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proposition: x∈The Jacobson radical <=> 1-xy is a unit in commutative ring A for all y∈A

I have proved (=>)

I don't figure out a detail of the proof of (<=).

Here is the proof on book:

suppose x∉m for some maximal ideal m.

then m and x generate the unit ideal (1). (*Why?*It's the only thing I confused about in this proof)

.......

Here is what I thought:

(m) is still (m)

(x)=Ax=mx+(A-m)x=m+(A-m)x

so I only need to prove that (A-m)x=A-m

Then I stuck at here .

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    $\begingroup$ Isn't it just that since $ m $ is maximal, the only ideal containing $ m\cup\{x\} $ is the unit ideal? $\endgroup$
    – Unwisdom
    Feb 25, 2014 at 22:59

1 Answer 1

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By definition $\mathfrak{m} \subset A$ is maximal if and only if for all other ideals $I$ for which $ \mathfrak{m} \subseteq I \subseteq A$ holds, either $I = \mathfrak{m}$ or $I = A$.

The ideal $(\mathfrak{m}, x)$, generated by $\mathfrak{m}$ and $x$ indeed sits in between $\mathfrak{m}$ and $A$ as above, i.e. $$ \mathfrak{m} \subseteq (\mathfrak{m}, x) \subseteq A. $$ However, it cannot be equal to $\mathfrak{m}$, as $x \not \in \mathfrak{m}$. Therefore $(\mathfrak{m},x) = A$.

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