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For $A$, a commutative ring with identity, show $J(A)=\{x\in A:xy-1 \in A^\times, \forall y \in A\}$, $J(A)$ being the Jacobson radical. I should add here that the Jacobson radical is the intersection of all the maximal ideals of $A$.

Firstly, $J(A)$ is itself an ideal (proof omitted).

Now, assume $x \in J(A)$, and consider the natural map $\phi:A \to A/J(A)$, where $x$ is mapped to the zero element in the quotient ring. Now, supposed $y$ is any element from $A$. Then $$\phi(xy-1)=\phi(x)\phi(y)-1=-1=\phi(-1)$$ So, $xy-1=-1$ in $A$, so $xy-1$ is a unit in $A$.

Now, for the other direction ( this is where I am having trouble ). Lets, assume that $ x\in A:xy-1 \in A^\times, \forall y \in A$. Now supposed that $x \notin J(A)$ and consider the same map as above. Then we must have that, $$\phi((xy-1)(xy-1)^{-1})=(\phi(x)\phi(y)-\phi(1))\phi(xy-1)^{-1}=1$$ $$\iff \phi(x)\phi(y)\phi(xy-1)^{-1}-\phi(xy-1)^{-1}=1$$ This needs to be true for all $y$, so I take $y=1$, so the expression reduces to $$\iff \phi(x)\phi(x-1)^{-1}-\phi(x-1)^{-1}=1$$ I think this should lead to a contradiction, since we can see that $x\ne1$, or that would imply $1=0$. But I cannot see how to conclude that this always is a contradiction? Am I on the right track here?

Update

After reviewing and thinking a little more, I don't seem to be using any specific properties of the Jacobson radical. I am thinking I must need to use the property of $J(A)$ being a maximal ideal itself in a commutative ring, so $J(A)$ would be a prime ideal. This would imply that $A/J(A)$ is a field. Still now quite sure how to use this.

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    $\begingroup$ I am thinking I must need to use the property of $J(A)$ being a maximal ideal itself in a commutative ring, so $J(A)$ would be a prime ideal. This would imply that $A/J(A)$ is a field. Still now quite sure how to use this.. Don't bother, because that isn't a property of the Jacobson radical, and such a line of reasoning would be incorrect. $\endgroup$ – rschwieb May 19 at 18:14
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    $\begingroup$ A proof is given here $\endgroup$ – rschwieb May 19 at 18:19
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Let $x \in J(A)$. Suppose, for contradiction, that $1 - ax$ is not a unit for some $a \in A$. Then $(1-ax)$ is a proper ideal (since it does not contain $1$), so it is contained in some maximal ideal $\mathfrak{m}$ of $A$. But since $x \in J(A)$, we have that $x \in \mathfrak{m}$ so $ax \in \mathfrak{m}$, and hence $1 = (1 - ax) + ax \in \mathfrak{m}$, which is a contradiction.

Now suppose that $x \not \in J(A)$. Then there is some maximal ideal $\mathfrak{m}$ such that $x \not \in \mathfrak{m}$, so $\mathfrak{m} + (x)$ is an ideal strictly containing $\mathfrak{m}$, so by maximality of $\mathfrak{m}$, we have $\mathfrak{m} + (x) = A$. Thus, there exist $m \in \mathfrak{m}$ and $a \in A$ such that $m + ax = 1$, so $1 - ax = m \in \mathfrak{m}$, which means that $1 - ax$ is not a unit.

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You don't want to look at $A/J(A)$, you want to look at $A/\mathfrak{m}$ for each maximal ideal $\mathfrak{m}$. For the first part: $\phi(xy - 1) = \phi(-1)$ does not mean that $xy - 1 = -1$. Instead, assume that $xy - 1$ is not invertible, then $xy - 1$ will be contained in a maximal ideal $\mathfrak{m}$.

Conversely, suppose that $x$ does not belong to some maximal ideal $\mathfrak{m}$, then $x$ is a unit in $A/\mathfrak{m}$.

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  • $\begingroup$ It is quite troubling if what you say is true, here, that $\phi(xy-1)=\phi(-1)$ yes i realize now that should only be true for an injective map, which we are not guaranteed here. $\endgroup$ – jeffery_the_wind May 20 at 1:33

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