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Let $A$ be a ring and $\frak{m}$ a maximal ideal of $A$. Let $\kappa$ be the field $A/\frak{m}$.

How to show that the $\kappa$-linear natural map $$ \mathfrak{m}/\mathfrak{m}^{2}\rightarrow\mathfrak{m}A_{\mathfrak{m}}/\mathfrak{m}^{2}A_{\mathfrak{m}} $$ is an isomorphism?

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1 Answer 1

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First, there is a natural homomorphism $A\rightarrow A_{\mathfrak{m}}$ where $a\in A\mapsto a/1\in A_{\mathfrak{m}}$. From this map, we can define a homomorphism $$\mathfrak{m}/\mathfrak{m}^2\rightarrow\mathfrak{m}A_{\mathfrak{m}}/\mathfrak{m}^2A_{\mathfrak{m}}.$$ Given by: $$a+\mathfrak{m}^2\in\mathfrak{m}/\mathfrak{m}^2\mapsto a/1+\mathfrak{m}^2A_{\mathfrak{m}}.$$ Now, you have to show the following:

1) The map is well-defined.

2) The map is surjective.

3) The map is injective.

1) Well-Defined: Suppose that $a+\mathfrak{m}^2=b+\mathfrak{m}^2$. Then, $a-b\in\mathfrak{m}^2$. But then, $a/1-b/1=(a-b)/1\in\mathfrak{m}^2A_{\mathfrak{m}}$.

2) Surjective: Suppose $a/b+\mathfrak{m}^2A_{\mathfrak{m}}\in\mathfrak{m}A_\mathfrak{m}/\mathfrak{m}^2A_\mathfrak{m}.$ This means that $a/b\in\mathfrak{m}A_\mathfrak{m}$, so $a\in\mathfrak{m}$ and $b\not\in\mathfrak{m}$. Now, consider the image of $a+\mathfrak{m}^2$, which is $a/1+\mathfrak{m}^2A_{\mathfrak{m}}$. The difference between these two elements is $\frac{a(b-1)}{b}+\mathfrak{m}^2A_{\mathfrak{m}}$. Since $\mathfrak{m}$ is maximal and $b\not\in\mathfrak{m}$, $b-1\in\mathfrak{m}$ (otherwise the larger ideal $\langle b\rangle+\mathfrak{m}$ would not be the entire ring).

3) Injective: Suppose that $a/1+\mathfrak{m}^2A_{\mathfrak{m}}=b/1+\mathfrak{m}^2A_{\mathfrak{m}}$ where $a,b\in\mathfrak{m}$. Then, $a-b\in\mathfrak{m}^2A_{\mathfrak{m}}$, so $a-b=c/d$ where $c\in\mathfrak{m}^2$ and $d\not\in\mathfrak{m}$. Then there is some $e\in A\setminus\mathfrak{m}$ such that $e[d(a-b)-c]=0$. Therefore, $ed(a-b)=ec$. Since $\mathfrak{m}$ is maximal, it is prime, and since $e,d\not\in\mathfrak{m}$, then $ed\not\in\mathfrak{m}$. Now, since $A/\mathfrak{m}$ is a field and $ed\not\in\mathfrak{m}$, there is some $f\in A$ such that $edf=1+g$ where $g\in\mathfrak{m}$. By multiplying both sides by $f$, we have $fed(a-b)=fec$. Therefore, $$ (a-b)=fec-g(a-b). $$ Since $c\in\mathfrak{m}^2$, $g\in\mathfrak{m}$ and $(a-b)\in\mathfrak{m}$, $fec-g(a-b)\in\mathfrak{m}^2$, so $a-b\in\mathfrak{m}^2$, so $a+\mathfrak{m}^2=b+\mathfrak{m}^2$.

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