# The Jacobson radical and quasi-regular elements in polynomial rings — trouble understanding a proof.

The concept of the Jacobson radical is quite new to me and I have to give a talk about a certain paper concerning it. I agreed to do this to find motivation to finally study this concept properly, as the paper was advertised to be accessible. Part of it is, but I ran into trouble when the author actually started using facts about the Jacobson radical. The paper is On the Jacobson Radical of Polynomial Rings by J. Krempa (1974).

Rings aren't necessarily unital here. An element $r$ of a ring $R$ is said to be quasi-regular when there is an element $s\in R$ such that $$r+s-rs=0.$$ (At least this seems to be the definition the author is using. Jacobson calls such elements right quasi-regular.) $R_n$ will denote the ring of $n\times n$-matrices over $R.$ A ring is nil iff all of its elements are nilpotent. A ring is radical iff it's equal to its Jacobson radical.

The aim of the paper is to prove the following statement.

Let $R$ be any ring. Then $R[x]$ is radical iff for all $n$ the ring $R_n$ is nil.

The proof starts with the following lemma.

For any ring $R$ and any $n\geq 1,$ there exists a monomorphism $f_n:R[[x]]\longrightarrow R_n[[x]]$ such that

(i) unimportant (ensuring that when $R$ is an algebra, the monomorphism respects the additional structure);

(ii) $f_n(R[x])=f_n(R[[x]])\cap R_n[x];$

(iii) If $p\in R[x]$ and $\deg p\leq k n,$ then $\deg f_n(p)\leq k.$

The proof of this lemma relies on two quite simple tricks. First, we send an element $p\in R[[x]]$ to an infinite matrix over $R$ like here. Then we "cut" the matrix into $n\times n$ blocks, obtaining an infinite matrix over $R_n.$ Then we notice that the resulting matrix is the image of a power series $r\in R_n[[x]],$ and we put $$f_n(p)=r.$$

The restriction of $f_n$ to $R[x]$ is a monomorphism to $R_n[x].$

Now, there's a second lemma, the one I'm having trouble with.

Let $R$ be any ring. Then if $R_n$ is a nil ring, any polynomial of degree $\leq n$ is quasi-regular in $R[x].$

The author takes any $p\in R[x]$ of degree $\leq n.$ He notes that $\deg f_n(p)\leq 1.$ And he says that $f_n(p)$ is quasi-regular in $R_n[x],$ which he doesn't explain and I don't understand. We have $f_n(p)=ax+b,$ where $a,b\in R_n[x]$ are nilpotent. I know that nilpotent elements are always quasi-regular, but in a noncommutative ring, the sum of two nilpotents doesn't have to be nilpotent. $(1)$ Why is $f_n(p)$ quasi-regular in $R_n[x]?$

The author denotes by $q$ the element of $R_n[x]$ such that $f_n(p)+q-f_n(p)q=0.$ Then the paper says this.

Since $R[[x]]$ is radical, there exists $p'\in R[[x]]$ such that $p+p'+pp'=0.$

$(2)$ Why is $R[[x]]$ radical? $(3)$ And why does it imply that $p$ is quasi-regular? From this Wikipedia page, I know that the Jacobson radical is the set of all elements $a$ of $R$ such that for any $r\in R,$ $ar$ is quasi-regular. So a radical ring is a ring in which any product of two elements is quasi-regular. But what about the elements which aren't products? There can be such elements in non-unital rings...

After that the author notes that $$f_n(p)+f_n(p')-f_n(p)f_n(p')=0,$$ and this is what follows:

The ring $R_n[[x]]$ also being radical, the element $f_n(p)$ has exactly one quasi-inverse. Therefore $q=f_n(p')\in f_n(R[x]),$ i.e. $p'\in R[x],$ since $f_n$ is a monomorphism of $R[x].$

I'm completely confused at this moment. I'm not even sure what the author means by "quasi-inverse". He seems to have changed the definition of "quasi-regular" to a one-sided version. $(4)$ Does it apply also to "quasi-inverse"?

Then it is said that from this lemma the non-obvious implication of the theorem follows.

EDIT Thanks to rschwieb's reference, I've been able to find answers to questions $(3)$ and $(4)$.

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It looks like it will take some thinking. I have a feeling that the problem of showing the product/sum of noncommuting nilpotent elements is circumvented by the matrix ring being nil, because its entries will record long sums/products of noncommuting elements. Is the author using induction in lemma 2? – rschwieb May 9 '12 at 19:22
@rschwieb I don't think he does. At least he doesn't explicitly say he does. – user23211 May 9 '12 at 19:26

This is not really an explanation about the paper, but it is meant to help with the notion of quasiregularity here.

If you denote $r\circ s:=r+s-rs$, then you can show that $\circ$ is a binary operation on $R$ which is associative and has unity element $0$. (I read about this in Heatherly and Tucci's paper The circle semigroup of a ring, but I think Jacobson is usually credited for the idea.) If the ring has identity, then $1$ actually acts as an absorbing element! That is $a\circ 1=1\circ a=1$ for all $a\in R$. This is quite the reverse casting of 0 and 1 compared to the original operation!

Of course we can consider right and left ("quasi"-)inverses in this operation $\circ$. If you check, the definition you wrote for quasiregular was: $r\circ s=0$, that is, $s$ was a right inverse for $r$ (remember that 0 is the identity). As with any associative binary operation, if $r$ has any left inverse, then it has to be equal to $s$, but it is also possible that it has no left inverses.

Lam puts some exercises about this in First course in Noncommutative rings (and so you would also be able to see it in his Exercises in classical ring theory also)

• when $R$ has $1$, the circle semigroup is monoid isomorphic to the semigroup of the ring
• if a right ideal consists entirely of right quasiregular elements, then the elements are also left-quasiregular.
• $\{a\in R\mid aR \text{ is right quasiregular}\}=rad(R)$
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How are rings without identity and their radical important? It's a silly question to ask, because I am already convinced of their utility when there is a unit, but it seems to me we can easily go from a ring without identity to one with one, by adjoining a unit, just add $\mathbb{Z}$, which also explains the circle product and which brings the radical theory back to that of rings. I am really confused. – Olivier Bégassat May 9 '12 at 22:29
Jacobson meticulously explored many ring/module theoretic problems in great generality, especially without the convenience of an identity. Several natural structures (Lie algebras, Baer and Rickart rings from functional analysis) are interesting without identity. Adjoining a unit does not trivialize the theory of rings without unity. Also see this:. I prefer rings with unity but I'm weary of the "adjoin 1 and forget" mantra. – rschwieb May 9 '12 at 23:41
@OlivierBégassat The theory of ring radicals is simply simpler without he requirement that rings contain a unity. It's convenient to consider ideals to be subrings. If I remember correctly, in one proof of Wedderburn's theorem, the difficult part is to prove that there is a unity in a certain ring, which makes the rest of the proof quite simple. This is what my mentor told me, and other ring theorists in my department have expressed a similar opinion. Another example is that it would be impossible to say "nil ring" if all rings had identity elements. – user23211 May 9 '12 at 23:53
@rschwieb I'm going through the exercises along with Jacobson's book and it appears that two of my questions are going to be answered. – user23211 May 9 '12 at 23:54
I have added two answers. I would be very grateful if you'd take a look at them. I seem to be missing something... – user23211 May 17 '12 at 20:14

This is the proof of the theorem in the paper, using lemmas proved in my other answer.

Theorem. Let $R$ be any ring. Then $R[x]$ is a radical ring iff for all $n$ we have that $R_n$ is nil.

Proof. To prove $\Longrightarrow,$ we use this fact. Let $R[x]$ be radical. By the linked theorem, for every $n$ we have that $(R[x])_n$ is radical, and therefore $R_n[x]$ is radical by Fact 5. It follows from Fact 4 that $R_n$ is nil for every $n$.

Conversely, suppose that $R_n$ is nil for all $n$. Let $p\in R[x], \;\deg p\leq n$. Then $\deg f_n(p)\leq 1.$ By Fact 3, $f_n(p)$ is quasi-regular in $R_n[x],$ that is we have a $q\in R_n[x]$ such that $$f_n(p)+q-f_n(p)q=0.$$

By Fact 2, $R[[x]]$ is radical. Therefore, there exists $p'\in R[[x]]$ such that $$p+p'-pp'=0.$$ Then also $$f_n(p)+f_n(p')-f_n(p)f_n(p')=0.$$ But $R_n[[x]]$ is also radical by Fact 2. Therefore, $q=f_n(p'),$ whence $p'\in R[x].$ Therefore $R[x]$ is radical.

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This post contains the necessary lemmas. The proof of the main theorem is in my other answer.

I still don't require rings to have identity elements. I will first define a radical ring.

Definition. A ring $R$ is radical iff every element of $R$ is quasi-regular, that is has a quasi-inverse. Equivalently, $R$ is radical iff every element of $R$ has a left (right) quasi-inverse.

Notation. Every ring is an algebra over $\mathbb Z,$ and as such can be embedded in a standard way in a unital algebra over $\mathbb Z.$ I will denote this algebra by $R^*.$

Fact 1. An element $p\in R[[x]]$ for some unital ring $R$ is invertible iff its constant coefficient is invertible in $R.$

Proof. We use the embedding given here. An upper-triangular infinite matrix over $R$ with all entries on the diagonal equal to $a\in R$ is invertible in the ring of all infinite matrices iff $a$ is invertible in $R.$ It is a matter of a simple calculation to prove that if $A\in M_\triangledown(R)$ (as defined on the linked page) is invertible in $M_\infty(R),$ then it is also invertible in $A\in M_\triangledown(R).$

Fact 2. Let $R_n$ be a nil ring for some $n$. Then $R[[x]]$ is a radical ring.

Proof. If for some $n$ we have that $R_n$ is nil, then $R$ is nil, because $R$ can be embedded in $R_n.$ Let

$$x=a_0+a_1x+a_2x^2+\ldots\in R[[x]].$$

Since $a_0$ is nilpotent, we have that $1-a_0$ is invertible in $R^*.$ Then $$1-x=1-(a_0+a_1x+a_2x^2+\ldots)=(1-a_0)-a_1x-a_2x^2-\ldots$$

has a constant coefficient that is invertible in $R^*.$ Therefore, by Fact 1, $1-x$ is invertible in $R^*[[x]],$ and so $x$ is quasi-regular.

Fact 3. If $R$ is a nil ring, and $f\in R[x]$ such that $\deg f\leq 1,$ then $f$ is quasi-regular.

Proof. Let $f=a+bx\in R[x].$ Since $a$ is nilpotent, $1-a$ is a unit in $R^*.$ Since $R$ is an ideal in $R^*$, we have that $$(1-a)^{-1}b\in R,$$ and so $(1-a)^{-1}b$ is nilpotent. Therefore $$1-f=(1-a)-bx=(1-a)(1 - (1-a)^{-1} b x)$$ is a product of two units in $R^*,$ and therefore a unit. (Thanks to Matt E for pointing out a mistake in this part of the proof.) We note that this would not work for a higher degree of $f$.

Fact 4. If $R[x]$ is a radical ring, then $R$ is nil.

Proof. Let $r\in R.$ Since $R[x]$ is radical, there is $f\in R^*[x]$ such that $$(1-rx)f(x)=1.$$ But also $$(1-rx)(1+rx+r^2x^2+\ldots)=1.$$ But then we must have $$R^*[x]\ni f=1+rx+r^2x^2+\ldots.$$ Hence $r^n=0$ for some $n$, and $r$ is nilpotent.

Fact 5. For any ring $R,$ we have $R_n[x]\cong (R[x])_n.$

Proof. $(R[x])_n$ is an algebra over the center of $R[x]$ with entry-wise scalar multiplication. Of course, $x$ is an element of that center. Let

$$\begin{pmatrix} (r_0^{11}+r_1^{11}x+\ldots+r_{k_{11}}^{11}x^{k_{11}}) & \ldots & (r_0^{1n}+r_1^{1n}x+\ldots+r_{k_{1n}}^{1n}x^{k_{1n}}) \\ \vdots & \ddots & \vdots \\ (r_0^{n1}+r_1^{n1}x+\ldots+r_{k_{n1}}^{n1}x^{k_{n1}}) & \ldots & (r_0^{nn}+r_1^{nn}x+\ldots+r_{k_{nn}}^{nn}x^{k_{nn}}) \end{pmatrix} \in (R[x])_n,$$

where all $k_{ij}$ are the indices of the maximal non-zero coefficients in the polynomials (if the polynomials are nonzero). We can take $k=\max_{i,j}k_{ij}.$ Then we can write $k$ instead of $k_{ij}$ everywhere above, taking the redundant $r_{ij}$ to be zero. Then the above is equal to

$$\sum_{i=1}^k \begin{pmatrix} r_i^{11}x^i & \ldots & r_i^{1n}x^i \\ \vdots & \ddots & \vdots \\ r_i^{n1}x^i & \ldots & r_i^{nn}x^i \end{pmatrix} = \sum_{i=1}^k \begin{pmatrix} r_i^{11} & \ldots & r_i^{1n} \\ \vdots & \ddots & \vdots \\ r_i^{n1} & \ldots & r_i^{nn} \end{pmatrix} x^i,$$

which is nothing less than a polynomial in $R_n[x].$ I don't know how to argue that this ends the proof but I dare say it is obvious that it does. It's clear that considering the last sum to be in $R_n[x]$ is licit.

These are all of the lemmas needed to understand the proof in the paper I believe. I will post the proof shortly, but in another answer because this one is already barely possible to type in.

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Dear ymar, In the proof of Fact 2, I don't think it's the nilpotence of $b$ that you use, but rather the nilpotence of $(1-a)^{-1} b$. (This element is nilpotent, since it belongs to the nil ring $R$.) Regards, – Matt E May 17 '12 at 2:42
@MattE Thank you for the comment. I see why $(1-a)^{-1}b$ being nilpotent works. But why doesn't my argument do? – user23211 May 17 '12 at 7:19
Dear ymar, Perhaps I'm just being stupid, but since $R$ is not necessarily commutative, $b$ being nilpotent doesn't imply that $(1-a)^{-1}b$ is nilpotent (without the nil hypothesis on the entire ring). Now $(1-a) + b x = (1-a)(1 + (1-a)^{-1} b x),$, and so to show that $(1-a) + b x$ is invertible, you need to show that $1 + (1-a)^{-1}b x$ is invertible, which needs nilpotence of $(1-a)^{-1} b$. Regards, – Matt E May 17 '12 at 11:42
@MattE I've corrected the proof. Is is fine now? I hope it is because it's very difficult to edit it because of all the latex! – user23211 May 17 '12 at 19:55
@MattE It's me who was being stupid... For some reason I believed that the sum of a unit and a nilpotent is always a unit in any ring, which is clearly not true because for example $$\begin{pmatrix}1&0\\1&1\end{pmatrix} + \begin{pmatrix}0&1\\0&0\end{pmatrix}= \begin{pmatrix}1&1\\1&1\end{pmatrix}.$$ Thank you very much for pointing this out. – user23211 May 17 '12 at 19:57