1 Introduction↩︎

An algebra is called a Novikov algebra if the following identities hold: \[(x,y,z)=(y,x,z), \qquad (xy)z=(xz)y,\] where \((x,y,z)=(xy)z-x(yz)\).

Novikov algebras first arose in the work of I. M. Gel’fand and I. Ya. Dorfman [1] in the context of the study of Hamiltonian operators in the formal calculus of variations. In the work of A. A. Balinskii and S. P. Novikov [2] (see also the work of S. P. Novikov [3]), these algebras arose in the study of local translation-invariant Lie algebras.

The theory of finite-dimensional Novikov algebras began with the work of E. I. Zel’manov [4], in which semisimple finite-dimensional Novikov algebras over a field of characteristic 0 were described: it turned out that all such algebras are commutative, that is, they are direct sums of fields. In the works of J. M. Osborn [5] and X. Xu [6], simple finite-dimensional Novikov algebras over an algebraically closed field of characteristic \(p>2\) were described; see also the work of V. N. Zhelyabin and A. S. Zakharov [7]. A classification of simple and semisimple finite-dimensional Novikov algebras over an arbitrary field of characteristic \(p>0\) was obtained by V. N. Zhelyabin and A. P. Pozhidaev [8].

Bimodules over simple Novikov algebras of characteristic 0 with an idempotent were studied by J. M. Osborn [9], and the final classification was obtained by X. Xu [10]. Bimodules over finite-dimensional simple algebras were considered by E. I. Zel’manov and J. M. Osborn [11]. A classification of finite-dimensional irreducible bimodules over a finite-dimensional simple Novikov algebra over an algebraically closed field of characteristic \(p>0\) was obtained by X. Xu [6].

Right and left modules over Novikov algebras have not been studied previously. However, A. M. Slinko and I. P. Shestakov [12] introduced the concept of a right representation for an arbitrary variety of algebras, including investigating right representations of alternative and right-alternative algebras.

The author previously studied the radicals of Novikov algebras. The Baer radical in Novikov algebras was constructed and studied in [13]; the Andrunakievich radical in Novikov algebras was described in [14]; and the radicals of Lie-solvable Novikov algebras were studied in [15].

In this paper, we study right representations of Novikov algebras. We prove that the class of Novikov algebras that do not have irreducible representations is radical. The constructed radical is described as the intersection of the largest ideals contained in the kernels of all irreducible representations. In addition, the concept of a primitive Novikov algebra is introduced and it is proved that the constructed radical is the intersection of all primitive ideals of the algebra.

2 Preliminaries↩︎

All algebras in this paper are considered over a field \(F\) of characteristic different from 2.

In any Novikov algebra \(A\), the following identities hold: [16] \[(x,y,z)t = (xt,y,z) = (x,yt,z), \quad x[y,z]=(x,z,y)-(x,y,z)\] for any \(x,y,z\in A\), where \([a,b]=ab-ba\) is the commutator of \(a\) and \(b\).

Furthermore, the following identity holds in any Novikov algebra [16]: \[[z,y]x=\frac{1}{2}[z,yx]+\frac{1}{2}[zx,y].\] From this equality it follows that for any ideal \(I\), the space \([I,I]\) generated by all commutators is an ideal of the algebra.

We present the definition of the right representation, following A. M. Slinko and I. P. Shestakov [12], but restricting the original theory to the case of algebras over the field \(F\). By \(T(V)\) we denote the tensor algebra of a vector space \(V\): \[T(V)=F \oplus V^{(1)}\oplus\dots\oplus V^{(n)}\oplus\dots,\] where \(V^{(n)} = V\otimes \dots \otimes V\) (the tensor product is taken \(n\) times). Every linear mapping \(\varphi:V\to W\) uniquely defines a homomorphism of tensor algebras \(T(\varphi):T(V)\to T(W)\) by the rule \(T(\varphi)(1)=1\) and \(T(v_1\otimes\dots\otimes v_n)=\varphi(v_1)\otimes \dots\otimes \varphi(v_n)\). It follows that if \(B\) is an associative unital algebra, then any linear map \(\varphi:V\to B\) induces a homomorphism of associative algebras \(T(\varphi):T(V)\to B\), where \(T(\varphi)(1)=1\) and \(T(\varphi)(v_1\otimes\dots\otimes v_n)=\varphi(v_1)\dots\varphi(v_n)\).

Let \(\mathcal{M}\) be the variety of algebras over a field \(F\), and \(\mathcal{M}_X\) be the free algebra in \(\mathcal{M}\) on countably many generators \(X=\{x_1,x_2,\dots\}\). Let \(F(x_1,\dots,x_n)\in T(\mathcal{M}_X)\). If \(A\) is an algebra of the variety \(\mathcal{M}\), then by \(F(a_1,\dots,a_n)\) we denote the element of \(T(A)\) obtained from \(F(x_1,\dots,x_n)\) by substituting \(a_i\) for \(x_i\).

Let \(\varphi:\mathcal{M}_X\to A\) be an algebra homomorphism such that \(\varphi(x_i)=a_i\) for \(i=1,\dots,n\). Then it is easy to see that \(F(a_1,\dots,a_n)=T(\varphi)(F(x_1,\dots,x_n))\).

Let \(\rho:A\to \mathrm{End}(V)\) be a linear map from an algebra \(A\) to the endomorphism algebra of a vector space \(V\), and \(T(\rho): T(A)\to\mathrm{End}(V)\) be the induced homomorphism of algebras. An element \(F(x_1,\dots,x_n)\in T(\mathcal{M}_X)\) is called an identity of the map \(\rho\) if \(T(\rho)(F(a_1,\dots,a_n))=0\) for any \(a_1,\dots,a_n\in A\).

Now let \(R:\mathcal{M}_X\to \mathrm{End}(\mathcal{M}_X)\) be the map, acting according to the rule \(R(f)=R_f\), where \(R_f\) is the operator of right multiplication by an element \(f\in\mathcal{M}_X\). Let \(T(R): T(\mathcal{M}_X)\to \mathrm{End}(\mathcal{M}_X)\) be the induced homomorphism of the tensor algebra. We denote the kernel of this homomorphism by \(I_{\mathcal{M}}\). Elements of the ideal \(I_{\mathcal{M}}\) will be called R-identities of the variety \(\mathcal{M}\).

Let \(A\) be the algebra of the variety \(\mathcal{M}\) and \(M\) a vector space. A linear map \(\rho:A\to \mathrm{End}(M)\) is called a right \(\mathcal{M}\)-representation of \(A\) if all R-identities of the variety \(\mathcal{M}\) are identities of the map \(\rho\). In this case, the space \(M\) with the action \(m\cdot a = m\rho(a)\), \(m\in M\), \(a\in A\), is called a right \(A\)-module in the variety \(\mathcal{M}\).

A left representation and a left module can be defined similarly. Note that, since the variety of Novikov algebras is not closed under the action of anti-isomorphism, the properties of right and left representations can differ significantly. Throughout what follows, we will only consider right modules and right representations; therefore, we will omit the word right.

The kernel of the representation \(\rho\) of an algebra \(A\) is denoted by \(\mathrm{Ker}_{\rho}(A)\). The largest ideal of \(A\) contained in \(\mathrm{Ker}_{\rho}(A)\) is denoted by \(K_{\rho}(A)\) and is called the quasikernel of the representation \(\rho\). A representation is called faithful if \(\mathrm{Ker}_{\rho}(A)=0\) and almost faithful if \(K_{\rho}(A)=0\).

An \(A\)-module \(M\) is called irreducible if it contains no nonzero proper submodules and \(MA\neq 0\). The representation corresponding to an irreducible module is also called irreducible.

In the study of structure theory, the concepts of prime and semiprime algebra are important.

Definition. An algebra is called prime if, for any two of its ideals \(I,J\), \(IJ=0\) implies \(I=0\) or \(J=0\). An algebra is called semiprime if for any of its ideals \(I\), the fact that \(I^2=0\) implies \(I=0\).

3 Irreducible Novikov Representations↩︎

From the identities of Novikov algebras it follows that the ideal of R-identities of the variety of Novikov algebras contains the following elements: \[\begin{gather} x_1\otimes x_2 - x_2\otimes x_1,\\ x_1\otimes x_2\otimes x_3 - (x_1x_2)\otimes x_3 - x_3\otimes x_1 \otimes x_2 + x_3\otimes (x_1x_2),\\ x_1\otimes x_2\otimes x_3 - (x_1x_2)\otimes x_3 - (x_1x_3)\otimes x_2 + (x_1x_3)x_2. \end{gather}\]

It means that every Novikov \(A\)-module \(M\) satisfies the following identities for \(m\in M\), \(a,b,c\in A\): \[\begin{gather} (ma)b=(mb)a,\\ (m,a,b)c=(mc,a,b)=(m,ac,b),\\ m[a,b]=(m,b,a)-(m,a,b). \end{gather}\]

The last identity follows from the presence of the element \(x_1\otimes x_2 - x_2\otimes x_1\) in the ideal of R-identities after expanding all brackets.

The paper [12] introduced the concepts of left and right associativity of a module, but they coincide in the case of right-commutative algebras. A Novikov \(A\)-module \(M\) is called associative if \((m,a,b)=0\) for any \(m\in M\) and \(a,b\in A\).

Let \(M\) be a Novikov \(A\)-module and \(B\) a subspace of \(A\). Then the subspace \(MB=\{\sum m_ib_i\mid m_i\in M, b_i\in B\}\) is a submodule of \(M\).

In what follows, we will need the fact that the zero quasikernel implies that an algebra is prime.

Lemma 3.1. If a representation \(\rho\) of a Novikov algebra \(A\) is irreducible then the algebra \(A/K_{\rho}(A)\) is prime.

Proof. Let \(B,C\) be ideals of the algebra \(A\), \(BC\subseteq K_{\rho}(A)\). Suppose that \(B\nsubseteq K_{\rho}(A)\) and \(C\nsubseteq K_{\rho}(A)\). Denote by \(M\) the module corresponding to the representation \(\rho\). Consider the submodule \(MB\). Since \(B\nsubseteq K_{\rho}(A)\), we have \(B\nsubseteq \mathrm{Ker}_{\rho}(A)\) and \(MB\neq 0\). Since \(M\) is irreducible, this means that \(MB=M\). Similarly, \(MC=M\). Then there exist \(b\in B\) and \(c\in C\) such that \((Mb)c\neq 0\). But \(((Mb)c)a = ((Ma)b)c\subseteq (Mb)c\), so \((Mb)c\) is a nonzero submodule of \(M\) and \((Mb)c=M\). Note that \(M(BC)=0\). Then \[M = (Mb)c = (((Mb)c)b)c = ((M,b,c)b)c = (M,b^2c,c) = 0,\] it is a contradiction. The lemma is proved.

There is a well known description of irreducible modules over an associative algebra as quotient regular modules by a maximal right ideal, which satisfies some additional condition. A similar construction also exists in Novikov algebras.

Lemma 3.2.

Let \(A\) be a Novikov algebra and \(M\) an irreducible \(A\)-module. Then one of the following holds:

• module \(M\) is associative and \(M\) is isomorphic to the quotient module \(A/I\) for the maximal two-sided ideal \(I\) of \(A\), where \(A/I\) is a field;

• module \(M\) is not associative and isomorphic to the quotient module \(A/I\) for the maximal right ideal \(I\) of the algebra \(A\), and there exist elements \(e,f\in A\) such that \(a-(a,e,f)\in I\) for any \(a\in A\). Moreover, the element \(e\) is determined by the element \(f\), and for \(f\) one can take any element such that \((A,A,f)\) is not embeddable in \(I\).

Proof. First, note that the set \(S=\{u\in M\mid uA = 0\}\) is an \(A\)-submodule. Since module \(M\) is irreducible then \(S=0\).

If \((M,A,A)=0\), then module \(M\) is associative. In this case, \(mA\) is a submodule of \(M\) for any \(m\in M\). Since \(MA=M\), there exists an \(m\in M\) such that \(mA=M\). The mapping \(\psi:A\to M\), acting according to the rule \(\psi(a)=ma\), is an epimorphism of modules, therefore the module \(M\) is isomorphic to the quotient module \(A/I\) for the right maximal ideal \(I\) of \(A\). Since \(A\) is a Novikov algebra, the space \([A,A]\) is an ideal. Moreover, \(A[A,A] \subseteq (A,A,A)\subseteq I\), so \([A,A]\subseteq K_{\rho}(A)\), where \(\rho\) is the right representation corresponding to the module \(A/I\). Thus, the algebra \(A/K_{\rho}(A)\) is associative and commutative, so \(I\subseteq K_{\rho}(A)\). This means that \(I=K_{\rho}(A)\), i.e. \(I\) is a two-sided ideal and the algebra \(A/I\) is a field.

Let \((M,A,A)\neq 0\). Then there exists an \(m\in M\) such that \((m,A,A)\neq 0\), which implies that there exists an \(f\in A\) such that \((m,A,f)\neq 0\). Since \((m,A,f)a = (m,Aa,f)\subseteq (m,A,f)\), then \((m,A,f)\) is a nonzero \(A\)-submodule of \(M\), that is, \((m,A,f)=M\).

Consider the map \(\psi:A\to M\) acting according to the rule \(\psi(c)=(m,c,f)\). Note that \[\psi(cd)=(m,cd,b)=(m,c,b)d=\psi(c)d,\] hence \(\psi\) is a homomorphism of \(A\)-modules. The kernel of this homomorphism \(I=\mathrm{Ker}(\psi)\) is a right \(A\)-submodule of the regular module \(A\). It means that \(I\) is a right ideal of the algebra \(A\). Since \(\psi(A)=(m,A,f)=M\), then, by the module homomorphism theorem, \(M\simeq A/I\). As in the classical case, it is easy to see that \(I\) is a maximal right ideal.

The equality \((m,A,f)=M\) implies the existence of an \(e\in A\) such that \(m=(m,e,f)\). But then for any \(a\in A\) we have \[(m,a,f)=((m,e,f),a,f) = ((m,e,f)a)f-(m,e,f)(af) = (m,(e,a,f),f),\] whence \((m,a-(a,e,f),f) = 0\) and \(a-(a,e,f)\in\mathrm{Ker}(\psi)=I\). The lemma is proved.

Obviously, any module satisfying the first option in Lemma 2 is irreducible. Let us prove this for the second option.

Lemma 3.3. Let \(A\) be a Novikov algebra and \(I\) be its maximal right ideal. If there exist elements \(e,f\in A\) such that \(a-(a,e,f)\in I\) for any \(a\in A\), then the quotient module \(A/I\) is an irreducible Novikov \(A\)-module.

Proof. Let \(I\) be a maximal right ideal of a Novikov algebra \(A\), and there exist \(e,f\in A\) such that \(a-(a,e,f)\in I\) for any \(a\in A\). Let us prove the irreducibility of the module \(A/I\). It is easy to see that \(A/I\) does not contain proper nonzero submodules (since \(I\) is maximal). It remains to prove that \((A/I)A\neq 0\). Let us suppose that \((A/I)A=0\). Let \(a\in A\). Then \(0=((a+I),e,f) = a+I\) and \(a\in I\), it is a contradiction with the fact that \(I\neq A\). The lemma is proved.

Example 3.4. ([6]) Let \(A=L(y_{-1},y_0,\dots,y_{p^{n}-2})\) be a finite-dimensional simple Novikov algebra over an algebraically closed field \(F\) of characteristic \(p>2\) with the following multiplication table: \[y_iy_j=\begin{pmatrix} i+j+1\\ j \end{pmatrix}y_{i+j}+\delta_{i,-1}\delta_{j,-1}ay_{p^n-2}+\delta_{i,-1}\delta_{j,0}by_{p^n-2}.\] Consider the subspace \(I=L(y_0,y_1,\dots,y_{p^n-2})\). It is easy to see that \(I\) is a right ideal. Then \((y_{-1},y_{-1},y_1) = ay_{p^n-2}y_{1}-y_{-1}y_0 = -y_{-1}-by_{p^n-2}\). Thus \(y_{-1}-(y_{-1},y_{-1},-y_1) = by_{p^n-2}\in I\). Then \(x-(x,y_{-1},-y_1)\in I\) for any \(x\in A\), that is, \(A/I\) is an irreducible non-associative Novikov \(A\)-module. If \(\rho\) is the corresponding representation, then \(K_{\rho}(A)=0\), but \(\mathrm{Ker}(\rho) = L(y_1,\dots,y_{p^n-2},y_{-1})\neq 0\). Moreover, \(I\) is the unique maximal right ideal.

Thus, we obtain two observations. First, even a simple non-commutative finite-dimensional Novikov algebra has non-associative modules (unlike, for example, the case of alternative algebras). Second, the intersection of kernels of irreducible representations need not be an ideal even in a finite-dimensional Novikov algebra (again, unlike all classical cases).

Let us adapt the concept of an irreducible Novikov module to the case of an associative right-commutative algebra.

Lemma 3.5. Let \(A\) be an associative Novikov algebra. Then every irreducible Novikov \(A\)-module \(M\) is an associative \(A\)-module.

Proof. If \(M\) is not an associative \(A\)-module, then by Lemma 3.2 there exists a maximal right ideal \(I\) in \(A\) and elements \(e,f\in A\) such that \(a-(a,e,f)\in I\) for any \(a\in A\). But \((a,e,f)=0\), therefore \(a\in I\) for any \(a\in A\). This contradicts the fact that \(I\neq A\). The lemma is proved.

Recall that a right ideal of an associative algebra \(A\) is called modular if there exists \(e\in A\) such that \(a - ea\in I\) for any \(a\in A\). If \(A\) is a Novikov algebra, then it satisfies the condition \(A[A,A]=0\), so \(e[A,A]=0\), whence \([A,A]\subseteq I\). Thus, in an associative Novikov algebra, a modular right ideal is a two-sided ideal whose quotient is a field. In connection with this and Lemma 3.2, we can give the following definition.

Definition. A right ideal \(I\neq A\) of a Novikov algebra \(A\) is called \(\textit{\boldsymbol{modular}}\) if either there exist \(e,f\in A\) such that \(a-(a,e,f)\in I\) for any \(a\in A\), or if \(I\) is a two-sided ideal such that \(A/I\) is a commutative algebra with a unit.

Thus, the concept of a modular ideal is defined for both the Novikov algebra and the associative algebra, and the two definitions coincide at the intersection of these classes.

To avoid further confusion, we need to kill an ambiguity of the phrase maximal modular right ideal.

Lemma 3.6. Let \(I\) be a modular right ideal of the Novikov algebra \(A\). Then it can be embedded in some maximal right ideal of \(A\), which is also modular.

Proof. Consider the set \(X\) of right ideals of \(A\) that contain \(I\) but not \(e\). By Zorn’s lemma, \(X\) contains a maximal element \(J\). Let \(K\) be a right ideal of \(A\) and \(J\subsetneq K\). Then \(e\in K\) and \((a,e,f)=(e,a,f)\in K\), which means \(a - (a,e,f) = a\in K\), i.e. \(K=A\). Hence \(J\) is a maximal right ideal of \(A\). The lemma is proved.

4 The Jacobson Radical of Novikov Algebras↩︎

There are several approaches to constructing the Jacobson radical in associative algebras. It is currently unclear whether all approaches will lead to the same radical in Novikov algebras. Some approaches differ in different classes of algebras: for example, the concept of quasiregularity in Jordan algebras differs slightly from the classical one. However, the language of irreducible modules is the same in any variety.

Denote by \(\mathcal{J}\) the class of Novikov algebras that do not have irreducible modules.

Lemma 4.1. The class \(\mathcal{J}\) is closed under extensions and homomorphic images.

Proof. Let \(A\) be a Novikov algebra in \(\mathcal{J}\) and let \(I\) be a nonzero ideal of \(A\). Suppose that \(A/I\) has an irreducible module \(M\). This means that \(A/I\) has a maximal modular right ideal \(\overline{J}\). It is easy to see that the full preimage \(J\) of this right ideal in \(A\) is a maximal modular right ideal of \(A\). Thus, the class \(\mathcal{J}\) is closed under homomorphic images.

Now let \(J\) be an ideal of \(A\) such that \(A/J\) and \(J\) belong to \(\mathcal{J}\). Assume that \(A\notin\mathcal{J}\). By Lemma 3.2, two cases are possible.

1) The algebra \(A\) contains a maximal right ideal \(I\) such that \(a-(a,e,f)\in I\) for every \(a\in A\) and the module \(A/I\) is not associative. Then \((A/I,A,A)\neq 0\). Note that \((I+J)/J\) is a right modular ideal of \(A/J\). By Lemma 3.6 and Lemma 3.3, this means that \(I+J=A\).

Assume that \((A/I,A,J)\neq 0\). This means that there exists \(m\in A/I\) such that \((m,A,J)\neq 0\), whence there exists \(y\in J\) such that \((m,A,y)\neq 0\). Since \((m,A,y)\) is a nonzero \(A\)-submodule in \(M\), it follows that \((m,A,y)=M\). Consider the map \(\psi:A\to M\) with the condition \(\psi(c)=(m,c,y)\). Note that \(\mathrm{Ker}(\psi)=I\). The existence of an element \(x\in A\) such that \(m=(m,x,y)\) implies (as in the proof of Lemma 3.2) that \((m,a,y)=((m,(a,x,y),y),\) whence \(a-(a,x,y)\in I\). We have \(x=x_i+x_j\), where \(x_i\in I\) and \(x_j\in J\). Let \(a\in J\). Then \((a,x_i,y)\in I\cap J\), so \(a-(a,x_j,y)\in I\cap J\). Thus, \(I\cap J\) is a proper modular right ideal of \(J\), a contradiction by Lemma 3.6 and Lemma 3.3.

Thus, \((A/I,A,J)=0\), that is, \((A,A,J)\subseteq I\). This means that we can assume that \(f\in I\) and \(e\in J\). In particular, \((J,J,J)\subseteq I\cap J\). This means that the \(J\)-module \(J/(I\cap J)\) is an associative \(J\)-module. If \(J^2\subseteq I\), then \((A,A,A)\subseteq I\), a contradiction. Thus, \(J/(I\cap J)\) is a nontrivial associative \(J\)-module. By the homomorphism theorem, \(J/(I\cap J)\simeq (I+J)/I = A/I\), so \(A/I\) is a nontrivial associative \(J\)-module, and \(J/I\cap J\) is an irreducible \(A\)-module.

Let \(K\) be the maximal ideal of the algebra \(A\) contained in \(I\cap J\). Since the \(A\)-module \(J/I\cap J\) is irreducible, then either \([J,J]\subseteq I\) or \([J,J]+I\cap J = J\). We have \(J[J,J]\subseteq I\cap J\). Since \(J^2\nsubseteq I\), then \([J,J]\subseteq I\cap J\), that is, \([J,J]\subseteq K\). Thus, the algebra \(J/K\) is associative and commutative, so \((I\cap J)/K\) is a two-sided ideal. This means that \(I\cap J = K\), that is, \(I\cap J\) is a two-sided ideal in \(A\). Note that there are no non-zero ideals between \(I\cap J\) and \(J\) since the module \(A/I\) is irreducible. Then \(J/(I\cap J)\) is a minimal ideal of \(A/(I\cap J)\). Since \(J^2\) is not embedded in \(I\), \(J/(I\cap J)\) is a simple algebra [17]. Its commutativity was proved above, so \(J/(I\cap J)\) is a field. It is a contradiction.

2) \(A\) has a maximal ideal \(I\) such that \(A/I\) is a field. If \(J\subseteq I\), then \(A/I\) is an irreducible \(A/J\)-module, it is a contradiction. This means that \(I+J=A\). Then the algebra \(A/I=(I+J)/I\) is isomorphic to the algebra \(J/I\cap J\). Thus, as in the previous case, \(J/I\cap J\) is an irreducible \(J\)-module, it is a contradiction. The lemma is proved.

Definition. Let \(\mathcal{M}\) be a class of algebras that is stable under taking homomorphic images and ideals (for the purposes of this paper, we can assume that \(\mathcal{M}\) is the class of Novikov algebras). A subclass \(\mathcal{R}\) of \(\mathcal{M}\) is called radical (in the Kurosh-Amitsur sense) if the following conditions are satisfied:

(A) A homomorphic image of an algebra in \(\mathcal{R}\) also lies in \(\mathcal{R}\).

(B) Every algebra \(A\) in \(\mathcal{M}\) has an \(\mathcal{R}\)-ideal \(\mathcal{R}(A)\) containing all \(\mathcal{R}\)-ideals of \(A\).

(C) If \(\mathcal{R}(A)\) is the ideal from condition (B), then there are no nonzero \(\mathcal{R}\)-ideals in the algebra \(A/\mathcal{R}(A)\).

Throughout the above, by an \(\mathcal{R}\)-ideal of an algebra \(A\) we mean an ideal of an algebra \(A\) that belongs (as an algebra) to the class \(\mathcal{R}\). An algebra \(A\) is called \(\mathcal{R}\)-semisimple if \(\mathcal{R}(A)=0\).

It is well known that the class \(\mathcal{R}\) is radical if and only if conditions (A), (D), and (E) are satisfied, where

(D) The class \(\mathcal{R}\) is closed under extensions.

(E) If \(\{I_i\}\) is a chain of \(\mathcal{R}\)-ideals in \(A\) whose union coincides with \(A\), then \(A\) lies in the class \(\mathcal{R}\).

Theorem 4.2. The class \(\mathcal{J}\) is radical in the class of Novikov algebras.

Proof. Let us prove condition (E). Let \(J_i\in\mathcal{J}\) be a chain of ideals in \(A\), where \(\bigcup\limits_{i\in\Gamma} J_i = A\). Suppose that \(A\notin\mathcal{J}\). By Lemma 3.2, two cases are possible.

1) There exist a maximal right ideal \(I\) of \(A\) and elements \(e,f\in A\) such that \(a - (a,e,f)\in I\) for any \(a\in A\). Then there exists an \(m\in\Gamma\) such that \(e,f\in J_m\). Then \(a - (a,e,f)\in I\cap J_m\) for any \(a\in J_m\), hence \(I\cap J_m\) is a modular ideal in \(J_m\). By Lemma 3.3 and Lemma 3.6, this means that \(I\cap J_m=J_m\), that is, \(J_m\subseteq I\). Then \(A/I\) is an irreducible \(A/J_m\)-module, it is a contradiction.

2) There exists a maximal ideal \(I\) in \(A\), where \(A/I\) is a field. Let \(e\in A\) be an arbitrary preimage of the identity in \(A/I\). Then \(e\in J_m\) for some \(m\). Let \(a,b\in A\). Then \[[a,b]+I\cap J_m = [ea,b] + I\cap J_m = -[e,ba] + 2[e,b]a + I\cap J_m = 0 + I\cap J_m,\] that is, \([A,A]\subseteq I\cap J_m\) and the algebra \(A/I\cap J_m\) is commutative. Then the algebra \(J_m/(I\cap J_m)\) is also commutative and isomorphic to \((I+J_m)/I = A/I\). This means that \(J_m/(I\cap J_m)\) is an irreducible \(J_m\)-module, it is a contradiction.

Conditions (A) and (D) were proved in Lemma 4.1. The theorem is proved.

Theorem 4.2 means that in any Novikov algebra \(A\), there exists a largest \(\mathcal{J}\)-ideal \(\mathcal{J}(A)\), and the algebra \(A/\mathcal{J}(A)\) does not contain nonzero \(\mathcal{J}\)-ideals. The radical \(\mathcal{J}\) will be called (in accordance with the classical case) the Jacobson radical.

5 Primitive Novikov Algebras↩︎

A Novikov algebra \(A\) is called primitive if it has a maximal modular right ideal \(I\) that does not contain nonzero two-sided ideals of \(A\). An ideal \(I\) of \(A\) is called primitive if \(A/I\) is primitive.

Lemma 5.1. A Novikov algebra \(A\) has an almost faithful irreducible module if and only if \(A\) is primitive. Moreover, a primitive algebra is prime.

Proof. Let \(M\) be an almost faithful irreducible \(A\)-module. By Lemma 3.2, \(M\) is isomorphic to the quotient module \(A/I\) for the maximal right ideal \(I\) of \(A\). Let \(J\) be a two-sided ideal contained in \(I\). Then \(MJ = 0\), which implies that \(J=0\) since \(M\) is almost faithful. Thus, \(A\) is primitive.

Conversely, let \(A\) be primitive and \(I\) be the maximal modular right ideal that does not contain nonzero two-sided ideals of \(A\). Then \(A/I\) is an irreducible \(A\)-module by Lemma 3.3. Let \(\rho\) be the corresponding representation, \(x\in K_{\rho}(A)\). By Lemma 3.2, either \(I=0\) (and \(A\) is an almost faithful irreducible \(A\)-module), or there exist \(e,f\in A\) such that \(x-(x,e,f)\in I\), so that \(x-(ex)f+e(xf)\in I\). Since \(x\in K_{\rho}(A)\), we have \(ex\in I\) and \(xf\in K_{\rho}(A)\), whence \(e(xf)\in I\). Thus, \(x\in I\), that is, \(K_{\rho}(A)\subseteq I\). Since \(I\) does not contain nonzero two-sided ideals of \(A\), we have \(K_{\rho}(A)=0\).

By what was proved above and Lemma 3.1, the primitive algebra is prime. The lemma is proved.

Semisimple (by Jacobson) algebras in the associative case are sometimes called semiprimitive. This definition is also justified for Novikov algebras.

Lemma 5.2. A primitive Novikov algebra is \(\mathcal{J}\)-semisimple.

Proof. Let \(A\) be a primitive Novikov algebra. By Lemma 5.1, \(A\) has a maximal modular right ideal \(I\) that does not contain nonzero two-sided ideals of \(A\). By Lemma 3.2, two cases are possible.

1) The module \(M\) is isomorphic to \(A/I\), where \(I\) is a two-sided ideal, and \(A/I\) is a field. Then \(I=0\) and \(A\) is a field.

2) The module \(M\) is isomorphic to \(A/I\), where \(I\) is the maximal right ideal of \(A\), and \(a-(a,e,f)\in I\) for any \(a\in A\). The corresponding representation is almost faithful, so \(I\) does not contain nonzero two-sided ideals. Suppose that \(\mathcal{J}(A)\neq 0\). Then \(\mathcal{J}(A)+I=A\).

Suppose \((A,A,\mathcal{J}(A))\nsubseteq I\). Then, by Lemma 3.2, we can assume that \(f\in \mathcal{J}(A)\). Note that since \((a,e,f)=(e,a,f)\), then we can assume that \(e\in \mathcal{J}(A)\). This means that \(a-(a,e,f)\in\;I\cap \mathcal{J}(A)\) for any \(a\in\mathcal{K}(A)\). Thus, \(I\cap\mathcal{J}(A)\) is a modular right ideal of \(\mathcal{J}(A)\), which embeds in the maximal modular right ideal by Lemma 3.6. By Lemma 3.3, there exists an irreducible \(\mathcal{J}(A)\)-module, it is a contradiction.

Thus, \((A,A,\mathcal{J}(A))\subseteq I\). Then \((A,A,\mathcal{J}(A))=0\) and \(\mathcal{J}(A)\) is an associative Novikov algebra. In particular, \(\mathcal{J}(A)[\mathcal{J}(A),\mathcal{J}(A)]=0\). By Lemma 3.1, \(A\) is prime, so \(\mathcal{J}(A)\) is commutative. Then \(I_1=I\cap\mathcal{J}(A)\) is a two-sided ideal of \(\mathcal{J}(A)\). But then \(AI_1=II_1+\mathcal{J}(A)I_1 = II_1+I_1\mathcal{J}(A)\subseteq I_1+II_1\). It remains to note that \(II_1\subseteq I\) and \(II_1\subseteq \mathcal{J}(A)\), so \(AI_1\subseteq I_1\) and \(I_1\) are two-sided ideals of \(A\) contained in \(I\). In particular, \(\mathcal{J}(A)I\subseteq \mathcal{J}(A)\cap I = 0\). Thus, \(I_1=0\). But then \(AI \subseteq I^2+\mathcal{J}(A)I \subseteq I\), that is, \(I\) is a two-sided ideal of \(A\) and \(I=0\). But then \(A=\mathcal{J}(A)\), it is a contradiction. The lemma is proved.

Lemma 5.3. Let \(A\) be a Novikov algebra, \(I\) be an ideal in \(A\), and \(J\) be the maximal modular right ideal of \(I\). Then \(J\) is a right ideal of \(A\).

Proof. By definition, two cases are possible.

1) \(J\) is a two-sided ideal in \(I\), where \(I/J\) is a field. Then \(J\) is an ideal in \(A\).

2) There exist elements \(e,f\in I\) such that \(a-(a,e,f)\in J\) for any \(a\in I\). Suppose that \(JA\nsubseteq J\). Then \(Jx\nsubseteq J\) for some \(x\in A\). Let \(j\in J, i\in I\). Then \((jx)i = (ji)x \in Jx\), so \(Jx\) is a right ideal of \(I\). Then \(J+Jx\) is a right ideal of \(I\) that strictly contains \(J\), that is, \(J+Jx=I\). In particular, \(e=c+dx\), where \(c,d\in J\). Then for any \(a\in I\) we have \(a-(a,e,f) = a-(a,c,f)-(a,dx,f)\in J\). Note that \((a,c,f)=(c,a,f)\in J\), \((a,dx,f) = (d,ax,f)\in J\). Thus, \(a\in J\) for any \(a\in I\), it is a contradiction. Hence, \(JA\subseteq J\) and \(J\) is a right ideal of \(A\). The lemma is proved.

To obtain the final result, we need to learn how to lift maximal modular right ideals from some ideal to the entire algebra.

Lemma 5.4. Let \(B\) be an ideal of the alternative algebra \(A\). Then the set of maximal modular right ideals of \(B\) coincides with the set of right ideals of the form \(I\cap B\), where \(I\) is the maximal modular right ideal of \(A\) that does not contain \(B\).

Proof Let \(J\) be the maximal modular right ideal of \(B\). By Lemma 5.3, \(J\) is a right ideal of \(A\). Two cases are possible.

1) \(J\) is a two-sided ideal and the algebra \(B/J\) is a field. Let \(e\in B\) be the preimage of the unit of the algebra \(B/J\). Consider the right ideal \(J_1=\{a-ea\mid a\in A\}\) and the right ideal \(J_2=J_1+J\). If \(e\in J_2\), then \(e=a-ea+b\) for some \(a\in A\), \(b\in J\). Then \(e^2=ea-e(ea)+eb\). Since \(ea\in B\), we have \(ea-e(ea)\in J\), whence \(e^2=ea-e(ea)+eb\in J\), a contradiction. Thus, \(e\notin J_2\). Then the set of right ideals containing \(J_2\) and not containing \(e\) contains a maximal element \(J_0\). If \(J_0\subsetneq K\), where \(K\) is a right ideal, then \(e\in K\). But \(a - ea \in J_1 \subseteq K\) for any \(a \in A\), therefore \(a \in K\) and \(K = A\). This means that \(J_0\) is a maximal right ideal of \(A\). Since \(a - ea \in J_0\) for any \(a \in A\) and \(e \notin J_0\), the module \(A/J_0\) is irreducible. From Lemma 3.2 it follows that \(J_0\) is a maximal modular right ideal of \(A\).

It remains to note that \(B \cap J_0\) is a right ideal of \(A\) satisfying the condition \(J \subseteq B \cap J_0 \subsetneq B\), whence \(B \cap J_0=J\) since \(J\) is maximal in \(B\).

2) There exist elements \(e,f\in B\) such that \(a-(a,e,f)\in J\) for any \(a\in B\). Consider the right ideal \(J_1=\{a-(a,e,f)\mid a\in A\}\) and the right ideal \(J_2=J_1+J\). If \(e\in J_2\), then \(e=a-(a,e,f)+b\) for some \(a\in A\), \(b\in J\). Then \(e^2 = ae-(ae,e,f)+be\). Since \(ae\in B\), we have \(ae-(ae,e,f)\in J\). Since \(J\) is a right ideal, we have \(be\in J\). Thus, \(e^2\in J\), it is a contradiction. Thus, \(e\notin J_2\). Then the set of right ideals containing \(J_2\) and not containing \(e\) contains a maximal element \(J_0\). As in the previous case, \(J_0\) is a maximal right ideal of \(A\). Since \(a-(a,e,f)\in J_0\) for any \(a\in A\), \(J_0\) is a maximal modular right ideal of \(A\). It remains to note that \(B\cap J_0\) is a right ideal of \(A\) satisfying the condition \(J\subseteq B\cap J_0\subsetneq B\), whence \(B\cap J_0=J\) since \(J\) is maximal in \(B\). The lemma is proved.

In some classical cases (associative, alternative), the Jacobson radical coincides with the intersection of the kernels of all irreducible representations. In the case of Novikov algebras, a different, slightly weaker result holds.

Theorem 5.5. The Jacobson radical of a Novikov algebra coincides with the intersection of the quasi-kernels of all its irreducible representations.

Proof. Denote by \(L\) the intersection of \(K_{\rho}(A)\) over all irreducible representations \(\rho\) of the Novikov algebra \(A\). For any such \(\rho\), by Lemma 5.1, \(\mathcal{J}(A/K_{\rho}(A))=0\). Therefore, \(\mathcal{J}(A)\subseteq K_{\rho}(A)\) by Theorem 4.2 due to the well-known properties of the radical. Thus, \(\mathcal{J}(A)\subseteq L\).

Suppose that \(\mathcal{J}(A)\subsetneq L\). Then there exists an irreducible \(L\)-module. By Lemma 3.2, there exists a maximal right modular ideal \(I\) of \(L\). By Lemma 5.4, \(I=J\cap L\) for some maximal modular right ideal \(J\) of \(A\), it is a contradiction. Thus, \(\mathcal{J}(A)=L\). The theorem is proved.

From Example 3.4 it follows that the intersection of kernels of irreducible representations need not even be a subalgebra, much less coincide with \(\mathcal{J}(A)\), even in the finite-dimensional simple algebra.

By Lemma 5.1, there is a one-to-one correspondence between the primitive ideals of \(A\) and the ideals of \(K_{\rho}(A)\) with respect to the irreducible representations of \(A\). Thus, the following corollary holds.

Corollary 5.6. In every Novikov algebra, the Jacobson radical coincides with the intersection of all its primitive ideals. Every \(\mathcal{J}\)-semisimple Novikov algebra is a subdirect product of primitive algebras.

Note that in the finite-dimensional case, the Jacobson radical is a solvable ideal, as the following assertion shows.

Proposition 5.7. Let \(A\) be a finite-dimensional Novikov algebra. Then \(\mathcal{J}(A)\) coincides with the largest solvable ideal \(N(A)\).

Proof. Let \(C\) be a simple finite-dimensional Novikov algebra and \(I\) be the maximal right ideal of \(C\). If \((C/I)C=0+I\), then \(C^2\subseteq I\), that is, \(C^2\neq C\). But then \(C^2=0\) since \(C\) is simple, therefore \((C/I)C\neq 0\), which means the module \(C/I\) is irreducible. Thus, a simple finite-dimensional Novikov algebra has an irreducible module, and hence a classically semisimple nonzero finite-dimensional Novikov algebra also has an irreducible module.

Let \(A\) be a finite-dimensional Novikov algebra. From Lemma 3.2, it is clear that \(N(A)\) has no irreducible right modules, hence \(N(A)\subseteq\mathcal{J}(A)\). Consider the algebra \(B=\mathcal{J}(A)/N(A)\). The algebra \(B\) is a semiprime \(\mathcal{J}\)-radical Novikov algebra. Then \(B\) is classically semisimple, and hence (if \(B\neq 0\)) it has an irreducible right module. Since \(B\) is \(\mathcal{J}\)-radical, this means that \(B=0\). The proposition is proved.

Definition. A radical class \(\mathcal{R}\) is called hereditary if for any algebra \(A\) and any of its ideals \(I\), the equality \[\mathcal{R}(I)=\mathcal{R}(A)\cap I\] holds.

Theorem 5.8. The Jacobson radical of Novikov algebras is hereditary.

Proof. It suffices to prove that the ideal of a \(\mathcal{J}\)-semisimple algebra is a \(\mathcal{J}\)-semisimple algebra, and the ideal of a \(\mathcal{J}\)-radical algebra is a \(\mathcal{J}\)-radical algebra.

Let \(A\) be a \(\mathcal{J}\)-semisimple algebra and \(I\) be an ideal of \(A\). Since \(I\neq\mathcal{J}(I)\), the algebra \(I/\mathcal{J}(I)\) is \(\mathcal{J}\)-semisimple. It is easy to see that an algebra with zero multiplication is \(\mathcal{J}\)-radical, so the algebra \(I/\mathcal{J}(I)\) is semiprime. Then \(\mathcal{J}(I)\) is a \(\mathcal{J}\)-ideal of \(A\) (by [13]), that is, \(\mathcal{J}(I)=0\).

Let \(A\) be a Novikov algebra and \(I\) be an ideal of \(A\). Suppose that \(I\neq\mathcal{J}(I)\). Then the algebra \(I\) has an irreducible module \(M\). By Lemma 3.2, \(M\) is isomorphic to the quotient module \(I/J\) for some right maximal modular ideal \(J\) of \(I\). By Lemma 5.4, the algebra \(A\) contains a maximal modular right ideal. By Lemma 3.3, this means that there exists an irreducible \(A\)-module, that is, \(A\neq\mathcal{J}(A)\). The theorem is proved.

Alexander Panasenko
Sobolev Institute of Mathematics
Acad. Koptyug ave. 4, 630090 Novosibirsk, Russia
e-mail: a.panasenko@g.nsu.ru

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