Skip to main content
MathOverflow

Return to Answer

added 235 characters in body
Source Link
Floresza
Floresza
  • 161
  • 5

First consider the case in which $(R, \mathfrak m)$ is local. IfIf we assume that $R/\mathfrak m$$\text{Spec}(R)$ is projectiveconnected, then it is a summand of $R$ is always a field (by consideringnote the natural surjection $R\rightarrow R/\mathfrak m$spectrum of a local ring is always connected). Now $R = I\oplus R/\mathfrak m$ for some (necessarily finitely generated) ideal $I$ and $I$ This is such that $I = \mathfrak m I$equivalent to the nonexistence of nontrivial idempotents. By Nakayamma's lemma, $I = 0$ and $R$ Indeed, hence $R = R/\mathfrak m$ and $R$ isover a field.

In the general casecommutative ring any, simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$. Write $R = R/\mathfrak m \oplus I$, for some finitely generated ideal $I$ of $R$. By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$ and. This implies $I$ is generated by an idempotent, since I can be taken to be finitely generated. .

Note if we assume thathence $\text{Spec}(R)$ is connected$I = 0$, thensince $R$$R/\mathfrak m$ is always a fieldnonzero.

First consider the case in which $(R, \mathfrak m)$ is local. If $R/\mathfrak m$ is projective, then it is a summand of $R$ (by considering the natural surjection $R\rightarrow R/\mathfrak m$). Now $R = I\oplus R/\mathfrak m$ for some (necessarily finitely generated) ideal $I$ and $I$ is such that $I = \mathfrak m I$. By Nakayamma's lemma, $I = 0$ and $R$, hence $R = R/\mathfrak m$ and $R$ is a field.

In the general case any, simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$. By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$ and $I$ is generated by an idempotent, since I can be taken to be finitely generated. .

Note if we assume that $\text{Spec}(R)$ is connected, then $R$ is always a field.

If we assume that $\text{Spec}(R)$ is connected, then $R$ is always a field (note the spectrum of a local ring is always connected). This is equivalent to the nonexistence of nontrivial idempotents. Indeed, over a commutative ring any simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$. Write $R = R/\mathfrak m \oplus I$, for some finitely generated ideal $I$ of $R$. By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$. This implies $I$ is generated by an idempotent, hence $I = 0$, since $R/\mathfrak m$ is nonzero.

added 235 characters in body
Source Link
Floresza
Floresza
  • 161
  • 5

Over a commutative ring $R$, any simple module hasFirst consider the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$. Hence we may localize at $\mathfrak m$ and assumecase in which $R$$(R, \mathfrak m)$ is local. If $R/\mathfrak m$ is projective, then it is a summand of $R$ (by considering the natural surjection $R\rightarrow R/\mathfrak m$). Now $R = I\oplus R/\mathfrak m$ for some (necessarily finitely generated) ideal $I$ and $I$ is such that $I = \mathfrak m I$. By Nakayamma's lemma, $I = 0$ and $R$, hence $R = R/\mathfrak m$ and $R$ is a field.

In the general case any, simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$. By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$ and $I$ is generated by an idempotent, since I can be taken to be finitely generated. .

Note if we assume that $\text{Spec}(R)$ is connected, then $R$ is always a field.

Over a commutative ring $R$, any simple module has the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$. Hence we may localize at $\mathfrak m$ and assume $R$ is local. If $R/\mathfrak m$ is projective, then it is a summand of $R$ (by considering the natural surjection $R\rightarrow R/\mathfrak m$). Now $R = I\oplus R/\mathfrak m$ for some (necessarily finitely generated) ideal $I$ and $I$ is such that $I = \mathfrak m I$. By Nakayamma's lemma, $I = 0$ and $R$, hence $R = R/\mathfrak m$ and $R$ is a field.

First consider the case in which $(R, \mathfrak m)$ is local. If $R/\mathfrak m$ is projective, then it is a summand of $R$ (by considering the natural surjection $R\rightarrow R/\mathfrak m$). Now $R = I\oplus R/\mathfrak m$ for some (necessarily finitely generated) ideal $I$ and $I$ is such that $I = \mathfrak m I$. By Nakayamma's lemma, $I = 0$ and $R$, hence $R = R/\mathfrak m$ and $R$ is a field.

In the general case any, simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$. By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$ and $I$ is generated by an idempotent, since I can be taken to be finitely generated. .

Note if we assume that $\text{Spec}(R)$ is connected, then $R$ is always a field.

added 132 characters in body
Source Link
Floresza
Floresza
  • 161
  • 5

For a slightly more down-to-earth explanation, notice overOver a commutative ring $R$, any simple $R$-modulemodule has the form $R/\mathfrak m$ for somea maximal ideal $\mathfrak m$. Hence we may localize at $\mathfrak m$ and assume $R$ is local. If $R/\mathfrak m$ is projective, then it is a summand of some free $R$-module (by considering the natural surjection $R\rightarrow R/\mathfrak m$). Since free Now $R$-modules are torsion-free,$R = I\oplus R/\mathfrak m$ for some $\mathfrak m = 0$, hence(necessarily finitely generated) ideal $R$$I$ and $I$ is fieldsuch that $I = \mathfrak m I$. In particular By Nakayamma's lemma, $R/\mathfrak m$ =$I = 0$ and $R$, hence $R = R/\mathfrak m$ and $R$ is injective!a field.

For a slightly more down-to-earth explanation, notice over a commutative ring $R$, any simple $R$-module has the form $R/\mathfrak m$ for some maximal ideal $\mathfrak m$. If $R/\mathfrak m$ is projective, then it is a summand of some free $R$-module. Since free $R$-modules are torsion-free, $\mathfrak m = 0$, hence $R$ is field. In particular, $R/\mathfrak m$ = $R$ is injective!

Over a commutative ring $R$, any simple module has the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$. Hence we may localize at $\mathfrak m$ and assume $R$ is local. If $R/\mathfrak m$ is projective, then it is a summand of $R$ (by considering the natural surjection $R\rightarrow R/\mathfrak m$). Now $R = I\oplus R/\mathfrak m$ for some (necessarily finitely generated) ideal $I$ and $I$ is such that $I = \mathfrak m I$. By Nakayamma's lemma, $I = 0$ and $R$, hence $R = R/\mathfrak m$ and $R$ is a field.

Source Link
Floresza
Floresza
  • 161
  • 5