Skolem-Noether algebras

TL;DR

This paper investigates Skolem-Noether algebras (SN algebras), characterizing their properties, providing examples, and establishing that many classes of algebras, including semilocal and polynomial algebras, are SN.

Contribution

It extends the classical Skolem-Noether theorem to broader classes of algebras and explores conditions under which an algebra is SN, including polynomial and free algebras.

Findings

Every semilocal and finite-dimensional algebra is SN.

UFDs, polynomial, and free algebras are SN.

SN property is preserved under power series extension.

Abstract

An algebra $S$ is called a Skolem-Noether algebra (SN algebra for short) if for every central simple algebra $R$, every homomorphism $R\to R\otimes S$ extends to an inner automorphism of $R\otimes S$. One of the important properties of such an algebra is that each automorphism of a matrix algebra over $S$ is the composition of an inner automorphism with an automorphism of $S$. The bulk of the paper is devoted to finding properties and examples of SN algebras. The classical Skolem-Noether theorem implies that every central simple algebra is SN. In this article it is shown that actually so is every semilocal, and hence every finite-dimensional algebra. Not every domain is SN, but, for instance, unique factorization domains, polynomial algebras and free algebras are. Further, an algebra $S$ is SN if and only if the power series algebra $S[[\xi]]$ is SN.

Equations

$$L_w(x)=wx\text{mbox}and{R}_z(x)=xz.$$

$$\sum_{j}w_{j}r_{1}z_{j}=1\quad\mbox{and}\quad\sum_{j}w_{j}r_{k}z_{j}=0\,\,\,\mbox{if $k>1$.}$$

$$\sum _i{r}_i\otimes t_i=\sum _j{p}_j\otimes s_j$$

$$\tilde{\phi }={\mathrm{id}}^{e-1}\otimes \phi :R^{\otimes e}\to R^{\otimes e}\otimes S.$$

$$R^{\otimes e}\cong M_{(\mathrm{deg}R{)}^e}(F),$$

$$c=\sum _{i_2,\dots ,i_e,j}{a}_{i_2,\dots ,i_e,j}\otimes r_{i_2}\otimes \cdots \otimes r_{i_e}\otimes s_j$$

$$\sum _{i_2,\dots ,i_e,j}(x{a}_{i_2,\dots ,i_e,j}-a_{i_2,\dots ,i_e,j}x)\otimes r_{i_2}\otimes \cdots r_{i_e}\otimes s_j=0.$$

$$\hat{\phi }={\mathrm{id}}^{e-2}\otimes \phi :R^{\otimes (e-1)}\to R^{\otimes (e-1)}\otimes S$$

$$1\otimes x=a(x\otimes 1)a^{-1}$$

$$1\otimes x=$$

$$=$$

$$\sum _j(p_{j}x-x{p}_j)\otimes s_j=0.$$

$$\varphi\big{(}(s_{ij})\big{)}=c\big{(}\sigma(s_{ij})\big{)}c^{-1}$$

$$b_{kl}{c}_k=1\otimes s_{kl}.$$

$$\phi (x)=cx{c}^{-1}$$

$$\phi (x)=\sum _i{f}_i(x)\otimes s_i$$

$$f_i=\sum _j{L}_{w_{ij}}{R}_{z_{ij}}.$$

$$\phi (x)$$

$$=\sum _i\sum _j(w_{ij}\otimes s_i)x{z}_{ij}.$$

$$\sum _{k=1}^d{c}_{k}xy{r}_k=\sum _{k=1}^d\phi (x)c_{k}y{r}_k.$$

$$\sum _j\sum _{k=1}^d{c}_{k}x{w}_j{r}_k{z}_j=\sum _j\sum _{k=1}^d\phi (x)c_k{w}_j{r}_k{z}_j,$$

$$\sum_{k=1}^{d}c_{k}x\Big{(}\sum_{j}w_{j}r_{k}z_{j}\Big{)}=\sum_{k=1}^{d}\varphi(x)c_{k}\Big{(}\sum_{j}w_{j}r_{k}z_{j}\Big{)}.$$

$$\sum _j{w}_j\phi (z_j)c_1$$

$$=\sum _j\sum _l{w}_j(r_l\otimes s_{1l})z_j$$

$$\displaystyle=\sum_{l}\Big{(}\sum_{j}w_{j}r_{l}z_{j}\Big{)}\otimes s_{1l}$$

$$=1\otimes s_{11}.$$

$$\sum _{k,l}{r}_l{r}_k\otimes s_{kl}=1\otimes 1.$$

$$\phi :R\to (R\otimes S_1)\times (R\otimes S_2).$$

$$\phi (x)=({\phi }_1(x),{\phi }_2(x))$$

$$c=(c_1,c_2)\in (R\otimes S_1)\times (R\otimes S_2)$$

$$\Phi :R\otimes K\to R\otimes S$$

$$\Phi (x\otimes k)=\phi (x)(1\otimes k).$$

$$1\otimes (1-ux)=1\otimes 1-1\otimes ux=1\otimes 1-(1\otimes u)(1\otimes x)$$

Full text

1Supported by the Slovenian Research Agency grant P1-0288.

2Supported by the Faculty Research Development Fund (FRDF) of The University of Auckland (project no. 3709120).

3Supported by the Marsden Fund Council of the Royal Society of New Zealand. Partially supported by the Slovenian Research Agency grants P1-0222, L1-6722, J1-8132.

4Supported by the University of Auckland Doctoral Scholarship.

1. Introduction

Our main motivation for this work is the celebrated Skolem-Noether theorem. We will state its version as given, for example, in [Her68]. But first, a word on conventions. All our algebras are assumed to be unital algebras over a fixed field $F$, subalgebras are assumed to contain the same unity, and all homomorphisms send $1$ to $1$.

Theorem 1.1**.**

(Skolem-Noether)* Let $A$ be simple artinian algebra with center $F$. If $R$ is a finite-dimensional simple $F$-subalgebra of $A$ and $\varphi$ is an $F$-algebra homomorphism from $R$ into $A$, then there exists an invertible element $c\in A$ such that $\varphi(x)=cxc^{-1}$ for all $x\in R$. (In other words, $\varphi$ can be extended to an inner automorphism of $A$.)*

Recall that an algebra is said to be central if its center consists of scalar multiples of unity. As usual, we will use the term central simple algebra for an algebra that is central, simple, and also finite-dimensional.

Definition 1.2**.**

An algebra $S$ is a Skolem-Noether algebra (SN algebra for short) if for every central simple algebra $R$ and every homomorphism $\varphi:R\to R\otimes S$ there exists an invertible element $c\in R\otimes S$ such that $\varphi(x)=cxc^{-1}$ for every $x\in R$. (Here, $R$ is identified with $R\otimes 1$).

The Skolem-Noether theorem, together with the well-known fact that the class of central simple algebras is closed under tensor products, implies that every central simple algebra $S$ is an SN algebra. A partial converse is also true: the assertion that central simple algebras are SN algebras implies an important special case of the Skolem-Noether theorem where $A$ is a central simple algebra and $R$ is its central simple subalgebra. This is because, under these assumptions, $A$ is isomorphic to $R\otimes S$ where $S$ is also a central simple subalgebra of $A$ [Bre14, Corollary 4.49].

SN algebras naturally arise from the problem of understanding automorphism groups of tensor products of algebras. Unlike the case of derivations on tensor products [Bre17], the general solution to this problem seems far out of reach. For instance, while automorphisms of univariate and bivariate polynomial algebras are well understood [Jun42], already the trivariate case is wild [SU03]. In another direction, functional analysts consider the question when the flip automorphism $A\otimes A\to A\otimes A$ is (approximately) inner for operator algebras $A$, see [Sak75, ER78, Izu17]. In this paper we settle the following special case of the above problem. If $S$ is an SN algebra and $R$ is a central simple algebra, then automorphisms of $R\otimes S$ are just compositions of inner automorphisms and automorphisms of $S$; see Proposition 3.3. While the class of SN algebras looks restrictive, our main results show that various classical and important families of algebras satisfy the SN property, for example semilocal (in particular artinian and finite-dimensional) algebras, unique factorization domains, free algebras, etc.

Some of the readers might be interested only in the case where $R=M_{n}(F)$, the algebra of $n\times n$ matrices with entries in $F$. Let us therefore mention that since $M_{n}(F)\otimes S$ can be identified with $M_{n}(S)$, the condition that $S$ is an SN algebra implies that every homomorphism from $M_{n}(F)$ into $M_{n}(S)$ can be extended to an inner automorphism of $M_{n}(S)$. Moreover, we show in Proposition 2.1 that the latter condition implies the SN property. However, this does not lead to any simplifications of our proofs, so we persist with central simple algebras as in Definition 1.2.

Main results and guide to the paper

The short Section 2 on preliminaries includes Proposition 2.1: $S$ is an SN algebra if and only if all homomorphisms $M_{n}(F)\to M_{n}(S)$ extend to inner automorphisms. Section 3 positions SN algebras into a wider context of automorphisms of tensor products. For instance, Proposition 3.3 proves that given an SN algebra $S$ and a central simple algebra $R$, every automorphism of $R\otimes S$ is the composition of an inner automorphism and an automorphism of $S$. In particular, this applies to matrix algebras over SN algebras.

We then identify classes of algebras which satisfy the SN property. In Section 4 we derive Lemma 4.1, which is the main technical tool for proving subsequent results. Section 5 culminates in Theorem 5.5 showing that semilocal algebras are SN. Hence all artinian algebras and thus all finite-dimensional algebras are SN. Section 6 refines the latter result. Namely, every homomorphism from a central simple subalgebra $R$ of a finite-dimensional algebra $A$ into $A$ extends to an inner automorphism of $A$ (see Theorem 6.1). In Section 7 we give examples of domains which are SN algebras, such as unique factorization domains (UFDs) and free algebras, see Corollary 7.2 and Corollary 7.4. Section 8 uses the Quillen-Suslin theorem to prove that matrix algebras over polynomial algebras are SN. The paper concludes with Section 9, where we show that an algebra $S$ is SN if and only if the formal power series algebra $S[[\xi]]$ is SN.

2. Preliminaries

The purpose of this section is to introduce the notation and terminology, and prove a proposition that yields a characterization of SN algebras.

Let $R$ be a central simple algebra. Given $w,z\in R$, we define the left and right multiplication operators $L_{w},R_{z}:R\to R$ by

[TABLE]

As is well-known, every linear map from $R$ into $R$ can be written as a sum of maps of the form $L_{w}R_{z}$, $w,z\in R$ [Bre14, Lemma 1.25]. Accordingly, given a basis $\{r_{1},\dots,r_{d}\}$ of $R$, there exists $w_{j},z_{j}\in R$ such that $h=\sum_{j}L_{w_{j}}R_{z_{j}}$ satisfies $h(r_{1})=1$ and $h(r_{k})=0$, $k\neq 1$. That is,

[TABLE]

We will be mostly concerned with tensor product algebras $R\otimes S$. Here $R,S$ are algebras over a field $F$ and the tensor product is taken over $F$. As usual, we identify $R$ by $R\otimes 1$, and, accordingly, often write $r\otimes 1\in R\otimes 1$ simply as $r$. Let us point out an elementary fact that will be used freely without further reference. If the $r_{i}$’s are linearly independent elements in $R$, then for all $p_{j}\in R$ and $s_{j},t_{i}\in S$,

[TABLE]

implies that each $t_{i}$ lies in the linear span of the $s_{j}$’s [Bre14, Lemma 4.9]. Similarly, assuming that the $t_{i}$’s are linearly independent, it follows from (2.2) that each $r_{i}$ lies in the linear span of the $p_{j}$’s.

By rad$(S)$ we denote the Jacobson radical of the algebra $S$. Recall that $S$ is called a semilocal algebra if $S/{\rm rad}(S)$ is a semisimple algebra, i.e., isomorphic to a finite direct product of simple artinian algebras. In the special case where $S/{\rm rad}(S)$ is a division algebra, $S$ is called a local algebra. Finally, we say that $S$ is a stably finite algebra if for all $n\geq 1$ and all $x,y\in M_{n}(S)$, $xy=1$ implies $yx=1$.

To conclude the section we give an alternative characterization of the SN property. In order to show that $S$ is an SN algebra it suffices to verify the condition of Definition 1.2 for $R=M_{n}(F)$, i.e., all $F$-algebra homomorphisms $M_{n}(F)\to M_{n}(S)$ are given by conjugation.

Proposition 2.1**.**

Let $S$ be an algebra and suppose that for every $n\in\mathbb{N}$ and a homomorphism $\varphi:M_{n}(F)\to M_{n}(S)$ there exists $c\in M_{n}(S)$ such that $\varphi(x)=cxc^{-1}$ for every $x\in R$. Then $S$ is an SN algebra.

Proof.

Let $R$ be a central simple algebra and $\varphi:R\to R\otimes S$ a homomorphism. Let $e\in\mathbb{N}$ be the exponent of $R$, i.e., the order of $R$ as an element of the Brauer group of $F$ [GS06, Definition 4.5.12]. Let

[TABLE]

Since

[TABLE]

by assumption there exists $c\in R^{\otimes e}\otimes S$ such that $\tilde{\varphi}(x)=cxc^{-1}$ for every $x\in R^{\otimes e}$.

If $e>1$, we can write $c$ as

[TABLE]

for some $a_{i_{2},\dots,i_{e},j},r_{i}\in R$ and $s_{j}\in S$ where $\{r_{i}\}_{i}\subset R$ and $\{s_{j}\}_{j}\subset S$ are linearly independent sets. If $x=x_{1}\otimes 1\otimes\cdots\otimes 1$ for $x_{1}\in R$, then $\tilde{\varphi}(x)c-cx=0$ becomes

[TABLE]

Since the elements $r_{i_{2}}\otimes\cdots r_{i_{e}}\otimes s_{j}$ form a linearly independent set in $R^{\otimes(e-1)}\otimes S$, we conclude that $xa_{i_{2},\dots,i_{e},j}=a_{i_{2},\dots,i_{e},j}x$ for all $a_{i_{2},\dots,i_{e},j}\in R$ and $x\in R$. As $R$ is central we have $a_{i_{2},\dots,i_{e},j}\in F$ and therefore $c\in 1\otimes R^{\otimes(e-1)}\otimes S\cong R^{\otimes(e-1)}\otimes S$. Consequently the homomorphism

[TABLE]

satisfies $\hat{\varphi}(x)=cxc^{-1}$ for every $x\in R^{\otimes(e-1)}$. Continuing by induction we conclude that $c\in R\otimes S$ and $\varphi(x)=cxc^{-1}$ for all $x\in R$, so $S$ is an SN algebra. ∎

While Proposition 2.1 seemingly facilitates demonstrating that $S$ is an SN algebra, it does not simplify our proofs in the sequel.

3. SN algebras and automorphisms

In this section we give a few motivating results and prove that every automorphism of a matrix algebra over an SN algebra $S$ is an inner automorphism composed with an automorphism of $S$, see Corollary 3.4.

We begin with a proposition which justifies the requirement in Definition 1.2 that the algebra $R$ is central simple.

Proposition 3.1**.**

Let $R$ be a subalgebra of an algebra $S$. If the homomorphism $x\otimes 1\mapsto 1\otimes x$ from $R=R\otimes 1$ into $R\otimes S$ can be extended to an inner automorphism of $R\otimes S$, then $R$ is a central simple algebra.

Proof.

By assumption, there exists an invertible element $a\in R\otimes S$ such that

[TABLE]

for all $x\in R$. Let us write $a=\sum_{i=1}^{m}u_{i}\otimes v_{i}$ and $a^{-1}=\sum_{j=1}^{n}w_{j}\otimes z_{j}$. Accordingly,

[TABLE]

This implies that every $x\in R$ lies in the linear span of all $v_{i}z_{j}$, $i=1,\dots,m$, $j=1,\dots,n$. Thus, $R$ is finite-dimensional. On the other hand, (3.1) implies that for every nonzero $x\in R$, $1$ lies in $RxR$. This means that $R$ is simple. Finally, if $z$ lies in the center of $R$, then $1\otimes z=a(z\otimes 1)a^{-1}=z\otimes 1$, which readily implies that $z$ is a scalar multiple of $1$, as desired. ∎

The question of when the automorphism $x\otimes y\mapsto y\otimes x$ of $R\otimes R$ is inner was initiated by Sakai [Sak75] in the C∗-algebra context, and investigated further by Bunce [Bun74]. The following corollary is an extension of [Bun74, Theorem 2].

Corollary 3.2**.**

Let $R$ be an arbitrary algebra. The homomorphism $x\otimes 1\mapsto 1\otimes x$ from $R=R\otimes 1$ into $R\otimes R$ can be extended to an inner automorphism of $R\otimes R$ if and only if $R$ is a central simple algebra.

Proof.

If $R$ is a central simple algebra, then so is $R\otimes R$ [Bre14, Corollary 4.44], and so every homomorphism from $R$ into $R\otimes R$ can be extended to an inner automorphism by the Skolem-Noether theorem. The converse follows from Proposition 3.1. ∎

The next proposition yields another motivation for considering SN algebras.

Proposition 3.3**.**

Let $R$ be a central simple algebra and let $S$ be an SN algebra. Then every automorphism $\varphi$ of $R\otimes S$ is the composition of an inner automorphism and an automorphism of the form ${\rm id}_{R}\otimes\sigma$ where $\sigma$ is an automorphism of $S$.

Proof.

By assumption, the restriction of $\varphi$ to $R$ can be extended to an inner automorphism $x\mapsto cxc^{-1}$ of $R\otimes S$. Considering the automorphism $x\mapsto c^{-1}\varphi(x)c$ we thus see that there is no loss of generality in assuming that $\varphi$ acts as the identity on $R$. Note that the proposition will be proved by showing that $\varphi$ maps $1\otimes S$ into itself. Pick $s\in S$. We can write $\varphi(1\otimes s)$ as $\sum_{j}p_{j}\otimes s_{j}$ where the $s_{j}$’s are linearly independent. Since $1\otimes s$ commutes with $x\otimes 1$ for every $x\in R$ it follows that so does $\varphi(1\otimes s)$. This implies that

[TABLE]

As the $s_{j}$’s are linearly independent it follows that $p_{j}x-xp_{j}=0$ for each $j$ and each $x\in R$. Hence, since $R$ is central, each $p_{j}$ is a scalar multiple of $1$. Consequently, $\varphi(1\otimes s)\in 1\otimes S$. ∎

If $R=M_{n}(F)$, then $R\otimes S$ can be identified with $M_{n}(S)$, and the proposition gets the following form.

Corollary 3.4**.**

If $S$ is an SN algebra, then every automorphism $\varphi$ of $M_{n}(S)$ is of the form

[TABLE]

where $c$ is an invertible element in $M_{n}(S)$ and $\sigma$ is an automorphism of $S$.

This result is known in the case where $S$ is either an artinian algebra [BO81, Theorem 3.13], a UFD [Isa80, Corollary 15], or a commutative local algebra (see, e.g., [Kov73, p. 163]). As we will see, all these algebras are SN algebras. On the other hand, [Isa80] shows that the commutative domain $\mathbb{Z}[\sqrt{-5}]$ does not satisfy the conclusion of Corollary 3.4. We give an algebra with the same property in Example 7.6.

4. Basic lemma

All our main results will be derived from the following technical lemma. Its proof will use some ideas from the proof of the (special case of) Skolem-Noether theorem given in [Bre14, pp. 13–14].

Lemma 4.1**.**

Let $R$ be a central simple algebra with basis $\{r_{1},\dots,r_{d}\}$ and let $S$ be an arbitrary algebra. Then $\varphi:R\to R\otimes S$ is a homomorphism if and only if there exist $c_{1},\dots,c_{d}\in R\otimes S$ such that

1. (a)

$\varphi(x)=\sum_{k=1}^{d}c_{k}xr_{k}$* for all $x\in R$,* 2. (b)

$\varphi(x)c_{k}=c_{k}x$* for all $x\in R$, and* 3. (c)

$\sum_{k=1}^{d}c_{k}r_{k}=1.$**

Moreover, writing $c_{k}=\sum_{l=1}^{d}r_{l}\otimes s_{kl}$, we have that for each $k$ and $l$ there exists $b_{kl}\in R\otimes S$ such that

[TABLE]

Accordingly, if $S$ is stably finite and there exist $k$ and $l$ such that $s_{kl}$ is invertible in $S$, then $c=c_{k}$ is invertible in $R\otimes S$ and

[TABLE]

for all $x\in R$.

Proof.

Since $R$ is finite-dimensional, there exist finitely many $s_{i}\in S$ and linear maps $f_{i}:R\to R$ such that

[TABLE]

for all $x\in S$. By [Bre14, Lemma 1.25] there exist $w_{ij},z_{ij}\in R$ such that

[TABLE]

Consequently, for every $x\in R$ we have

[TABLE]

Writing each $z_{ij}$ as a linear combination of $r_{1},\dots,r_{d}$ we see that $\varphi$ is of the form described in (a).

We now use the multiplicativity of $\varphi$, i.e., $\varphi(xy)=\varphi(x)\varphi(y)$ for all $x,y\in R$. In view of (a) we can rewrite this as

[TABLE]

Pick $w_{j},z_{j}\in R$ such that (2.1) holds. Setting $y=w_{j}$ in (4.1), multiplying the identity, so obtained, from the right by $z_{j}$, and then summing up over all $j$ we get

[TABLE]

that is,

[TABLE]

By (2.1) this reduces to $c_{1}x=\varphi(x)c_{1}$. Of course, the same proof applies to every $c_{k}$, so (b) holds. Finally, (c) follows from $\varphi(1)=1$.

A direct verification shows that (a), (b), and (c) imply that $\varphi$ is a homomorphism.

Let us write $c_{1}=\sum_{l}r_{l}\otimes s_{1l}$, and let $w_{j},z_{j}$ be as above. Using (b) we obtain

[TABLE]

Thus, $b_{11}=\sum_{j}w_{j}\varphi(z_{j})$ satisfies $b_{11}c_{1}=1\otimes s_{11}$. Similarly we find other $b_{kl}$’s.

Finally, assume that $s_{kl}$ is invertible in $S$ for some $k$ and $l$. Then $(1\otimes s_{kl}^{-1})b_{kl}$ is a left inverse of $c_{k}$. If $S$ is stably finite, then the result by Montgomery [Mon83, Theorem 1] implies that this element is also a right inverse. Therefore, (b) shows that $c=c_{k}$ satisfies $\varphi(x)=cxc^{-1}$ for all $x\in R$. ∎

We continue with a simple application of Lemma 4.1, showing that local algebras are SN. This result will be generalized to semilocal algebras (with a considerably more involved proof) in the next section, see Theorem 5.5.

Corollary 4.2**.**

Every local algebra is an SN algebra.

Proof.

Let $r_{k},s_{kl}$ be elements from Lemma 4.1. From (c) it follows that

[TABLE]

This implies that $1$ lies in the linear span of $s_{kl}$. Consequently, at least one $s_{kl}$ does not lie in rad$(S)$. Since $S$ is local it follows that $s_{kl}$ is invertible in $S$. As $S$ is stably finite [Lam01, Theorem 20.13], the last assertion of Lemma 4.1 shows that there exists $c\in R\otimes S$ such that $\varphi(x)=cxc^{-1}$ for all $x\in R$. ∎

5. Semilocal algebras

The main result of this section is Theorem 5.5 showing that semilocal algebras are SN. We begin with a simple lemma.

Lemma 5.1**.**

If $S_{1}$ and $S_{2}$ are SN algebras, then so is their direct product $S_{1}\times S_{2}$.

Proof.

Recall that $R\otimes(S_{1}\times S_{2})$ can be identified with $(R\otimes S_{1})\times(R\otimes S_{2})$. Take a homomorphism

[TABLE]

Writing

[TABLE]

it is immediate that $\varphi_{i}$ is a homomorphism from $R$ into $R\otimes S_{i}$, $i=1,2$. By assumption, there exist $c_{i}\in R\otimes S_{i}$ such that $\varphi_{i}(x)=c_{i}xc_{i}^{-1}$ for all $x\in R$, $i=1,2$. Hence,

[TABLE]

satisfies $\varphi(x)=cxc^{-1}$ for all $x\in R$. ∎

As mentioned in the introduction, the Skolem-Noether theorem implies that every central simple algebra is an SN algebra. With a little extra effort we can extend this to semisimple algebras.

Lemma 5.2**.**

Every semisimple algebra is an SN algebra.

Proof.

In view of Lemma 5.1 it suffices to consider the case where $S$ is simple artinian. Let $R$ be a central simple algebra. The algebra $R\otimes S$ is then simple [Bre14, Theorem 4.42]. We claim that it is also artinian. Indeed, considering $R\otimes S$ as a left $S$-module in the natural way we see that it is isomorphic to the left $S$-module $S^{d}$ where $d$ is the dimension of $R$, and that a descending chain of left ideals of $R\otimes S$ is also a descending chain of left $S$-submodules. The desired conclusion thus follows from the fact that $S^{d}$ is artinian.

Let $K$ be the center of $S$. The center of $R\otimes S$ is equal to $1\otimes K$ [Bre14, Corollary 4.32], which we identify with $K$. Consider $R\otimes K$ as an algebra over $K$ in the usual way. Clearly, it is finite-dimensional and, again by [Bre14, Theorem 4.42], simple.

Now, given a homomorphism $\varphi:R\to R\otimes S$, we define

[TABLE]

by

[TABLE]

Note that $\Phi$ is a $K$-algebra homomorphism. The Skolem-Noether theorem thus tells us that there exists $c\in R\otimes S$ such that $\Phi(x\otimes k)=c(x\otimes k)c^{-1}$ for all $x\in R$, $k\in K$. Setting $k=1$ we get the desired conclusion. ∎

Our goal is to show that semilocal algebras are SN algebras by reducing the general case to the semisimple case. We will actually prove a general reduction theorem whose possible applications are not limited to semilocal algebras. To this end, we need the following lemma. From its nature one would expect that it is known, but we were unable to find a good reference. We include a short proof for the sake of completeness.

Lemma 5.3**.**

If $R$ is a central simple algebra, then ${\rm rad}(R\otimes S)=R\otimes{\rm rad}(S)$ for every algebra $S$.

Proof.

As an ideal of $R\otimes S$, ${\rm rad}(R\otimes S)$ is necessarily of the form $R\otimes I$ for some ideal $I$ of $S$ [Bre14, Theorem 4.42]. We will show that $I\subseteq{\rm rad}(S)$, by making use of the following characterization of rad$(A)$: $v\in{\rm rad}(A)$ if and only if $1-vx$ is invertible for every $x\in A$. Take $u\in I$. Since $1\otimes u\in{\rm rad}(R\otimes S)$ it follows that

[TABLE]

is invertible in $R\otimes S$ for every $x\in S$. However, this is possible only if $1-ux$ is invertible, implying that $u\in{\rm rad}(S)$. Thus, $I\subseteq{\rm rad}(S)$, and so ${\rm rad}(R\otimes S)\subseteq R\otimes{\rm rad}(S)$.

As the lemma is well-known if $R=M_{n}(F)$ (see, e.g., [Lam01, pp. 57-58]), we will establish the converse inclusion by reducing the general case to this one. Take a splitting field $K$ for $R$ which is a finite separable extension of $F$ (see, e.g., [GS06, Proposition 4.5.4]). Then $K\otimes R$ may be identified with $M_{n}(K)$ for some $n\geq 1$, and, therefore, $K\otimes R\otimes S$ may be identified with $M_{n}(K\otimes S)$. Thus, by what we pointed out at the beginning of the paragraph, we have

[TABLE]

By [Lam01, Theorem 5.17], ${\rm rad}(K\otimes S)=K\otimes{\rm rad}(S)$, so that

[TABLE]

According to [Lam01, Theorem 5.14],

[TABLE]

and hence, by (5.1),

[TABLE]

Since both $R\otimes S$ and $M_{n}(K)\otimes{\rm rad}(S)$ readily contain $R\otimes{\rm rad}(S)$, it follows that $R\otimes{\rm rad}(S)\subseteq{\rm rad}(R\otimes S)$. ∎

We can now prove the announced reduction theorem.

Theorem 5.4**.**

If an algebra $S$ is stably finite and $S/{\rm rad}(S)$ is an SN algebra, then $S$ is an SN algebra.

Proof.

Let $R$ be central simple and write

[TABLE]

Take a homomorphism $\varphi:R\to R\otimes S$. We define

[TABLE]

by

[TABLE]

Since $(R\otimes S)/J$ is canonically isomorphic to $R\otimes(S/{\rm rad}(S))$, and $S/{\rm rad}(S)$ is assumed to be an SN algebra, it follows that there exists an invertible element $a\in(R\otimes S)/J$ such that

[TABLE]

As $J$ is, by Lemma 5.3, the Jacobson radical of $R\otimes S$, it follows that there exists an invertible element $b\in R\otimes S$ such that $a=b+J$. Obviously, we have

[TABLE]

that is,

[TABLE]

Replacing the role of $\varphi$ by the homomorphism $x\mapsto b^{-1}\varphi(x)b$ we see that without loss of generality we may assume that $b=1$. Thus,

[TABLE]

Now apply Lemma 4.1. Picking a basis $\{r_{1},\dots,r_{d}\}$ of $R$, we can thus find $s_{kl}\in S$, $k,l=1,\dots,p$, such that

[TABLE]

and our goal is to show that at least one $s_{kl}$ is invertible in $S$.

Let $\lambda_{k}\in F$ be such that $1=\sum_{k=1}^{d}{\lambda_{k}}r_{k}$. Then

[TABLE]

Using (5.2) and (5.3) we thus obtain

[TABLE]

We may assume that $\lambda_{1}\neq 0$. Choose $w_{j},z_{j}\in R$ that satisfy (2.1). Denoting the expression in (5.4) by $\rho(x)$, we have

[TABLE]

As $\rho$ maps into $J$ it follows that

[TABLE]

Since the $r_{k}$’s are linearly independent, we must have $s_{k1}-\lambda_{k}\lambda_{1}\cdot 1\in{\rm rad}(S)$ for each $k$. In particular, $s_{11}=\lambda_{1}^{2}\cdot 1+u$ for some $u\in{\rm rad}(S)$. Since $\lambda_{1}\neq 0$ it follows that $s_{11}$ is invertible, as desired. ∎

We are now in a position to give our main result.

Theorem 5.5**.**

Every semilocal algebra $S$ is an SN algebra.

Proof.

Since $S$ is stably finite [Lam01, Theorem 2.13] and the algebra $S/{\rm rad}(S)$ is semisimple, the theorem follows from Lemma 5.2 and Theorem 5.4. ∎

Corollary 5.6**.**

Every artinian algebra is an SN algebra.

6. Finite-dimensional algebras

Corollary 5.6 shows that every finite-dimensional algebra is an SN algebra. The next result gives a strengthening of this property.

Theorem 6.1**.**

Let $A$ be a finite-dimensional algebra and let $R$ be its central simple subalgebra. Then every homomorphism from $R$ into $A$ can be extended to an inner automorphism of $A$.

Proof.

Assume first that $R=M_{n}(F)$. Then $A$ contains a set of $n\times n$ matrix units and is therefore isomorphic to $M_{n}(S)\cong R\otimes S$ for some subalgebra $S$ of $A$ [Bre14, Lemma 2.52]. Since $S$ is also finite-dimensional, the desired conclusion follows from Corollary 5.6.

Now let $R$ be an arbitrary central simple algebra. We may assume that the field $F$ is infinite, for otherwise $R\cong M_{n}(F)$ by Wedderburn’s theorem on finite division rings. Let $\varphi$ be a homomorphism from $R$ into $A$. Take a splitting field $K$ for $R$. Identifying $K\otimes R$ with $M_{n}(K)$, $n\geq 1$, it follows from the preceding paragraph that there exists $b\in K\otimes A$ such that

[TABLE]

for all $y\in K\otimes R$. In particular,

[TABLE]

for all $x\in R$. Writing $b=\sum_{i=1}^{m}k_{i}\otimes a_{i}$ with the $k_{i}$’s linearly independent, it follows that

[TABLE]

and so $\varphi(x)a_{i}=a_{i}x$ for all $x\in R$ and every $i$. Hence we see that it suffices to show that span${}_{F}\{a_{1},\dots,a_{m}\}$ contains an element which is invertible in $A$.

As a finite-dimensional algebra, $A$ can be considered as a subalgebra of $M_{N}(F)$ for some $N\geq 1$. Take the polynomial

[TABLE]

Note that $K\otimes A$ can be viewed as a subalgebra of $M_{N}(K)$. Since $b$ is invertible in $K\otimes A$, we know that span${}_{K}\{a_{1},\dots,a_{m}\}$ contains an invertible element in $K\otimes A$. This clearly implies that $f$ is a nonzero polynomial. As $F$ is infinite, there exist $\lambda_{i}\in F$ such that $f(\lambda_{1},\dots,\lambda_{m})\neq 0$. That is, span${}_{F}\{a_{1},\dots,a_{m}\}$ contains an element $c$ which is invertible in $M_{N}(F)$. However, since we are in finite dimensions, $c^{-1}$ is a polynomial in $c$. Thus, $c$ is invertible in $A$. ∎

Using the standard homomorphism construction we will now see that Theorem 6.1 can be used for showing that all derivations from $R$ into any $R$-bimodule $M$ are inner (in accordance with the conventions mentioned at the very beginning of the paper, we assume that our bimodules are unital). This is, of course, a well-known result. Another way of stating it is that central simple algebras are separable.

Corollary 6.2**.**

Every derivation from a central simple algebra $R$ into an arbitrary $R$-bimodule $M$ is inner.

Proof.

Let $d:R\to M$ be a derivation. As a finite-dimensional subspace of $M$, $d(R)$ generates a finite-dimensional subbimodule of $M$. Therefore, there is no loss of generality in assuming that $M$ itself is finite-dimensional.

Let $\widetilde{A}$ be the set of all matrices of the form $\left[\begin{smallmatrix}x&u\\ 0&x\end{smallmatrix}\right]$, where $x\in R$ and $u\in M$. Note that $\widetilde{A}$ is a (finite-dimensional!) algebra under the standard matrix operations. Let $\widetilde{R}$ be its subalgebra consisting of all matrices of the form $\left[\begin{smallmatrix}x&0\\ 0&x\end{smallmatrix}\right]$, $x\in R$. Of course, $\widetilde{R}\cong R$.

Define $\varphi:\widetilde{R}\to\widetilde{A}$ by

[TABLE]

One immediately checks that $\varphi$ is a homomorphism. By Theorem 6.1 there exists an invertible element $c=\left[\begin{smallmatrix}t&v\\ 0&t\end{smallmatrix}\right]\in\widetilde{A}$ such that $\varphi(\tilde{x})=c\tilde{x}c^{-1}$ for all $\tilde{x}\in\widetilde{R}$. Consequently, $\varphi(\tilde{x})c=c\tilde{x}$, that is,

[TABLE]

for all $x\in R$. This yields

[TABLE]

for all $x\in R$. Since $R$ is central, the first identity shows that $t\in F$. Moreover, $t\neq 0$ for $c$ is invertible. Hence $w=t^{-1}v$ satisfies $d(x)=wx-xw$ by the second identity. ∎

7. Domains

In this section we give classes of domains which are SN algebras. For instance, UFDs and free algebras are SN algebras (Corollaries 7.2 and 7.4).

As the coordinate ring of an elliptic curve demonstrates (see Example 7.6), not every commutative domain is an SN algebra. However, the following proposition shows that every domain $S$ embedded into a division algebra satisfies a certain weaker condition.

Proposition 7.1**.**

Let $R$ be a central simple algebra, and let an algebra $S$ be a domain which can be embedded into a division algebra $D$. If $\varphi$ is a homomorphism from $R$ into $R\otimes S$, then there exists $c\in R\otimes S$ which is invertible in $R\otimes D$, in fact

[TABLE]

for some nonzero $s\in S$ and $b\in R\otimes S$, such that

[TABLE]

for all $x\in R$. Moreover, if $\{r_{1},\dots,r_{d}\}$ is a basis of $R$, $b=\sum_{k=1}^{d}r_{k}\otimes s_{k}$ for some $s_{k}\in S$, and $c=\sum_{l=1}^{d}r_{l}\otimes t_{l}$ for some $t_{l}\in S$, then

[TABLE]

for all $k$ and $l$.

Proof.

Not every $c_{k}$ from Lemma 4.1 can be [math] (in view of (c)), and so $s_{kl}\neq 0$ for some $k$ and $l$. Set $c=c_{k}$ and $s=s_{kl}$. By the lemma we have $\varphi(x)c=cx$ for all $x\in R$ and $bc=1\otimes s$ for some $b\in R\otimes S$. Of course, $s$ is invertible in $D$. Therefore $(1\otimes s^{-1})b$ is a left inverse of $c$ in $R\otimes D$. By [Mon83, Theorem 1], a left inverse in $R\otimes D$ is also a right inverse, so $c^{-1}=(1\otimes s^{-1})b$.

Now take a basis $\{r_{1},\dots,r_{d}\}$ of $R$, and let us write $b=\sum_{k=1}^{d}r_{k}\otimes s_{k}$ and $c=\sum_{l=1}^{d}r_{l}\otimes t_{l}$. Then

[TABLE]

for all $x\in R$. Pick $w_{j},z_{j}\in R$ satisfying (2.1). We have

[TABLE]

Since the left hand side, i.e. $\sum_{j}w_{j}\varphi(z_{j})$, lies in $R\otimes S$, so does the right hand side. This readily yields that $t_{1}s^{-1}s_{k}\in S$. ∎

Corollary 7.2**.**

Every UFD is an SN algebra.

Proof.

Let $S$ be a UFD and let $R$, $\varphi$, $b$, $c$, $s$, $s_{k}$, $t_{l}$ be as in Proposition 7.1. Since $S$ is a UFD, $t_{1},\dots,t_{d}$ have a greatest common divisor $e$ and $c$ can be replaced with $(1\otimes e^{-1})c$, we can without loss of generality assume that $t_{1},\dots,t_{d}$ are coprime. Then it suffices to prove that $s^{-1}s_{k}\in S$ for every $k$. Since $t_{l}s^{-1}s_{k}\in S$ for every $k,l$, we see that $s$ divides $t_{l}s_{k}$ for every $l,k$. Let $p$ be a prime such that $p^{n}$ divides $s$. Suppose that $p^{n}$ does not divide $s_{k_{0}}$ for some $k_{0}$. Since $p^{n}$ divides $t_{l}s_{k_{0}}$ for every $l$, we conclude that $p$ divides $t_{l}$ for every $l$, which contradicts the assumption about $t_{l}$ being coprime. Hence $s$ divides $s_{k}$ for every $k$. ∎

We now move to the noncommutative setting. Since embeddings of noncommutative domains into division rings can be ill-behaved or nonexistent, one needs stronger assumptions than in Corollary 7.2. Let $S$ be an arbitrary ring. The inner rank of $A\in S^{m\times n}$ is the least $r$ such that $A=BC$ for some $B\in S^{m\times r}$ and $C\in S^{r\times n}$. We write $\rho A=r$. For example, if $S$ is a division ring, then $\rho A$ is just the rank of $A$. We say that $S$ is a Sylvester domain [Coh06, Section 5.5] if for any $P\in S^{\ell\times m}$ and $Q\in S^{m\times n}$ such that $PQ=0$, it follows that $\rho P+\rho Q\leq m$.

We say that an element $s\in S$ right divides $a\in S$ if $a=a^{\prime}s$ for some $a^{\prime}\in S$. If $S$ is a domain and $a,b\in S\setminus\{0\}$, then $s$ is a highest common right factor (HCRF) of $a$ and $b$ if $s$ right divides $a,b$ and every $s^{\prime}\in S$ that right divides $a,b$ also right divides $s$. We say that $S$ is an HCRF domain if every pair of nonzero elements in $S$ admits a HCRF. Special examples of HCRF domains are filtered rings satisfying the 2-term weak algorithm [Coh06, Section 2.8] or more generally, 2-firs with right ACC1 (ascending chain condition on principal right ideals) [Coh06, Exercise 3.2.1].

Theorem 7.3**.**

If $S$ is an HCRF domain and a Sylvester domain, then $S$ is an SN algebra.

Proof.

Since $S$ is a Sylvester domain, it admits a universal skew field of fractions $D$ and this embedding preserves the inner rank by [Coh06, Theorem 7.5.13]. Let $R$ be a central simple algebra and $\varphi:R\to R\otimes S$ a homomorphism. By Proposition 7.1 there exists $c\in R\otimes S$ invertible in $R\otimes D$ such that $\varphi(x)=cxc^{-1}$ for all $x\in R$. Furthermore, if $\{r_{1},\dots,r_{d}\}$ is a basis of $R$ and

[TABLE]

for $t_{l}\in S$ and $u_{k}\in D$, then $t_{l}u_{k}=s_{lk}\in S$ for all $1\leq l,k\leq d$ (here $u_{k}=s^{-1}s_{k}$ from Proposition 7.1). Since $S$ is an HCRF domain, we can assume that $t_{1},\dots,t_{d}$ have no non-trivial common right factors (otherwise they have a nontrivial HCRF $e$ and we can replace $c$ with $c(1\otimes e^{-1})$). Fix $k$ such that $u_{k}\neq 0$. Then $(u_{k},-1)^{\rm t}\in D^{2}$ belongs to the right kernel of the matrix

[TABLE]

which is therefore of (inner) rank $1$ over $D$. Since the embedding $S\subseteq D$ is inner rank preserving, this matrix is also of inner rank 1 over $S$, so

[TABLE]

for some $v_{i},w_{j}\in S$. Since $w_{1}$ right divides $t_{l}$ for every $l$ and $t_{1},\dots,t_{d}$ are right coprime by assumption, we conclude that $w_{1}$ is invertible in $S$. By taking some $t_{l}\neq 0$ we get

[TABLE]

Consequently $c^{-1}\in R\otimes S$. ∎

Corollary 7.4**.**

Every free algebra $F\langle X\rangle$ is an SN algebra.

Proof.

A free algebra is a filtered ring with a weak algorithm [Coh06, Theorem 2.5.3], so it is a HCRF domain and a fir (free ideal ring) by [Coh06, Theorem 2.4.6] and hence a Sylvester domain by [Coh06, Proposition 5.5.1]. ∎

Theorem 7.3 has the following form for commutative rings.

Corollary 7.5**.**

Every Bézout domain is an SN algebra.

Proof.

Every Bézout domain is a GCD domain, which is just a commutative HCRF domain. Moreover, by [Coh06, Proposition 2.3.17] it is also a semifir and hence a Sylvester domain by [Coh06, Proposition 5.5.1]. Therefore Theorem 7.3 applies. ∎

In the next example we present a domain that is not an SN algebra; cf. [RZ61, Theorem 15].

Example 7.6**.**

Let $S=F[x,y]/(y^{2}-x^{3}-x)$. Then $S$ is a domain,

[TABLE]

is invertible as a matrix over the field of fractions of $S$ and

[TABLE]

Since every product of an entry in $a$ and an entry in $a^{-1}$ lies in $S$, it follows that

[TABLE]

is a well-defined homomorphism. Suppose there exists an invertible $c\in M_{2}(S)$ such that $\varphi(u)=cuc^{-1}$ for all $u\in M_{2}(F)$. Then $\gamma=\det(c)$ is invertible in $S$ and it is easy to see that this implies $\gamma\in F\setminus\{0\}$. Since $c^{-1}a$ commutes with every $u\in M_{2}(F)$ by the definition of $\varphi$, we have $c^{-1}a=fI_{2}$ for some $f\in S$. But then

[TABLE]

contradicts the irreducibility of $x$ in $S$.

8. Polynomial matrix algebras

In this section we prove that matrix algebras over polynomial algebras are SN.

Theorem 8.1**.**

$M_{n}(F[\xi_{1},\dots,\xi_{s}])$* is an SN algebra.*

Besides the Quillen-Suslin theorem, saying that over $F[\xi_{1},\dots,\xi_{s}]$ every finitely generated projective module is free, see [Qui76, Su76], the proof of this theorem is mostly based on the following simple factorization lemma.

Lemma 8.2**.**

Let $A$ be a commutative algebra, which is a domain with field of fractions $K$, $R$ a central simple algebra and $a\in R\otimes M_{n}(A)$. Suppose that $a$ is invertible in $R\otimes M_{n}(K)$ and that $a(x\otimes 1)a^{-1}\in R\otimes M_{n}(A)$ for all $x\in R$. Then for $c\in M_{n}(A)$ the following are equivalent:

1. (i)

There exists a factorization $a=u(1\otimes c)$ for some invertible $u\in R\otimes M_{n}(A)$; 2. (ii)

The left ideal

[TABLE]

is generated by $c$; 3. (iii)

The rows of $c$ form a basis of the $A$-module

[TABLE]

Proof.

(i)$\Rightarrow$(ii): Let $u\in R\otimes M_{n}(A)$ be invertible such that $a=u(1\otimes c)$. Then for any $m\in M_{n}(A)$ we have that $(1\otimes m)a^{-1}=(1\otimes mc^{-1})u^{-1}$ lies in $R\otimes M_{n}(A)$ if and only if $mc^{-1}\in M_{n}(A)$, i.e., $m\in M_{n}(A)c$.

(ii)$\Rightarrow$(i): Let $\{r_{1},\dots,r_{d}\}$ be a basis of $R$ and for fixed $i$ take $w_{k},z_{k}\in R$ such that $\sum_{k}w_{k}r_{j}z_{k}=\delta_{ij}$, see (2.1) at the beginning of Section 2. If $a=\sum_{j}r_{j}\otimes a_{j}$, then

[TABLE]

Since $i$ was arbitrary, this shows that all coefficients $a_{i}$ of $a$ lie in $I(a)$. Assuming that $I(a)=M_{n}(A)c$, we can thus factor $a=u(1\otimes c)$ for some $u\in R\otimes M_{n}(A)$. But then $u^{-1}=(1\otimes c)a^{-1}\in M_{n}(A)$, i.e., $u$ is invertible.

(ii)$\Rightarrow$(iii): Suppose $I(a)=M_{n}(A)c$. Since $a$ is invertible over $K$, there exists a nonzero $e\in A$ such that $e\cdot 1\in I(a)$. Hence the rows of $c$ are clearly linearly independent. Given any $r\in M(a)$, we can extend $r$ by zero to form a matrix $m\in I(a)$ which has $r$ as one of its rows. By assumption $m\in M_{n}(A)c$. In particular, $r$ is a linear combination of the rows of $c$, which shows that they form a basis of $M(a)$.

(iii)$\Rightarrow$(ii): Conversely, suppose that the rows of $c$ form a basis of $M(a)$. In particular, $c\in I(a)$. Moreover, for any $m\in I(a)$ the rows of $m$ lie in $M(a)$ and are, therefore, linear combinations of the rows of $c$, which implies that $m\in M_{n}(A)c$. ∎

Proof of Theorem 8.1.

Let $A:=F[\xi_{1},\dots,\xi_{s}]$, $K$ its field of fractions, $R$ a central simple algebra and $\varphi\colon R\to R\otimes M_{n}(A)$ a homomorphism. Since $M_{n}(K)$ is SN, there exists (after clearing denominators) $a\in R\otimes M_{n}(A)$, invertible in $R\otimes M_{n}(K)$, such that $\varphi(x)=a(x\otimes 1)a^{-1}$ for all $x\in R$.

Fix any prime ideal $P$ of $A$. Then $M_{n}(A_{P})/{\rm rad}(M_{n}(A_{P}))$ is canonically isomorphic to the simple algebra $M_{n}(F)\otimes A_{P}/PA_{P}$, see Lemma 5.3. Therefore, $M_{n}(A_{P})$ is semilocal and by Theorem 5.5 it is also an SN algebra.

It follows that there exists an invertible $u_{P}\in R\otimes M_{n}(A_{P})$ such that $\varphi(x)=u_{P}(x\otimes 1)u_{P}^{-1}$. Then $u_{P}^{-1}a$ commutes with all elements of $R\otimes 1$ and thus lies in $1\otimes M_{n}(A)$. This means, $a$ can be factored as $a=u_{P}(1\otimes c_{P})$ for some $c_{P}\in M_{n}(A)$. By Lemma 8.2, this implies that the $A_{P}$-module $A_{P}M(a)$ is free of rank $n$. Since the prime ideal $P$ was arbitrary, this shows that $M(a)$ is locally free of rank $n$. As being projective is a local property, this implies $M(a)$ is projective. By the Quillen-Suslin theorem $M(a)$ is free of rank $n$. We choose $c\in M_{n}(A)$ such that its rows form a basis of $M(a)$. Then again from Lemma 8.2, we get a factorization $a=u(1\otimes c)$ where $u\in R\otimes M_{n}(A)$ is invertible. Now $\varphi(x)=u(x\otimes 1)u^{-1}$ for all $x\in R$. ∎

9. Formal power series

The aim of this section is to show that the property of being an SN algebra transfers from $S$ to the formal power series algebra $S[[\xi]]$.

Theorem 9.1**.**

$S$* is an SN algebra if and only if $S[[\xi]]$ is an SN algebra.*

Proof.

$(\Rightarrow)$ Let $R$ be a central simple algebra and let $\varphi:R\to R\otimes S[[\xi]]$ be a homomorphism. Since $R$ is finite-dimensional, we can identify $R\otimes S[[\xi]]$ with $(R\otimes S)[[\xi]]$ and write

[TABLE]

where $\varphi_{i}:R\to R\otimes S$. Note that $\varphi_{0}$ is an algebra homomorphism. By assumption, there exists an invertible element $a\in R\otimes S$ such that $\varphi_{0}(x)=axa^{-1}$ for all $x\in R$. Considering the map $x\mapsto a^{-1}\varphi(x)a$ we see that without loss of generality we may assume that $\varphi_{0}(x)=x$ for all $x\in R$, so that

[TABLE]

Now apply Lemma 4.1. Thus, let $\{r_{1},\dots,r_{d}\}$ be a basis of $R$ and let $c_{1},\dots,c_{d}\in R\otimes S[[\xi]]$ be such that

[TABLE]

for all $x\in R$ and all $k$. Writing

[TABLE]

where $c_{kj}\in R\otimes S$, it follows from (9.1) and (9.2) that

[TABLE]

for all $x\in R$. Let us write $c_{k0}=\sum_{j}p_{kj}\otimes s_{kj}$ with the $s_{kj}$’s linearly independent. From (9.3) we infer that

[TABLE]

for all $x\in R$, yielding $xp_{kj}-p_{kj}x=0$. Since $R$ is central this means that each $p_{kj}$ is a scalar multiple of $1$. Accordingly, each $c_{k0}$ is of the form $1\otimes t_{k}$ for some $t_{k}\in S$. From the first identity in $\eqref{ena2}$ one easily deduces that $\sum_{k=1}^{d}r_{k}\otimes t_{k}=1\otimes 1$. Writing $1=\sum_{k=1}^{d}\lambda_{k}r_{k}$, where $\lambda_{k}\in F$, it follows that $t_{k}=\lambda_{k}1$. We may assume that $\lambda_{1}\neq 0$. Accordingly, $c_{10}$ is a nonzero scalar multiple of unity of $R\otimes S$, implying that $c_{1}$ is invertible in $(R\otimes S)[[\xi]]$. Applying $\eqref{ena2}$ we arrive at $\varphi(x)=c_{1}xc_{1}^{-1}$ for all $x\in R$.

$(\Leftarrow)$ Straightforward; more generally, the SN property is clearly preserved by retractions. Here an algebra $S^{\prime}$ is a retract of $S$ if $S^{\prime}\subset S$ and there exists a homomorphism $\pi:S\to S^{\prime}$ that restricts to the identity map on $S^{\prime}$. ∎