Random generation of associative algebras

There has been considerable interest in recent decades in questions of random generation of finite and profinite groups, and finite simple groups in particular. In this paper we study similar notions for finite and profinite associative algebras. Let $k=F_q$ be a finite field. Let $A$ be a finite dimensional, associative, unital algebra over $k$. Let $P(A)$ be the probability that two elements of $A$ chosen (uniformly and independently) at random will generate $A$ as a unital $k$-algebra. It is known that, if $A$ is simple, then $P(A) \to 1$ as $|A| \to \infty$. We extend this result to a large class of finite associative algebras. For $A$ simple, we find the optimal lower bound for $P(A)$ and we estimate the growth rate of $P(A)$ in terms of the minimal index $m(A)$ of any proper subalgebra of $A$. We also study the random generation of simple algebras $A$ by two elements that have a given characteristic polynomial (resp. a given rank). In addition, we bound above and below the minimal number of generators of general finite algebras. Finally, we let $A$ be a profinite algebra over $k$. We show that $A$ is positively finitely generated if and only if $A$ has polynomial maximal subalgebra growth. Related quantitative results are also established.


Introduction
In the past few decades there has been extensive research on random generation of finite and profinite groups with emphasis on finite simple groups.See for instance the survey articles [13,25] and the references therein.
The study of random generation of associative algebras is less well developed.Consider the algebra M n (q) of n × n matrices over a finite field F q .In 1995 it was shown by Peter Neumann and Cheryl Praeger [19] that the probability that two matrices in M n (q), chosen independently under the uniform distribution, generate M n (q) as a F q -algebra tends to 1 as |M n (q)| → ∞.See also the subsequent paper [10] by Kravchenko, Mazur and Petrenko for additional results on random generation of finite and infinite algebras.
One can refine this problem and consider random generation of an algebra by two elements that satisfy a certain property.A matrix in M n (q) is cyclic if its characteristic polynomial is equal to its minimal polynomial.Neumann and Praeger showed in [19] that almost all pairs of cyclic matrices in M n (q) will generate it as a F q -algebra.Amongst other results of this flavour, we show that -given a monic polynomial f of degree n over F qalmost all pairs of matrices in M n (q) with characteristic polynomial f will generate it as a F q -algebra.
In this paper we study random generation of finite and profinite associative algebras, and we obtain some new results also in the case of simple algebras.
Let k be a finite field, that is, k = F q for some prime power q.Unless otherwise stated, all algebras in this paper are assumed to be over k, and are associative and unital.
Subalgebras of a unital algebra are required to contain the multiplicative identity of the original algebra.We first focus on the study of finite algebras.Later on, we look at profinite algebras.
Let A be an associative, unital, finite-dimensional algebra over k (a.k.a. a finite algebra).Let A × denote the group of units of A. Let A N denote the set of nilpotent elements of A. The Jacobson radical J(A) of A is a nilpotent ideal of A. If J(A) is trivial then A is semisimple.
In this paragraph we summarise the Wedderburn-Malcev Principal Theorem (Theorems 5.3.20 and 5.3.21 of [23]).There exists a semisimple subalgebra S of A such that A = S ⊕ J(A) as vector spaces.If S ′ is another subalgebra of A satisfying A = S ′ ⊕ J(A) then S ′ is conjugate to S by an element of 1+ J(A).Wedderburn's little theorem (Theorem 7.1.11of [23]) states that all finite division algebras are fields.Combining this with another theorem of Wedderburn (Theorem 2.1.8 of [23]), it follows that there is an algebra isomorphism S ∼ = r i=1 M n i (q m i ) for some integers r, n 1 , ..., n r , m 1 , ..., m r that is unique up to permutation of the factors.
Denote n := min i=1,...,r {n i } and m := min i=1,...,r {m i }.Fix constants c > 1 and λ > 0. We say that A is bounded by (c, λ) if r ≤ λc min{m,n}/2 and dim J(A)/J(A) 2 ≤ log q λ + min{m, n} 2 log q c.A subset X of A is a generating set if the set of all monomials in the elements of X (including the trivial monomial) spans A as a k-vector space.We define P (A) to be the probability that two elements of A chosen uniformly at random will generate A as a (unital) k-algebra.That is, Theorem 1.1.Fix constants 1 < c < q and λ > 0. Let A be a finite algebra, say A = r i=1 M n i (q m i ) ⊕ J(A), that is bounded by (c, λ).Denote n := min i=1,...,r {n i } and m := min i=1,...,r {m i }.Then P (A) → 1 as n → ∞, as m → ∞ or as q → ∞.
It is not true in general that P (A) → 1 as |A| → ∞.For example, let A be as in the theorem above and suppose there exists a positive integer i ≤ r such that n i = 1 and m i = 2. Then A has a maximal subalgebra B satisfying A/B ∼ = k.Hence |B|/|A| = q −1 , so 1 − P (A) ≥ |B| 2 /|A| 2 = q −2 .Fixing q and letting |A| tend to infinity we see that P (A) ≤ 1 − q −2 is bounded away from 1.
Moreover, let A = k r for some r ∈ N. Then any maximal subalgebra B of A has codimension 1, and it is easy to see that P (A) → 0 as r → ∞.However, Theorem 1.1 implies the following known result.
This corollary is somewhat more general than the Neumann-Praeger result stated above, in the sense that it also deals with A = M n (q m ) as a F q -algebra, but it is obtained in [10] using different methods.
An equivalent formulation of Corollary 1.2 is as follows.Let A be a simple algebra and consider the free associative algebra k X 1 , X 2 .Then the probability that a randomly chosen k-algebra homomorphism k X 1 , X 2 → A is surjective tends to 1 as |A| → ∞.
It is well known that any finite simple algebra is 2-generated, see for instance Theorem 6.4 of [10].So it follows from Corollary 1.2 that there exists an absolute constant δ > 0 such that P (A) ≥ δ for all finite simple algebras A. In the following result, we find the best possible value for this constant.
Theorem 1.3.Let A be a finite simple algebra.Then P (A) ≥ 3/8, with equality if and For G a finite simple group, let P (G) be the probability that two randomly chosen elements of G will generate G.It is a consequence of Theorem 1.1 of [18] that P (G) ≥ 53/90, with equality if and only if G = A 6 .
For A simple and not a field, we investigate the growth rate of P (A) in more detail.Let m(A) be the minimal index (as an additive group) of any proper subalgebra of A.
Theorem 1.4.Let A be a finite simple algebra that is not a field.Then We will see in Section 5 that the constants in Theorem 1.4 are best possible.Note that Theorem 1.4 gives us an alternate proof of Corollary 1.2.Results of this flavour for finite simple groups were obtained by Liebeck and Shalev, see Theorems 1.5 and 1.6 in [14].
We next look at randomly generating a finite algebra by its nilpotent elements.Define P N (A) to be the probability that two nilpotent elements of A chosen uniformly at random will generate A as a k-algebra.That is, We prove an analogue of Theorem 1.1.
Note that Theorem 1.5 does not hold when n = 1.For example, let A 0 be a finite algebra and let A = F q m × A 0 for some m > 1.Let b be a prime divisor of m and consider the maximal subalgebra B = F q m/b × A 0 of A. Observe that all nilpotent elements of A are contained in B. So P N (A) = 0, regardless of the choice of q, m or A 0 .
Theorem 1.5 immediately implies the following.
Corollary 1.6.Let A be a finite simple algebra that is not a field.Then P N (A) → 1 as |A| → ∞.
We now consider random generation of a finite simple algebra by two elements that have a given characteristic polynomial.Let A = M n (q m ), let f be a monic polynomial of degree n over F q m and let A f be the set of elements of A with characteristic polynomial f .We define P f (A) to be the probability that two elements of A f chosen uniformly at random will generate A as a k-algebra.That is, Theorem 1.7.Let A be a finite simple algebra that is not a field, say A = M n (q m ) for n > 1.Let f be a monic polynomial of degree n over F q m .Then P f (A) → 1 as |A| → ∞.
By applying Theorem 1.7 to the case where f (X) = X n , we find an alternate proof of Corollary 1.6.
Note that Theorem 1.7 does not hold when A is a field.For example, let A = F q m for some m > 1.Let b be a prime divisor of m and consider the maximal subfield B = F q m/b of A. Let x ∈ B and let f be the polynomial X − x over F q m .Then A f = B f = {x}, and so P f (A) = 0 regardless of the choice of q or m.
We remark that Theorem 1.7 still holds, with essentially the same proof, if we replace P f (A) with P f,g (A), where g is another monic polynomial of degree n over F q m and P f,g (A) is the probability that a random element of A f × A g will generate A as a k-algebra.
We now consider random generation of a finite simple algebra by two matrices that have a given rank.Let α be a non-negative integer.Let A = M n (q m ) where n ≥ α and let A α be the set of matrices in A with rank α.We define P α (A) to be the probability that two elements of A α chosen uniformly at random will generate A as a k-algebra.That is, Theorem 1.8.Let A be a finite simple algebra that is not a field, say A = M n (q m ) for n > 1.Let α := α(n) be a positive integer.
It is not true that P α (A) always tends to 1 as |A| → ∞.This is an immediate consequence of Theorem 1.8(i).We can see this by taking α to be independent of n, and letting n tend to infinity whilst fixing q, m and p.
Let P × (A) to be the probability that two invertible elements of A chosen uniformly at random will generate A as a k-algebra.Theorem 1.8(ii) implies the following.
Next, we investigate the minimal number of generators d(A) of a finite algebra A. An obvious upper bound for d(A) is log q |A|− 1 (the −1 term arises from our convention that the multiplicative identity of A is automatically included in any generating set of A, and of course dim A = log q |A|).This upper bound is strict, and is realised in the case where J(A) has codimension 1 in A and J(A) 2 = 0.
In general d(A) often grows much slower than log q |A|.For example, if A is the direct product of finitely many copies of k then d(A) = log q log q |A| .In particular, if A = k r for some 1 < r ≤ q then d(A) = 1.Moreover, as remarked earlier, if A is simple then Theorem 1.10.Let A be a finite algebra, say A = S ⊕ J(A) where S = r i=1 S α i i , S i = M n i (q m i ) for each i and the S i 's are pairwise non-isomorphic.Let f (A, i) := m −1 i n −2 i log q α i m i , let f (A) := max i {f (A, i)} and let µ(A) be the minimal length of an unrefinable chain of S-subbimodules of J(A).Then In the final part of this paper we study positively finitely generated (profinite) algebras and related topics.For the theory of positively finitely generated groups see [17,2,20,22,5,15,9] and the references therein.
A profinite algebra is a topological algebra (over k) that is isomorphic to a projective limit of discrete finite algebras.Henceforth, let A be a profinite algebra.
For d ≥ 1 let P (A, d) be the probability that d randomly chosen elements of A generate A (topologically if A is infinite).We say that A is positively finitely generated (PFG) if P (A, d) > 0 for some d.We say that A has polynomial maximal subalgebra growth (PMSG) if the number m n (A) of index n (open) maximal subalgebras of A is bounded by n c for some fixed constant c.It was shown in [17] that, for profinite groups, PFG is equivalent to PMSG.Here we study these notions and related invariants for profinite algebras.We establish the following.
Theorem 1.11.Let A be a profinite algebra.Then A is PFG if and only if A has PMSG.Moreover, if A is infinite we have M * (A) ≤ d 0 (A) + 1.
The bound above is better than related bounds obtained for profinite groups.
For any real number η ≥ 1, define the Pomerance invariant of A by Clearly V η (A) ≥ d 0 (A), with equality for sufficiently large η.The case where η = e, which we denote by V (A) := V e (A), was studied by Pomerance [22] for finite abelian groups.
Next, define the Pak invariant E(A) of A to be the expected number of random elements of A chosen uniformly and independently which generate A (topologically).A similar invariant was introduced by Pak [20] for finite groups.
Our final main result establishes bounds on these invariants, and is a ring-theoretic analogue of results of Lubotzky [15] and Lucchini-Moscatiello [16] for finite groups.
Theorem 1.12.Let A be a finite algebra, say A/J(A) = r i=1 S i .Then In particular, the expected number of random elements of A which generate A is of the order of magnitude O(d(A) + log q log q |A|).
This paper is structured as follows.In Section 2 we present a classification of maximal subalgebras of a finite algebra A, then we introduce and investigate a related zeta function of A. In Sections 3, 4 and 5 we investigate P (A) and its growth rate.In particular, in Section 3 we prove Theorem 1.1 and Corollary 1.2, in Section 4 we prove Theorem 1.3 and in Section 5 we prove Theorem 1.4.In Sections 6, 7 and 8 we study random generation of a finite algebra by special elements.In Section 6 we prove Theorem 1.5 and Corollary 1.6, in Section 7 we prove Theorem 1.7 and in Section 8 we prove Theorem 1.8 and Corollary 1.9.In Section 9 we look at the minimal number of generators of a finite algebra, and prove Theorem 1.10.Finally, in Section 10 we investigate positively finitely generated profinite algebras, and prove Theorems 1.11 and 1.12.

Preliminaries
Recall that k = F q where q is a prime power.Let A be an (associative, unital) finite simple algebra (over k).By Wedderburn's Theorem, we can write A = M n (q m ) for some positive integers n and m.
Some remarks on notation.Let α = (α 1 , ..., α s ) be a composition of n (i.e.n = s i=1 α i where the α i 's are positive integers) and suppose s ≥ 2. Let P α (q m ) be the subalgebra of A that consists of all block upper triangular matrices with s blocks on the diagonal such that the i'th block has size α i .
Let r be a positive integer.There is a natural embedding of F q r in M r (q) via the left regular representation.If r divides n then this extends to an embedding of M n/r (q mr ) in M n (q m ).If r divides m then the subfield F q m/r of F q m extends naturally to a subalgebra M n (q m/r ) of M n (q m ).Let P(r) denote the set of prime divisors of r (not counting multiplicities).Let ω(r) := |P(r)|.
We define three sets of subalgebras of A; A subalgebra of A that is conjugate to an element of S1 (resp.S2, S3) is said to be of type (S1) (resp.(S2), (S3)).
Theorem 1.Let A be a finite simple algebra.With the above notation, S1 ∪ S2 ∪ S3 is a set of representatives of the conjugacy classes of maximal subalgebras of A Proof.Over any field k, Lemma 3.6 of Iovanov and Sistko [8] classifies maximal subalgebras of a simple k-algebra up to isomorphism.We adapt this result to the case where k = F q , and then we consider conjugacy classes.
Let B be a maximal subalgebra of A. If B is not simple then, by Lemma 3.6 of [8], B is conjugate to P l,n−l (q m ) for some positive integer l < n.Let l ′ < n be a positive integer.It is well known that P l,n−l (q m ) is conjugate to P l ′ ,n−l ′ (q m ) if and only if l = l ′ (see for instance §3 of [6]).
Henceforth let B be simple.By Lemma 3.6 of [8], there are two possibilities.Either for some prime divisor a of n.By the double centraliser theorem (Theorem 7.1.9 of [23]), Z(B) = F and [F : Recall from Wedderburn's little theorem that all finite division algebras are fields.It follows that B ∼ = M n/a (F ).Any subalgebra of A that is isomorphic to B is then conjugate to B by the Skolem-Noether theorem.
Now assume that Z(A) ⊇ Z(B).Then, by Lemma 3.6 of [8], b for some prime divisor b of m.Since A and B are both simple, it follows from Wedderburn's theorem that B ∼ = M n (q m/b ).
Let ι : B ֒→ A be inclusion.Observe that ι extends to a Z(A)-isomorphism ι * : We call S1 ∪ S2 ∪ S3 the standard set of representatives of the conjugacy classes of maximal subalgebras of A.
We now relax the assumption that A is simple.Let A be any finite algebra over k.By the Wedderburn-Malcev Principal Theorem, there exists a semisimple subalgebra S of A such that A = S ⊕ J(A).Decompose S = r i=1 S i where each S i is simple.Let i ∈ {1, ...r}.Write S i = M n i (q m i ) for some integers m i and n i .Let B i be the standard set of representatives of the conjugacy classes of maximal subalgebras of S i .If S j ∼ = S i for some j = i then let S ij denote the image of the diagonal embedding S i → S i × S j .
We define three sets of subalgebras of A; and A subalgebra of A that is conjugate to an element of T 1 (resp.T 2, T 3) is said to be of type (T 1) (resp.(T 2), (T 3)).
Theorem 2. Let A be a finite algebra.With the above notation, T 1 ∪ T 2 ∪ T 3 is a set of representatives of the conjugacy classes of maximal subalgebras of A.
Proof.By Theorems 2.5 and 3.10 of [8], every maximal subalgebra of A is conjugate to an element of T 1 ∪ T 2 ∪ T 3. It remains to check that all elements of T 1 ∪ T 2 ∪ T 3 are pairwise non-conjugate in A.
We first consider the case where A is semisimple, that is, J(A) = 0. Note that T 3 = ∅.It is easy to see that the elements of T 1 ∪ T 2 are pairwise non-conjugate as the simple components of A commute with each other.
We now consider the general case.That is, A is any algebra.
We call T 1 ∪ T 2 ∪ T 3 the standard set of representatives of the conjugacy classes of maximal subalgebras of A.
We now introduce a 'zeta function' of A. Let B be the standard set of representatives of the conjugacy classes of maximal subalgebras of A. For ǫ > 0, we define where Next, we prove a result which serves as a main tool in this paper.Recall the notation A = r i=1 M n i (q m i ) ⊕ J(A).Denote n := min i=1,...,r {n i } and m := min i=1,...,r {m i }.
Theorem 3. Fix constants λ > 0 and ǫ > 0. With the above notation, there exists Proof.Fix ǫ > 0. Let B be the standard set of representatives of the conjugacy classes of maximal subalgebras of A. Let B ∈ B.
We first consider the case where A is simple.That is, A = M n (q m ).Let Σ 1 (resp.Σ 2 , Σ 3 ) denote the contribution to the sum in (1) of the maximal subalgebras in S1 (resp.S2, S3).
We consider individually each of the possibilities that B is in S1, S2 or S3.Let B ∈ S1.That is, B = P l,n−l (q m ) for some positive integer l < n.Observe that |B| = q m(n 2 −l(n−l)) .Then That is, B = M n/a (q ma ) for some prime divisor a of n.Observe that |B| = q mn 2 /a .Then This completes the proof for the case where A is simple.We now consider the general case.That is, A = S ⊕ J(A) where S = r i=1 S i is semisimple and S i = M n i (q m i ) for each i.Let Ω 1 (resp.Ω 2 , Ω 3 ) denote the contribution to the sum in (1) of the maximal subalgebras in T 1 (resp.T 2, T 3). Let For simplicity, denote n 0 := n i 0 and m 0 := m i 0 .
Let c ∈ R such that 1 < c < q ǫ and let λ > 0. We impose the condition that A is bounded by (c, λ).That is, r ≤ λc min{m,n}/2 and dim J(A)/J(A) 2 ≤ log q λ + min{m, n} 2 log q c.Rearranging this second inequality gives us for some j ∈ {1, ..., r} and maximal subalgebra B j of S j .Then we have Let S op denote the opposite algebra of S. Observe that J(A)/H is a non-trivial simple S-bimodule and hence, by the equivalence of categories in Proposition 10.1 of [21], J(A)/H also has the structure of a non-trivial simple left S ⊗ k S op -module.Consider the k-algebra isomorphism S ⊗ k S op ∼ = 1≤i,j≤r M n i n j (q m i m j ).Then, by Proposition 2.3 of [21], any simple left module of S ⊗ k S op is isomorphic to (F q m i m j ) n i n j for some i, j ∈ {1, ..., r}.
Let H be the set of two-sided ideals of A that are maximal with respect to being properly contained in J(A).By the proof of Theorem 2.5 of [8], all ideals in H contain So Ω 3 → 0 as n → ∞, as m → ∞ or as q → ∞.This completes the proof.Proof.Write A = M n (q m ).Recall from the proof of Theorem 3 that ζ A (ǫ) → 0 as n → ∞, as m → ∞ or as q → ∞.The result follows immediately as |A| = q mn 2 .Lemma 5. Let A be a finite algebra and let S be a semisimple subalgebra of A such that A = S ⊕ J(A).Then A × = S × × J(A) and A N = S N × J(A), where × denotes Cartesian product of sets.
Proof.Let g ∈ A × .Write g = s + j and g −1 = s ′ + j ′ for s, s ′ ∈ S and j, j ′ ∈ J(A).Then and let α be the (nilpotency) index of x.Write x = s 1 + j 1 for s 1 ∈ S and j 1 ∈ J(A).Then 0 = x α = s α 1 + j ′ 1 , for some j ′ 1 ∈ J(A).Hence s 1 ∈ S N .Conversely, let y = s 2 + j 2 ∈ S N × J(A) and let β be the index of s 2 .Then y β ∈ J(A) and so y ∈ A N .
For positive integers u, v, define a function where F (u, 0) = 1.We will need the following elementary lemmas.
Proof.If u = 1 then F (u, v) = 0 and the inequality holds.So assume that u > 1.
Observe that u c − (u − 1) c ≥ 1. Rearranging, we have 1 The lower bound then follows immediately since u is arbitrary.
For the upper bound, observe that (1 ).Then we are done again since u is arbitrary.
Proof.If u = 1 then we are done.So assume that u > 1.Let x ∈ N. We first show that by induction on x.If x = 1 then it certainly holds.If x > 1 then x using the inductive hypothesis.Without loss of generality, assume that w ≤ v/2 (otherwise we swap w and v − w).Using (2), we have .
One can use Leibniz's alternating series test to show that F (u, v) converges towards a positive limit as v → ∞ and u is fixed.This limit is also known as φ(1/u), where φ denotes the Euler function.It is known that φ(1/u) is transcendental.For example, φ(1/2) ≈ 0.2888.Lemma 8. Let A be a finite simple algebra, say A = M n (q m ).Then Proof.It is easy to check that |A × |/|A| = q −mn 2 n−1 i=0 (q mn − q mi ) = F (q m , n).Observe that F (q m , n) is monatonically decreasing (resp.increasing) in n (resp.q m ).Then the result follows from the remark preceding this lemma.
Finally, we will need the elementary inequality for all integers x ≥ y ≥ 2.

Proof of Theorem 1.1 and Corollary 1.2
Let A be a finite algebra, say A = S ⊕ J(A) where S = r i=1 S i is semisimple and S i = M n i (q m i ) for each i.Denote n := min i=1,...,r {n i } and m := min i=1,...,r {m i }.Recall that φ(1/2) ≈ 0.2888.
We begin by considering two examples.Let p be a prime.Let A = M p (2 p ) and let So we see that the constants in the following lemma are best possible.
Proof.We first consider the case where A is simple.That is, A = M n (q m ).Note that A × = GL n (q m ).For simplicity, denote t := q m .Assume that B is of type (S1).That is, B ∼ = P l,n−l (t) for some positive integer l < n.Observe that |B × | = |(B/J(B)) × ||J(B)| by Lemma 5. Then applying Lemma 8 gives us Now assume that B is not of type (S1).Then B is simple by Theorem 1.Hence, by Lemma 8, we have This completes the proof for the case where A is simple.We now consider the general case.Recall that A = S ⊕ J(A) where S = r i=1 S i is semisimple.By Theorem 2, B is of type (T 1), (T 2) or (T 3).We consider each of these possibilities.
Let B be of type (T 1).That is, B ∼ = (B j × i =j S i ) ⊕ J(A) for some j ∈ {1, ..., r} and maximal subalgebra B j of S j .Applying Lemma 5 gives us Since S j is simple, Lemma 8 gives us Let B be of type (T 2).That is, B ∼ = ( i =j 0 S i ) ⊕ J(A) for some j 0 ∈ {1, ..., r}.Again using Lemma 5, we have Again using Lemma 8, we have Finally, let B be of type (T 3).That is, B ∼ = S ⊕ H where H is a two-sided ideal of A that is maximal with respect to the condition H ⊂ J(A).Then by Lemma 5 and since J(B) ∼ = H.This completes the proof of the lemma.
Let x, y ∈ A be chosen uniformly at random.If x, y = A then x and y are both contained in a maximal subalgebra B of A. For a given B, the probability that this occurs is |B| 2 /|A| 2 .Let Max A denote the set of maximal subalgebras of A. Then Let B be the standard set of representatives of the conjugacy classes of maximal subalgebras of A. ( If A is simple then, by Corollary 4, P (A) → 1 as |A| → ∞.This completes the proof of Corollary 1.2.Let c ∈ R such that 1 < c < q and let λ > 0. For the general case, we need the assumption that A is bounded by (c, λ).Then, by Theorem 3 (and its proof), P (A) → 1 as n → ∞, as m → ∞ or as q → ∞.This completes the proof of Theorem 1.1.
Let n = 1.Let G(A) be the set of generators of A as a k-algebra.That is, G(A) is the subset of A consisting of all elements whose minimal polynomial over k has degree m.It is a classical result, dating back to Gauss, that the number of monic irreducible polynomials over k of degree m is ν q (m).So |G(A)| = mν q (m).Hence . Then, by Equation ( 9) of [10], we have with equality if and only if q = 2.
If (m, n) = (2, 2) then we have just shown that the probability of two randomly chosen elements of A generating A as a F q 2 -algebra is strictly greater than 3/8.So, certainly, P (A) > 3/8 as a F q -algebra.

Proof of Theorem 1.4
Let A be a finite simple algebra, say A = M n (q m ).Recall that m(A) is the minimal index of any proper subalgebra of A. Note that m(A) is undefined if m = n = 1.

Lemma 10. Let C be the set of conjugacy classes of subalgebras of A that have index m(A).
If m > 1 then let p be the smallest prime divisor of m.Then m(A) and C are as follows: Proof.Let B be a subalgebra of A with index m(A).Then B is maximal, so we refer to the classification in Theorem 1.
We first assume that n = 1.There do not exist any subalgebras of A of type (S1) or (S2).So B ∼ = q m/p , where p is the smallest prime divisor of m.There is one conjugacy class of such a B. Now assume that n > 1. Observe that dim B divides dim A if B is of type (S2) or (S3), whilst 2 dim B > dim A if B is of type (S1).So B is of type (S1), that is, B is conjugate to P l,n−l (t) for some 1 ≤ l < n.We compute [A : B] = q ml(n−l) .Hence m(A) = q m(n−1) , which is realised when B is conjugate to P 1,n−1 (q m ) or to P n−1,1 (q m ).Finally, we note that P 1,n−1 (q m ) is not conjugate to P n−1,1 (q m ) unless n = 2 (in which case they are equal).
Henceforth assume that A is not a field.That is, n > 1.
Lemma 11.Let B be a subalgebra of A. (i) (ii) If m(A) 4/3 ≤ [A : B] < m(A) 5/3 then either n = 4, 5 or 6 and B is conjugate to P 2,n−2 (q m ) or P n−2,2 (q m ), or B is non-maximal in A and is not over F q m .Proof.Note that m(A) = q m(n−1) by Lemma 10 (since n > 1).We first consider the case where B is maximal.By Theorem 1, B is of type (S1), (S2) or (S3).We consider each of these possibilities.
Let B be of type (S1).That is, B is conjugate to P l,n−l (q m ) for some positive integer l < n.Observe that [A : B] = q ml(n−l) .If l = 1 or n − 1 then
We have shown that there exist no maximal subalgebras (and hence no subalgebras) B of A that satisfy m(A) < [A : B] < m(A) 4/3 .This proves (i).
Now assume (for a contradiction) that B is a F q m -subalgebra of A that is not maximal (as a F q -subalgebra) and satisfies m(A) 4/3 ≤ [A : B] < m(A) 5/3 .Let M be a maximal subalgebra of A that contains B. By the previous argument, either M ∼ = P 1,n−1 (q m ) or M ∼ = P 2,n−2 (q m ) and n = 4, 5 or 6.
Let M ∼ = P 2,n−2 (q m ) and n = 4, 5 or 6.It follows from Theorems 1 and 2 that the minimal index of a subalgebra of M is q m .Then [A : B] ≥ q 2m(n−2)+m ≥ m(A) 5/3 , which is a contradiction.
Let {B i | i = 1, ..., α} denote the set of maximal subalgebras of A. Let β be the number of maximal subalgebras of A with index m(A).We arrange the B i 's such that B i has index m(A) if and only if i ≤ β.
Let κ : A → R be defined by κ(A) Let x, y ∈ A be chosen uniformly at random.If x, y = A then x and y are both contained in a maximal subalgebra of A. For a given B i , the probability that this occurs using Corollary 13.
The theorem then follows from combining the inequalities ( 7) and ( 8) with Corollary 13 and Lemmas 14 and 15.
We conclude this section with the following estimate of the zeta function of A. Let ǫ > 0. By the same argument as in the proof of Lemma 14, it is easy to see that ρ(A) = o(m(A) ǫ/3 ).Combining this with Lemmas 10 and 11 gives us where δ : A → R is a function given by δ(A) = 1 if n = 2 and δ(A) = 2 otherwise.

Proof of Theorem 1.5 and Corollary 1.6
Let A be a finite algebra, say A = S ⊕ J(A) where S = r i=1 S i is semisimple and S i = M n i (q m i ) for each i.Denote n := min i=1,...,r {n i } and m := min i=1,...,r {m i }.If A is simple, note that n = 1 if and only if A is a field.Let T denote the group of scalar matrices of S × .Recall that φ(1/2) ≈ 0.2888.
Assume that n > 1.We will need the following lemma.
Lemma 16.Let B be a maximal subalgebra of A. Then Proof.We first consider the case where A is simple.That is, A = M n (q m ).For simplicity, denote t := q m .By Theorem 1, B is of type (S1), (S2) or (S3).We consider individually each of these possibilities.We will repeatedly use the fact that |A N | = t n 2 −n , which was proved in Theorem 1 of [7].
Let B be of type (S1).That is, B ∼ = P l,n−l (t) for some positive integer l < n.Observe that |B N | = |(B/J(B)) N ||J(B)| by Lemma 5. Then we have Let B be of type (S2).That is, B ∼ = M n/a (t a ) for some prime divisor a of n.Then Hence, by Lemma 9, we have for all B of type (S1) or (S2).
Let B be of type (S3).That is, by Lemma 9 and since n > 1.This completes the proof for the case where A is simple.
We now consider the general case.Recall that A = S ⊕ J(A) where S = r i=1 S i is semisimple.By Theorem 2, B is of type (T 1), (T 2) or (T 3).We consider each of these possibilities.
Finally, let B be of type (T 3).That is, B ∼ = S ⊕ H where H is a two-sided ideal of A that is maximal with respect to the condition H ⊂ J(A).Then by Lemma 5.This completes the proof of the lemma.
Let x, y ∈ A N be chosen uniformly at random.If x, y = A then x and y are both contained in a maximal subalgebra B of A. For a given B, the probability that this occurs is |B N | 2 /|A N | 2 .Let Max A denote the set of maximal subalgebras of A. Then Let B be the standard set of representatives of the conjugacy classes of maximal subalgebras of A. For a given B ∈ B, recall that there are (11) with Lemma 16 gives us If A is simple then, by Corollary 4, P N (A) → 1 as |A| → ∞.This completes the proof of Corollary 1.6.Let c ∈ R such that 1 < c < q 1/4 and let λ > 0. For the general case, we need the assumption that A is bounded by (c, λ).Then, by Theorem 3 (and its proof), P N (A) → 1 as n → ∞, as m → ∞ or as q → ∞.This completes the proof of Theorem 1.5.

Proof of Theorem 1.7
Let A be a finite simple algebra that is not a field.That is, A = M n (q m ) where n > 1.For simplicity, denote t := q m .Let f be a polynomial of degree n over ..f αs s where the f i 's are distinct and irreducible over F t .For each i, let d i be the degree of f i .Without loss of generality, we assume that f is monic.
For positive integers u, v, recall the definition and F (u, 0) = 1.We will need Theorem 2 of [24], which states that .
Lemma 17.Let B be a maximal subalgebra of A. There exists an absolute constant C > 0 such that Proof.By Theorem 1, B is of type (S1), (S2) or (S3).We consider individually each of these possibilities.If B f is empty then we are done, so assume otherwise.Let B be of type (S1).That is, B ∼ = P l,n−l (t) for some positive integer l ≤ n/2.Let Λ be the set of polynomials over F t that divide f and have degree l.We can assume that Λ is non-empty (as otherwise B f is empty).Observe that |Λ| ≤ n l .Consider a generic element f 0 ∈ Λ. Factorise ..f βs s where 0 ≤ β i ≤ α i for each i.Then using Lemmas 5, 7 and Theorem 2 of [24].For sufficiently large n, say n ≥ 200, observe that Let C = 3 • 199 199 3 2 199 .Then, using (13) and Lemma 9, we have Let B be of type (S2).That is, B ∼ = M n/a (t a ) for some prime divisor a of n.Let z ∈ B f .Recall that f is the characteristic polynomial of z as a n × n matrix over F t .Let g be the characteristic polynomial of z as a n/a×n/a matrix over F t a .Let Γ a := Gal(F t a /F t ) ∼ = Z a .
Without loss of generality, we rearrange the factors of f such that, for some positive integer c ≤ s, f i is reducible over F t a if and only if i ≤ c.
Let i ∈ {1, ..., s}.Let g i be a F t a -irreducible factor of f i .If i > c then f i = g i .If i ≤ c then, since a is prime, f i = σ∈Γa g σ i where the Γ a -conjugates of g i are all distinct.So the polynomials in the set {g σ i | i = 1, ..., c; σ ∈ Γ a } ∪ {g i | i = c + 1, ..., s} are all F t a -irreducible and distinct.Let p i be the greatest common divisor of f α i i and g.Note that f = σ∈Γa g σ by Lemma 5.1 of [19].So if i > c then p i = g α i /a i and if i ≤ c then p i = σ∈Γa (g σ i ) σγi where each σ γ i is a non-negative integer such that σ∈Γa σ γ i = α i .Given that f is fixed, observe that there are at most a n a possibilities for g (by allowing σ∈Γa σ γ i = α i to range over all partitions for each i ≤ c).
Let B be of type (S3).That is, by Theorem 2 of [24] and Lemma 6. Recall that |N A × (B × ) : Let x, y ∈ A f be chosen uniformly at random.If x, y = A then x and y are both contained in a maximal subalgebra B of A. For a given B, the probability that this occurs is Let B be the standard set of representatives of the conjugacy classes of maximal subalgebras of A. For a given B ∈ B, recall that there are (14) with Lemma 17 gives us for some absolute constant C > 0. Hence, by Corollary 4, P f (A) → 1 as |A| → ∞.
8. Proof of Theorem 1.8 and Corollary 1.9 Let A be a finite simple algebra, say A = M n (q m ), where n ≥ 2 and m ≥ 1.Let p be the smallest prime divisor of n.Let α := α(n) be a positive integer such that α ≤ n.
For simplicity, denote t := q m .It is a classical result, dating back to [12], that We first prove part (i) of the theorem.In part (i) we consider n, and hence α, to be fixed constants.Assume that n ≥ pα.
Let B be a subalgebra of A such that B ∼ = M n/p (t p ).Such a B exists and is maximal by Theorem 1.We claim that We first consider the case where α = 1.Then, using ( 16), we have Proof.Take the image/preimage of a generating set under the natural projection A → A/I.
We now characterise when d(A) ≤ 1. Recall that where µ is the Möbius function.
Lemma 20.Let A be a finite algebra.Then the following hold.(ii) We first consider the case where A is simple, say A = M n (q m ).Let d(A) = 1.Assume (for a contradiction) that n > 1.Let x be a generator of A. Let χ n (x) be the characteristic polynomial of X as a n × n matrix over F q m .If χ n (x) is F q m -irreducible then, by Theorem 2.1 of [19], dim k F q m x = mn and so k x is a proper subalgebra of A. If χ n (x) is F q m -reducible then x is contained in a parabolic subalgebra of A. This is a contradiction, hence n = 1.Conversely, let A = F q m for m > 1.Any generator of the multiplicative group A × then generates A as an algebra.Now consider the case where A = S α for simple S. Let d(A) = 1.Then S is a field by Lemma 19 and the above arguments.Write S = F q m .Recall from the proof of Theorem 1.3 that the number of generators of S as a k-algebra is mν q (m).By Theorem 6.3 of [10], A can be generated by 1 element if and only if α ≤ ν q (m).The converse follows immediately.
Next consider the case where A is semisimple, say A = r i=1 S α i i where the S i 's are pairwise non-isomorphic simple algebras.It follows from Proposition 2.12 of [10] that d(A) = max i=1,...,r {d(S α i i )}.The result then follows from Lemma 19 and the above arguments.
Finally, we consider the general case.If d(A) = 1 then d(A/J(A)) ≤ 1 by Lemma 19.This completes the proof.
Corollary 21.Let A be a finite simple algebra.Then Proof.By Theorem 6.4 of [10], A is 2-generated.The result then follows immediately from Lemma 20.
Let P (A, l) be the probability that l elements of A chosen uniformly at random will generate A as a k-algebra.Note that P (A, l) ≥ P (A, l 0 ) for all l ≥ l 0 .Recall our previous notation P (A) := P (A, 2).

Proof of Theorem 1.10.
Let C i be a list of the distinct cores of maximal subalgebras of A. For each i choose a maximal subalgebra B i of A with core C i .For each n ≥ 2 let c n (A) denote the number of maximal subalgebras of index n obtained in this way.
Consider X = A d as a probability space and the events X i = B d i < X.By Lemma 24 the events X i are pairwise independent.Let p i = [A : B i ] −d , the probability of the event X i .
By the Borel-Cantelli Lemma, if i p i = ∞, then, with probability 1, infinitely many events X i occur.This implies that a random d-tuple in A d generates A with probability 0, a contradiction to P (A, d) > 0. We conclude that i p i converges.Moreover, by the effective version of the Borel-Cantelli Lemma we have Now, by Lemma 23, there are at most 6.93n maximal subgroups of A of index n with a given core C i .This yields In particular, A has PMSG as required.We now move on to the proof of Theorem 1.12.Proposition 25.Let A be a finite algebra, say A = S ⊕ J(A) where S = r i=1 S i and S i = M n i (q m i ) for each i.Then, for all n > 1 we have m n (A) ≤ n 6.93r + r(r − 1)/2 + r 2 n d(A) .
Proof.Let B < A be a maximal subalgebra of index n and let C = B A be its core.
If B is of type (T 1) then C = i =j S i ⊕ J(A) for some 1 ≤ j ≤ r, so there are r possibilities for C. Given C, B/C < A/C ∼ = S j is a maximal subalgebra of index n, so by Lemma 23 there are at most 6.93n possibilities for B/C, hence for B given C. We conclude that there are at most 6.93rn possibilites for B of type (T 1).
Suppose B is of type (T 2).Then C = i =j 1 ,j 2 S i ⊕ J(A) for some 1 ≤ j 1 < j 2 ≤ r, so there are r(r − 1)/2 possibilities for C. As follows from the proof of Lemma 23, there are at most n possibilities for B given C. Hence there are at most nr(r − 1)/2 possibilities for B in this case.
Finally, let B be of type (T 3).Then C = i =j 1 ,j 2 S i ⊕ H for some (not necessarily distinct) integers 1 ≤ j 1 ≤ j 2 ≤ r and some two-sided ideal H of A that is maximal with respect to being contained in J(A).We first want to count the possibilities for H.
Observe that B × is a maximal subgroup of A × .Then A × acts primitively by leftmultiplication on the set of left cosets A × /B × (see for instance 1.7(b) of [3]).In other words, A × /B × is a primitive (left) A × -space.Applying 1.3 of [3] then tells us that the conjugacy class of B × in A × , and hence H, is uniquely determined by the isomorphism class of A × /B × as an A × -space.
Consider the (non-unital) quotient algebra V := J(A)/H.We equip V with the structure of an A-bimodule under the action v → ava ′ for a, a ′ ∈ A and v ∈ V .Note that V is a simple A-bimodule since B is a maximal subalgebra of A, and hence J(A) acts trivially on V (on both the left and the right) by Nakayama's lemma.So V 2 = 0.
Next consider the quotient algebra A/C =: A and the natural projection ρ : A → A. Let S be a maximal semisimple subalgebra of A. Since our field k is perfect, A/J(A) ∼ = S is separable.Observe that V is isomorphic to J(A) as a (non-unital) algebra.Then applying Proposition 11.7 of [21] (a version of Wedderburn's Principal theorem) gives us a semidirect product of algebras A = V ⋊ S (that is, (v, s)(v ′ , s ′ ) = (vs ′ + sv ′ , ss ′ ) for all v, v ∈ V and s, s ′ ∈ S, and (0, 1) is the unity element).
The core of A is trivial, and so S is isomorphic to either S j 1 (if j 1 = j 2 ) or to S j 1 × S j 2 (if j 1 = j 2 ).One can interpret V as the natural left S j 1 ⊗ (S j 2 ) op -module (refer to §10.1 of [21]).It follows that there are at most r 2 possibilities for the isomorphism class of A.
Taking the respective groups of units of A = V ⋊ S gives us a semidirect product of groups A × = V ⋊ S × (considering V to be its additive group).The natural projection ρ : A → A induces a (left) action of A × by permutations on V as follows.Let v ∈ V and a ∈ A × , say ρ(a) = (v ′ , s) for v ′ ∈ V and s ∈ S × , and define a • v := svs −1 + v ′ s −1 .It is then easy to see that the map V → A × /B × given by x + H → (x + 1)B × for x ∈ J(A) is an isomorphism of A × -spaces.
The isomorphism class of V as an A × -space is uniquely determined by the isomorphism class of A × along with a 1-cocycle A × → V , which arises from a derivation δ : A → V .Proof of Theorem 1.12.
If we do not specify a base, log refers to base 2. Set M (A) := sup n>1 log m n (A)/ log n, M * (A) := lim sup n>1 log m n (A)/ log n, which measure the degree of polynomial subgroup growth of A (and are infinite unless A has PMSG).Let d 0 (A) := min{d ≥ 1 | P (A, d) > 0}.
we can rearrange the factors of f such that f i is over F t 1/b if and only if i ≤ d, b divides s − d and, for every i = 1, ..., (s − d)/b, b−1 j=0 f d+i+j(s−d)/b = σ∈Γ b f σ d+i and α d+i = α d+i+(s−d)/b = ... = α d+i+(b−1)(s−d)/b .For i ∈ {1, ..., d+(s−d)/b}, define a polynomial h i by h i = f i if i ≤ d and h i = σ∈Γ b f i otherwise.Observe that the h i 's are all distinct and F t 1/b -irreducible.Then |B

Lemma 19 .
Let A be a finite algebra and let I be an ideal of A. Then d(A/I) ≤ d(A) ≤ d(A/I) + d(I).
Certainly A× is determined up to isomorphism by A and δ : A → V is determined by its values on the generators of A. By assumption, |V | = n.In summary, the number of possibilities for H is bounded above by r 2 n d(A) .For a given H, by Malcev's contribution to the Principal Theorem and since B ∩J(A) = J(B) = H, there are precisely |J(A)/H| = n possibilities for B. So there are at most r 2 n d(A)+1 possibilities for B of type (T 3).This completes the proof.Let x = (x d ) d∈N be a sequence of elements of A that are chosen randomly, uniformly and independently.Define a random variable τ A byτ A = min{d ≥ 1 | x 1 , ..., x d = A} ∈ N ∪ {+∞}.Recall that E(A) is the expected number of random elements of A chosen uniformly and independently which generate A. Observe thatE(A) = d≥1 dP (τ A = d) = d≥1 c≥d P (τ A = c) = d≥0 1 − P (A, d) .(20)Recall the definitionsM (A) = supn>1 log m n (A)/ log n and, for any real number η ≥ 1, V η (A) = min{d ≥ 1 : P (A, d) ≥ η −1 }.Let ζ denote the Riemann zeta function.
semisimple then the converse holds.Proof.(i) We have d(A) = 0 if and only if A does not have a maximal subalgebra if and only if A = k.