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Version 3 |
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| \PMlinkescapeword{order} |
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| \PMlinkescapeword{maps} |
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| \PMlinkescapeword{natural} |
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| \PMlinkescapeword{notation} |
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| Let $R$ be a ring and let $M$ be a right $R$-module. |
Let $R$ be a ring and let $M$ be a right $R$-module. |
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| An \emph{endomorphism} of $M$ is a $R$-module homomorphism from $M$ to itself. |
An \emph{endomorphism} of $M$ is a $R$-module homomorphism from $M$ to itself. |
| We shall write endomorphisms on the left, |
We shall write endomorphisms on the left, |
| so that $f : M \to M$ maps $x \mapsto f(x)$. |
so that $f : M \to M$ maps $x \mapsto f(x)$. |
| If $f,g : M \to M$ are two endomorphisms, we can add them: |
If $f,g : M \to M$ are two endomorphisms, we can add them: |
| $$ f+g : x \mapsto f(x) + g(x) $$ |
$$ f+g : x \mapsto f(x) + g(x) $$ |
| and multiply them |
and multiply them |
| $$ fg : x \mapsto f(g(x)) $$ |
$$ fg : x \mapsto f(g(x)) $$ |
| With these operations, the set of endomorphisms of $M$ becomes a ring, which we call |
With these operations, the set of endomorphisms of $M$ becomes a ring, which we call |
| the \emph{ring of endomorphisms of $M$}, written $\End_R(M)$. |
the \emph{ring of endomorphisms of $M$}, written $\End_R(M)$. |
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| Instead of writing endomorphisms as functions, it is often convenient |
Instead of writing endomorphisms as functions, it is often convenient |
| to write them multiplicatively: we simply write the application of the |
to write them multiplicatively: we simply write the application of the |
| endomorphism $f$ as $x \mapsto fx$. Then the fact that each $f$ is an $R$-module homomorphism can be expressed as: |
endomorphism $f$ as $x \mapsto fx$. Then the fact that each $f$ is an $R$-module homomorphism can be expressed as: |
| $$f(xr) = (fx)r$$ |
$$f(xr) = (fx)r$$ |
| for all $x \in M$ and $r \in R$ and $f \in \End_R(M)$. |
for all $x \in M$ and $r \in R$ and $f \in \End_R(M)$. |
| With this notation, it is clear that $M$ becomes an $\End_R(M)$-$R$-bimodule. |
With this notation, it is clear that $M$ becomes an $\End_R(M)$-$R$-bimodule. |
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| Now, let $N$ be a left $R$-module. We can construct the ring $\End_R(N)$ in the same way. |
Now, let $N$ be a left $R$-module. We can construct the ring $\End_R(N)$ in the same way. |
| There is a complication, however, if we still think of endomorphism as functions written |
There is a complication, however, if we still think of endomorphism as functions written |
| on the left. In order to make $M$ into a bimodule, we need to define an action of |
on the left. In order to make $M$ into a bimodule, we need to define an action of |
| $\End_R(N)$ on the right of $N$: say |
$\End_R(N)$ on the right of $N$: say |
| $$ x\cdot f = f(x) $$ |
$$ x\cdot f = f(x) $$ |
| But then we have a problem with the multiplication: |
But then we have a problem with the multiplication: |
| $$ x \cdot fg = fg(x) = f(g(x)) $$ |
$$ x \cdot fg = fg(x) = f(g(x)) $$ |
| but $$(x \cdot f) \cdot g = f(x) \cdot g = g(f(x))!$$ |
but $$(x \cdot f) \cdot g = f(x) \cdot g = g(f(x))!$$ |
| In order to make this work, we need to reverse the order of composition |
In order to make this work, we need to reverse the order of composition |
| when we define multiplication in the ring $\End_R(N)$ when it acts on the right. |
when we define multiplication in the ring $\End_R(N)$ when it acts on the right. |
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| There are essentially two different ways to go from here. One is to define the muliplication |
There are essentially two different ways to go from here. One is to define the muliplication |
| in $\End_R(N)$ the other way, which is most natural if we write the endomorphisms as functions |
in $\End_R(N)$ the other way, which is most natural if we write the endomorphisms as functions |
| on the right. This is the approach taken in many older books. |
on the right. This is the approach taken in many older books. |
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| The other is to leave the |
The other is to leave the |
| muliplication in $\End_R(N)$ the way it is, but to use the opposite ring to define the bimodule. |
muliplication in $\End_R(N)$ the way it is, but to use the opposite ring to define the bimodule. |
| This is the approach that is generally taken in more recent works. Using this approach, |
This is the approach that is generally taken in more recent works. Using this approach, |
| we conclude that $N$ is a $R$-$\End_R(N)^{\text{op}}$-bimodule. We will adopt this convention |
we conclude that $N$ is a $R$-$\End_R(N)^{\text{op}}$-bimodule. We will adopt this convention |
| for the lemma below. |
for the lemma below. |
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| Considering $R$ as a right and a left module over itself, |
Considering $R$ as a right and a left module over itself, |
| we can construct the two endomorphism rings $\End_R(R_R)$ and $\End_R({}_RR)$. |
we can construct the two endomorphism rings $\End_R(R_R)$ and $\End_R({}_RR)$. |
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| \begin{lemma} |
\begin{lemma} |
| Let $R$ be a ring with an identity element. |
Let $R$ be a ring with an identity element. |
| Then $R \isom \End_R(R_R)$ and $R \isom \End_R({}_RR)^{\text{op}}$. |
Then $R \isom \End_R(R_R)$ and $R \isom \End_R({}_RR)^{\text{op}}$. |
| \end{lemma} |
\end{lemma} |
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| \begin{proof} |
\begin{proof} |
| Define $\rho_r \in \End_R({}_RR)$ by $x \mapsto xr$. |
Define $\rho_r \in \End_R({}_RR)$ by $x \mapsto xr$. |
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| A calculation shows that $\rho_{rs} = \rho_s \rho_r$ (functions written on the left) |
A calculation shows that $\rho_{rs} = \rho_s \rho_r$ (functions written on the left) |
| from which it is easily seen that the map $\theta : r \mapsto \rho_r$ is a ring |
from which it is easily seen that the map $\theta : r \mapsto \rho_r$ is a ring |
| homomorphism from $R$ to $\End_R({}_RR)^{\text{op}}$. |
homomorphism from $R$ to $\End_R({}_RR)^{\text{op}}$. |
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| We must show that this is an isomorphism. |
We must show that this is an isomorphism. |
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| If $\rho_r = 0$, then $r = 1r = \rho_r(1) = 0$. So $\theta$ is injective. |
If $\rho_r = 0$, then $r = 1r = \rho_r(1) = 0$. So $\theta$ is injective. |
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| Let $f$ be an arbitrary element of $\End_R({}_RR)$, and let $r = f(1)$. |
Let $f$ be an arbitrary element of $\End_R({}_RR)$, and let $r = f(1)$. |
| Then for any $x \in R$, $f(x) = f(x1) = xf(1) = xr = \rho_r(x)$, so |
Then for any $x \in R$, $f(x) = f(x1) = xf(1) = xr = \rho_r(x)$, so |
| $f = \rho_r = \theta(r)$. |
$f = \rho_r = \theta(r)$. |
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| The proof of the other isomorphism is similar. |
The proof of the other isomorphism is similar. |
| \end{proof} |
\end{proof} |
