M$.
What this means is that $k: Potimes vmapsto P(f)(v)$ is well-defined on $M$. Now we check that $k$ and $h$ are inverse to eachother, but this is easy as we already proved that $Potimes v = 1otimes P(T)(v)$ in $M$.
So we proved that $Msimeq V$ as $k[x]$-modules : what does this have to do with your problem ?
Well for any ring $R$ and any $R$-module $K$, $Rotimes_R Ksimeq K$ as $R$-modules via $xotimes mmapsto xm$. So $Vsimeq k[x]otimes_{k[x]}V$ as $k[x]$-modules.
answered Dec 22 '18 at 13:29
MaxMax
15k11143
$endgroup$
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 17:17
$begingroup$
It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
$endgroup$
– Max
Dec 30 '18 at 18:51
|
show 1 more comment
$begingroup$
Let me simply follow what you tried and see where it doesn't work, if it's ever the case.
A first thing to ask yourself is "what is the structure of $k[x]$-module on the right ?". But $k[x]$ is a $k[x]$-module so this is essentially basic commutative algebra : this structure is simply $lambdacdot (aotimes b) =(lambda a)otimes b$. Let me call this module $N$.
The second thing to note is that the quotient is a quotient as $k[x]$-modules, so the $langle rangle$ denotes the generated sub-$k[x]$-module. Let me call $M$ the quotient module on the right, and $L$ the submodule by which we quotient. In short, $M=N/L$
With these things in mind, let's proceed. By induction, we prove that $x^notimes v = 1otimes T^n v$ in the module on the right.
Indeed $xotimes v - 1otimes Tv$ is in the kernel of the quotient map by definition; and then if it's true for $n$ we have $x^{n+1}otimes v = xcdot (x^notimes v) = xcdot (1otimes T^n v) = xotimes T^nv= 1otimes T^{n+1}v$.
Now from this it follows by linearity that for any $Pin k[x]$, $Potimes v = 1otimes P(T)(v)$.
This proves that $h: Vmapsto M, vmapsto 1otimes v$ is $k[x]$-linear.
Let's now define the converse. Define $f: k[x]times Vto V$, $(P,v)mapsto P(T)(v)$. It's easy to check that this is $k$-bilinear so factors through $k[x]otimes_k V = N$. Let's call $g$ the map $Nto V$ thay we get.
But now on $L$, $g(xotimes v -1otimes Tv) = g(xotimes v) - g(1otimes Tv) = Tv- Tv = 0$. So $g$ factors through $N/L= M$.
What this means is that $k: Potimes vmapsto P(f)(v)$ is well-defined on $M$. Now we check that $k$ and $h$ are inverse to eachother, but this is easy as we already proved that $Potimes v = 1otimes P(T)(v)$ in $M$.
So we proved that $Msimeq V$ as $k[x]$-modules : what does this have to do with your problem ?
Well for any ring $R$ and any $R$-module $K$, $Rotimes_R Ksimeq K$ as $R$-modules via $xotimes mmapsto xm$. So $Vsimeq k[x]otimes_{k[x]}V$ as $k[x]$-modules.
answered Dec 22 '18 at 13:29
MaxMax
15k11143
$endgroup$
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 17:17
$begingroup$
It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
$endgroup$
– Max
Dec 30 '18 at 18:51
|
show 1 more comment
$begingroup$
Let me simply follow what you tried and see where it doesn't work, if it's ever the case.
A first thing to ask yourself is "what is the structure of $k[x]$-module on the right ?". But $k[x]$ is a $k[x]$-module so this is essentially basic commutative algebra : this structure is simply $lambdacdot (aotimes b) =(lambda a)otimes b$. Let me call this module $N$.
The second thing to note is that the quotient is a quotient as $k[x]$-modules, so the $langle rangle$ denotes the generated sub-$k[x]$-module. Let me call $M$ the quotient module on the right, and $L$ the submodule by which we quotient. In short, $M=N/L$
With these things in mind, let's proceed. By induction, we prove that $x^notimes v = 1otimes T^n v$ in the module on the right.
Indeed $xotimes v - 1otimes Tv$ is in the kernel of the quotient map by definition; and then if it's true for $n$ we have $x^{n+1}otimes v = xcdot (x^notimes v) = xcdot (1otimes T^n v) = xotimes T^nv= 1otimes T^{n+1}v$.
Now from this it follows by linearity that for any $Pin k[x]$, $Potimes v = 1otimes P(T)(v)$.
This proves that $h: Vmapsto M, vmapsto 1otimes v$ is $k[x]$-linear.
Let's now define the converse. Define $f: k[x]times Vto V$, $(P,v)mapsto P(T)(v)$. It's easy to check that this is $k$-bilinear so factors through $k[x]otimes_k V = N$. Let's call $g$ the map $Nto V$ thay we get.
But now on $L$, $g(xotimes v -1otimes Tv) = g(xotimes v) - g(1otimes Tv) = Tv- Tv = 0$. So $g$ factors through $N/L= M$.
What this means is that $k: Potimes vmapsto P(f)(v)$ is well-defined on $M$. Now we check that $k$ and $h$ are inverse to eachother, but this is easy as we already proved that $Potimes v = 1otimes P(T)(v)$ in $M$.
So we proved that $Msimeq V$ as $k[x]$-modules : what does this have to do with your problem ?
Well for any ring $R$ and any $R$-module $K$, $Rotimes_R Ksimeq K$ as $R$-modules via $xotimes mmapsto xm$. So $Vsimeq k[x]otimes_{k[x]}V$ as $k[x]$-modules.
answered Dec 22 '18 at 13:29
MaxMax
15k11143
$endgroup$
Let me simply follow what you tried and see where it doesn't work, if it's ever the case.
A first thing to ask yourself is "what is the structure of $k[x]$-module on the right ?". But $k[x]$ is a $k[x]$-module so this is essentially basic commutative algebra : this structure is simply $lambdacdot (aotimes b) =(lambda a)otimes b$. Let me call this module $N$.
The second thing to note is that the quotient is a quotient as $k[x]$-modules, so the $langle rangle$ denotes the generated sub-$k[x]$-module. Let me call $M$ the quotient module on the right, and $L$ the submodule by which we quotient. In short, $M=N/L$
With these things in mind, let's proceed. By induction, we prove that $x^notimes v = 1otimes T^n v$ in the module on the right.
Indeed $xotimes v - 1otimes Tv$ is in the kernel of the quotient map by definition; and then if it's true for $n$ we have $x^{n+1}otimes v = xcdot (x^notimes v) = xcdot (1otimes T^n v) = xotimes T^nv= 1otimes T^{n+1}v$.
Now from this it follows by linearity that for any $Pin k[x]$, $Potimes v = 1otimes P(T)(v)$.
This proves that $h: Vmapsto M, vmapsto 1otimes v$ is $k[x]$-linear.
Let's now define the converse. Define $f: k[x]times Vto V$, $(P,v)mapsto P(T)(v)$. It's easy to check that this is $k$-bilinear so factors through $k[x]otimes_k V = N$. Let's call $g$ the map $Nto V$ thay we get.
But now on $L$, $g(xotimes v -1otimes Tv) = g(xotimes v) - g(1otimes Tv) = Tv- Tv = 0$. So $g$ factors through $N/L= M$.
What this means is that $k: Potimes vmapsto P(f)(v)$ is well-defined on $M$. Now we check that $k$ and $h$ are inverse to eachother, but this is easy as we already proved that $Potimes v = 1otimes P(T)(v)$ in $M$.
So we proved that $Msimeq V$ as $k[x]$-modules : what does this have to do with your problem ?
Well for any ring $R$ and any $R$-module $K$, $Rotimes_R Ksimeq K$ as $R$-modules via $xotimes mmapsto xm$. So $Vsimeq k[x]otimes_{k[x]}V$ as $k[x]$-modules.
answered Dec 22 '18 at 13:29
MaxMax
15k11143
answered Dec 22 '18 at 13:29
MaxMax
15k11143
answered Dec 22 '18 at 13:29
MaxMax
15k11143
answered Dec 22 '18 at 13:29
MaxMax
15k11143
15k11143
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 17:17
$begingroup$
It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
$endgroup$
– Max
Dec 30 '18 at 18:51
|
show 1 more comment
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 17:17
$begingroup$
It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
$endgroup$
– Max
Dec 30 '18 at 18:51
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Oh yes, i forgot about that last sentence ,yeah that really helps with the problem , thanks !!
$endgroup$
– Pedro Miguel Santos
Dec 22 '18 at 14:01
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Im not sure how u proved that the first homomorphism is $K[x]-linear$ why can you just put the polinomyal in the tensor product off $1 otimes_k v $, it doenst make much sense to me
$endgroup$
– Pedro Miguel Santos
Dec 30 '18 at 12:52
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
Well I proved right above that $Potimes v = 1otimes P(T)(v)$ just above, so $h: vmapsto 1otimes v$ will be $k[x]$-linear : indeed $h(Pcdot v) = h(P(T)(v))= 1otimes P(T)(v) = Potimes v= Pcdot (1otimes v) = Pcdot h(v)$
$endgroup$
– Max
Dec 30 '18 at 14:05
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
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– Pedro Miguel Santos
Dec 30 '18 at 17:17
$begingroup$
When you use ur 4th equal sign , wny can u pull the polinomyal out of the tensor product, isnt that tensor over $k$ ? shouldnt u only be able to pull out $a in k$
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– Pedro Miguel Santos
Dec 30 '18 at 17:17
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It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
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– Max
Dec 30 '18 at 18:51
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It's the definition of the action of $k[x]$ on $k[x]otimes_k V$ !
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– Max
Dec 30 '18 at 18:51
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$begingroup$
A translation, written up in MathJax, would be highly advisable. I would imagine the vast majority of users do not know Spanish. Even if they didn't, if you were to type it up so that it can be copied and pasted into e.g. Google Translate even that would be better than an image. Languages other than English are not strictly prohibited but it will give much less visibility to your question.
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– Matt Samuel
Dec 22 '18 at 12:13