Is there a brute force proof that the first-order characterization of a Jacobson radical is an ideal?

Question

Let rings be commutative and unital.

Is there a brute force proof that the first-order characterization of a Jacobson radical is an ideal?

Basically, for context, I'm having some trouble wrapping my head around the Jacobson radical and the three definitions I've seen for it (the first two shown here and the simple module definition). I have no intuition for what the Jacobson radical does or why we care about it, but I think that will come with time.

For one thing, it's surprising to me that the intersection of all maximal ideals has a first-order characterization in the first place, since it doesn't feel very first-order to me in a vague intuitive sense.

Anyway, as an exercise, I am sitting down and proving that the two simplest definitions (given below) are equivalent in the first place.

The very first thing that I tried to do is to show that the first-order characterization is actually an ideal. This turned out to be too hard so I just showed that the two definitions are equivalent (assuming that the proof in Appendix A works).

Being an ideal is also a first-order property, so there should be a way to prove that the Jacobson radical is an ideal without leaving first-order land.

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Consider the following two definitions, which both define the Jacobson radical.

1. The intersection of all maximal ideals, call it $I(R)$.
2. The set of all ring elements $x$ such that $\forall y \mathop. 1-xy \;\text{is a unit}$, call it $J(R)$.

The simplest way I can think of to show that $J(R)$ is an ideal is to show that it is exactly $I(R)$. See the proof in Appendix A.

Checking that $I(R)$ actually defines an ideal without knowing what it is is pretty hard.

Strong multiplicative closure is trivial. If $\forall y \mathop. 1-xy \;\text{is a unit}$ then it holds that $\forall y \mathop. 1 - xyr \;\;\text{is a unit}$ where $r$ is an arbitrary ring element.

Getting additive closure of $I(x)$, though is much harder. To prove to myself that it is an ideal, I ended up just showing that the first and second definitions are equal, from which I can lean on the known result that the intersection of a family of ideals is an ideal.

This naturally leads me to the question of whether there's a way to prove that the first-order characterization is an ideal that's more naive and brute force, and doesn't involve appealing to another equivalent characterization?

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Appendix A: $I(R)$ is equal to $J(R)$.

Lemma 101: For any element $c$ of $J(R)$ and any proper ideal $\alpha$, $(\alpha, c)$ is proper.

I will show that if $(\alpha, c)$ is not proper, then $c$ does not satisfy the defining formula of $J(R)$.

Suppose for contrapositive that $(\alpha, c)$ is not proper. Thus there exists $r \in \mathbb{R}$ and an $a \in \mathbb{\alpha}$ so that $a + rc = 1$. It follows that $a = 1 - rc$. However, $\alpha$ is proper and hence $a$ is not a unit. Thus it holds that $\exists y \mathop. 1 - yc \; \text{is not a unit}$, which is precisely the negation of (2), the defining formula of $J(R)$.

End of proof of Lemma 101.

Lemma 102: $J(R) \subset I(R)$.

Consider an element $x$ that is in $J(R)$. Consider an arbitrary maximal ideal $\alpha$. Since $(\alpha, x)$ is a proper ideal by Lemma 102, it follows that $\alpha = (\alpha, x)$ and thus $x \in \alpha$. Since $x$ is in every maximal ideal $\alpha$, $x$ is in the intersection of all maximal ideals.

End of proof of Lemma 102.

Lemma 103: $I(R) \subset J(R)$.

Suppose for contrapositive that $x$ is not in $J(R)$.

There exists a $w$ such that $1-wx$ is not a unit.

Let $\alpha$ be a maximal ideal containing $(1-wx)$. Every non-unit is contained in some maximal ideal, which can be shown by transfinite induction.

If $\alpha$ contained $x$, then it would also contain $wx$ and hence $1-wx+wx = 1$, making it a unit ideal.

Thus, there is at least one maximal ideal that does not contain $x$, as desired.

Answer 1

Suppose $x, y \in J(R)$. We want to show that $x + y \in J(R)$, meaning we want to show that for every $z \in R$ we have that $1 - z(x + y)$ is invertible. By hypothesis, $1 - zx$ is invertible, with some inverse $u$. This gives

$$u(1 - zx - zy) = 1 - uzy$$

and by hypothesis, $1 - uzy$ is also invertible, with some inverse $v$. This gives

$$uv(1 - zx - zy) = 1$$

so $1 - zx - zy$ is invertible with inverse $uv$, as desired.

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Some additional comments.

I have no intuition for what the Jacobson radical does or why we care about it, but I think that will come with time.

Loosely speaking, the Jacobson radical is one way of describing what the "negligible" elements of a ring $R$ are. The specific sense of "negligible" here is that these are elements that are "invisible" with respect to all quotients $R \to R/m$ by maximal ideals. In algebraic geometry terms these are the closed points of the affine scheme $\text{Spec } R$, so the Jacobson radical is describing the elements of $R$ which "can't be detected by closed points."

A nice special case to think about is that if $R$ is (commutative and) Artinian then the Jacobson radical agrees with the nilradical; these guys are "negligible" in a maybe more intuitive sense. But the Jacobson radical is also capable of detecting that, for example, the element $x \in k[[x]]$ in a formal power series ring ought to also qualify as "negligible" in some sense; it is in some sense "almost nilpotent."

As for what this has to do with the first-order characterization, if you haven't done this already, show as an exercise that if $x$ is nilpotent then $1 - xy$ is always invertible. So here there's a more direct connection to nilpotence; loosely speaking that $1 - xy$ condition says that $x$ is "very far from being invertible."

For one thing, it's surprising to me that the intersection of all maximal ideals has a first-order characterization in the first place, since it doesn't feel very first-order to me in a vague intuitive sense.

I agree that this is surprising! If I were to try to make it less surprising I think I'd start with the first-order definition $J(R)$ and try to work the other way. So, what can we say about elements $x \in J(R)$?

To me the first observation is that this condition that $1 - xy$ is always invertible naturally descends to quotients: if $x \in J(R)$ satisfies this property then so does its image in any quotient $R/I$. A second observation is that if $m$ is a maximal ideal, so that $R/m$ is a field, this condition is equivalent to the condition that $x = 0$, because any nonzero element of $R/m$ is invertible. So we've already learned that $x$ must lie in the intersection of all maximal ideals.

To me this argument suggests that the relationship between the two definitions of the Jacobson radical is analogous to the following: you can characterize a unit $u \in R$ either as an element with an inverse, which is the first-order characterization, or as an element which does not lie in any proper ideal $I$; equivalently, as an element which is nonzero in every nonzero quotient $R/I$. As in the above the key observation is that ring homomorphisms preserve units.