# 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.

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.

• 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. – rschwieb May 19 at 18:14
• A proof is given here – rschwieb May 19 at 18:19

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.
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 $$A/\mathfrak{m}$$.
• 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. – jeffery_the_wind May 20 at 1:33