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I have some confusion in Atiyah Commutative algebra

Here is an outline

Let $A$ be a ring and $\mathcal{m}$ a maximal ideal of $A$, such that every element of $1+\mathcal{m}$ is a unit in $A$. Then $A$ is a local ring.

Proof :Let $x\in A\setminus\mathcal{m}$. Since $m$ is maximal, the ideal generated by $x$ and $\mathcal{m}$ is $(1)$. Hence there exist $y \in A$ and $t \in m$ such that $xy +t=1$

My thinking : Here $(1)= A$ and $x\in A\setminus\mathcal{m}$. The ideal generated by $x$ and $\mathcal{m}$ is strictly larger than $\mathcal{m}$

$$ \implies (x)+ \mathcal{m}= (1)=A$$.

there exist $y \in A$ and $ t \in$ m such that $xy +t=-1$ since $-1 \in A$

My question : Can i take $-1$ instead of $1$ in $xy +t=1$ ? I mean $xy +t=-1$

jasmine
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1 Answers1

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Sure. The important point is that any unit will generate the entire ring, and knowing that $(x)+\mathfrak{m} = (1)=A$ implies that there exist some $y\in A$ and $t\in\mathfrak{m}$ such that $xy+t = u$ for some unit $u$, in which case $xy=u-t$. But recall that the condition you are given is that every element of $1+\mathfrak{m}$ is a unit, so really you are more interested in the fact that $$xyu^{-1}=uu^{-1}-tu^{-1}=1-t$$ so that $xyu^{-1}$ is a unit (since $1-t\in 1+\mathfrak{m}$), in which case $x$ is a unit. But since $x\in A-\mathfrak{m}$ was arbitrary, it follows from part i) of the same proposition (1.6) that $\mathfrak{m}$ is maximal, and that $A$ is a local ring.

(That $x$ is a unit follows from the fact that $xyu^{-1}$ has some inverse $z\in A$, and hence $yu^{-1}z$ is inverse to $x$.)

Horizon
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