Let ∑n=0∞anXn,∑n=0∞bnXn∈R⟦X⟧. Then we obtain

Also, we have

and

so ϕ ((∑n=0∞anXn) (∑n=0∞bnXn))=ϕ (∑n=0∞anXn)★ϕ (∑n=0∞bnXn). Thus ϕ is a ring homomorphism.

To show the second statement, we first suppose that ϕ is an isomorphism. Then for any n ∈ ℕ₀, there exists an element 1[n]q!Xn∈R⟦X⟧ such that ϕ (1[n]q!Xn)=Xn. Thus 1[n]q!∈R. For the converse, we suppose that 1[n]q!∈R for all n ∈ ℕ₀. If ϕ (∑n=0∞anXn)=0, then ∑n=0∞[n]q!anXn=0; so [n]_q!a_n = 0 for all n ∈ ℕ₀. Since 1[n]q!∈R for all n ∈ ℕ₀, a_n = 0 for all n ∈ ℕ₀. Hence ∑n=0∞anXn=0, which shows that ϕ is one-to-one. Let ∑n=0∞bnXn∈Gq(R). Then by the assumption, we can find an element ∑n=0∞1[n]q!bnXn∈R⟦X⟧ such that ϕ (∑n=0∞1[n]q!bnXn)=∑n=0∞bnXn. Hence ϕ is onto. Thus ϕ is an isomorphism.

Let R be a commutative ring with identity, ℤ the ring of integers, and q a prime power. Then R may be viewed as a ℤ-module in the usual sense. Recall that R is torsion-free if whenever na = 0 for n ∈ ℤ and a ∈ R, n = 0 or a = 0. As the q-analogue of ‘torsion-free’, we define the concept of ‘q-torsion-free. We say that R is q-torsion-free if whenever [nr]qa=0 for a ∈ R and n, r ∈ ℕ₀ with n ≥ r, a = 0. Clearly, if R is torsion-free, then R is q-torsion-free. The following examples show that R being q-torsion-free does not imply that R is torsion-free.

(1) ⇒ (2) Suppose that f=∑n=0∞anXn, g=∑n=0∞bnXn are nonzero elements of G_q(R) such that f ★ g = 0, and let m₁,m₂ be the smallest nonnegative integers such that a_m1 ≠ 0 and b_m2 ≠ 0. Then by comparing the coefficients of X^m₁+m₂ in f ★ g = 0, we have

[m1+m2m1]qam1bm2=0.

Since R is q-torsion-free, a_m1b_m2 = 0. Since R is an integral domain, a_m1 = 0 or b_m2 = 0, a contradiction. Thus G_q(R) is an integral domain.

(2) ⇒ (1) Let a, b ∈ R such that ab = 0. Then a, b ∈ G_q(R) such that a ★ b = 0. Since G_q(R) is an integral domain, a = 0 or b = 0. Thus R is an integral domain. Suppose that [nr]qa=0 for a ∈ R and n, r ∈ ℕ₀ with n ≥ r. Then aXr★Xn-r=[nr]qaXn=0. Since G_q(R) is an integral domain and X^n−r ≠ 0, aX^r = 0; so a = 0. Thus R is q-torsion-free.

Let R be a commutative ring with identity. Then Idem(R) denotes the set of idempotent elements of R.

Clearly, Idem(R) ⊆ Idem(G_q(R)). For the reverse containment, let f=∑i=0∞aiXi∈Idem(Gq(R)). We first claim that a₀a_n = 0 for all n ∈ ℕ. Since a02=a0 and 2a₀a₁ = a₁, 2a₀a₁ = a₀a₁; so a₀a₁ = 0. Suppose that a₀a₁ = · · · = a₀a_m = 0 for some positive integer m. Then by comparing the coefficients of X^m+1 in f ★ f = f, we obtain

By multiplying a₀ in both sides, 2a02am+1=a0am+1. Since a02=a0, a₀a_m+1 = 0. Hence by the induction, a₀a_n = 0 for all n ∈ ℕ.

We next show that f ∈ R. Suppose to the contrary that f ∉ R. Let k be the smallest positive integer such that a_k ≠ 0, and set g=∑i=k∞aiXi. Then f = a₀+g and a₀ ★ g = 0. Since f ★ f = f and a02=a0

, g ★ g = g. Hence by comparing the coefficients of X^k in g ★g = g, a_k = 0. This contradicts the choice of k. Thus f ∈ R, which means that f ∈ Idem(R).

The final result in this section gives an equivalent condition for a Gaussian series to be a unit.

(⇒) Suppose that ∑n=0∞anXn is a unit in G_q(R). Then there exists an element ∑n=0∞bnXn∈Gq(R) such that (∑n=0∞anXn)★(∑n=0∞bnXn)=1; so a₀b₀ = 1. Thus a₀ is a unit in R.

(⇐) To show that ∑n=0∞anXn is a unit in G_q(R), we construct a suitable element ∑n=0∞bnXn∈Gq(R) such that (∑n=0∞anXn)★(∑n=0∞bnXn)=1. Since a₀ is a unit in R, we can find an element b₀ ∈ R such that a₀b₀ = 1. For each n ∈ ℕ, set

Then it is easy to check that (∑n=0∞anXn)*(∑n=0∞bnXn)=1. Thus ∑n=0∞anXn is a unit in G_q(R).

Write f=∑i=0∞aiXi and g=∑i=m∞biXi with b_m ≠ 0, and let c ∈ (b_m) be a nonzero idempotent element. Then c = rb_m for some r ∈ R. Since f ★ g = 0, f ★ (r ★ g) = 0; so ca₀ = 0. Suppose that ca₀ = · · · = ca_n = 0 for some nonnegative integer n. By comparing the coefficients of X^m+n+1 in f ★ (r ★ g) = 0, we obtain

By multiplying c in both sides, we have

Since R is q-torsion-free and c is idempotent, ca_n+1 = 0. Hence by the induction, ca_k = 0 for all k ∈ ℕ₀. Thus c ★ f = 0.

Suppose to the contrary that there exist a nonzero element a ∈ R and nonnegative integers n > k such that [nk]qa=0. Then Xn-k★aXk=[nk]qaXn=0; so X^n−k is a zero-divisor of G_q(R). Note that b * X^n−k ≠ 0 for all b ∈ R{0}. This contradicts the assumption that McCoy’s theorem for G_q(R) holds. Thus R is q-torsion-free.

Note that for any nonzero element a ∈ R, [nk]qa=0; so R is not q-torsion-free. Thus by Theorem 3.2, McCoy’s theorem for G_q(R) does not hold.

Let R be a commutative ring with identity. Then R is said to be reduced if it has no nonzero nilpotent elements. The next result shows that if R is reduced, then the converse of Theorem 3.2 holds.

The “if” part is obvious. To show the converse, we suppose that f ★ g = 0. Then a₀b₀ = 0. We first claim that a_nb₀ = 0 for all n ∈ ℕ₀. Suppose that a₀b₀ = · · · = a_mb₀ = 0 for some nonnegative integer m. Then by calculating the coefficients of X^m+1 in f ★ g = 0, we obtain

By multiplying b₀ in both sides, am+1b02=0; so (a_m+1b₀)² = 0. Since R is reduced, a_m+1b₀ = 0. Hence by the induction, a_nb₀ = 0 for all n ∈ ℕ₀.

We next prove that a_ib_j = 0 for all i, j ∈ ℕ₀. Suppose that a_nb₀ = a_nb₁ = · · · = a_nb_m = 0 for all n ∈ ℕ₀ and some nonnegative integer m. If b_m+1 = 0, then a_nb_m+1 = 0 for all n ∈ ℕ₀; so we assume that b_m+1 ≠ 0. Then we obtain

so a₀b_m+1 = 0. If a₀b_m+1 = · · · = a_kb_m+1 = 0 for some nonnegative integer k, then a similar argument as in the proof of the previous claim shows that a_k+1b_m+1 = 0. Hence by the induction, a_nb_m+1 = 0 for all n ∈ ℕ₀. Thus by applying the induction again, a_ib_j = 0 for all i, j ∈ ℕ₀.

Let R be a commutative ring with identity and let a ∈ R. Then ann_R(a) denotes the annihilator of a, i.e., ann_R(a) := {r ∈ R| ra = 0}. Also, if I is an ideal of R, then Gq(I):={∑n=0∞anXn∣an∈I for all n∈ℕ0} is an ideal of G_q(R).

(1) If g=∑i=0∞biXi∈annGq(R)(f), then f ★ g = 0. By Theorem 3.4, a_ib_j = 0 for all i, j ∈ ℕ₀; so bj∈∩i=0∞annR(ai) for all j ∈ ℕ₀. Thus g∈Gq(∩i=0∞annR(ai)). To show the reverse inclusion, let h=∑i=0∞ciXi∈Gq(∩i=0∞annR(ai)). Then a_ic_j = 0 for all i, j ∈ ℕ₀. Thus by Theorem 3.4, f ★h = 0, which means that h ∈ ann_Gq(R)(f).

(2) If b∈∩i=0∞annR(ai), then b ★ f = 0 by (1). Since f is a regular element of G_q(R), b = 0. Thus ∩i=0∞annR(ai)=(0). Conversely, if g=∑i=0∞biXi is an element of G_q(R) with f ★ g = 0, then by (1), bj∈∩i=0∞annR(ai) for all j ∈ ℕ₀. Since ∩i=0∞annR(ai)=(0), b_j = 0 for all j ∈ ℕ₀. Hence g = 0, and thus f is a regular element of G_q(R).

Let R be a commutative ring with identity. Then Nil(R) denotes the set of nilpotent elements of R, and Z(R) means the set of zero-divisors of R. Also, a prime ideal of R is said to be divided if it is comparable to any principal ideal of R. We are closing this section by considering the case when R is not reduced.

Let f=∑n=0∞anXn∈Gq(R) with a₀ ∈ Z(R)Nil(R). We now construct a nonzero element g=∑n=0∞bnXn∈Gq(Nil(R)) such that f ★ g = 0. Since a₀ is a zero-divisor of R, there exists a nonzero element b₀ ∈ R such that a₀b₀ = 0. Since Nil(R) is a prime ideal of R and a₀ ∉ Nil(R), b₀ ∈ Nil(R). Also, since Nil(R) is divided and a₀ ∉ Nil(R), Nil(R) ⊊ (a₀). Note that −a₁b₀ ∈ Nil(R); so −a₁b₀ = a₀b₁ for some b₁ ∈ R. Since Nil(R) is a prime ideal of R and a₀ ∉ Nil(R), b₁ ∈ Nil(R). Suppose that b₀, …, b_n ∈ Nil(R) for some nonnegative integer n. Then -([n+11]qa1bn+⋯+[n+1n+1]qan+1b0)∈Nil(R). Since Nil(R) ⊊ (a₀), we obtain

for some b_n+1 ∈ R. Since Nil(R) is a prime ideal of R and a₀ ∉ Nil(R), b_n+1 ∈ Nil(R). By the induction, we obtain an infinite sequence (b_n)_n≥0 in Nil(R) such that [n0]qa0bn+⋯+[nn]qanb0=0 for all n ∈ ℕ₀. Thus by setting g=∑n=0∞bnXn∈Gq(Nil(R)), we deduce that f ★ g = 0.

Let a, b ∈ R. Then a, b ∈ G_q(R). Since G_q(R) is a right K-Hermite ring, we can find an element f ∈ G_q(R) and an invertible matrix (g11g12g21g22) over G_q(R) such that (ab) (g11g12g21g22)=(f0); so (ab) (g11(0)g12(0)g21(0)g22(0))=(f(0)0). Note that the determinant of (g11(0)g12(0)g21(0)g22(0)) is the constant term of the determinant of (g11g12g21g22); so by Proposition 2.6, the determinant of (g11(0)g12(0)g21(0)g22(0)) is a unit in R. Hence (g11(0)g12(0)g21(0)g22(0)) is an invertible matrix over R. Thus R is a right K-Hermite ring.

The next example shows that the Gaussian series ring over a right K-Hermite ring need not be a right K-Hermite ring.

(1) ⇒ (2) Let (f₁, …, f_m) ∈ G_q(R)^m be such that f₁ ★ G_q(R) + · · · + f_m ★ G_q(R) = G_q(R). Then we can find an element (g₁, …, g_m) ∈ G_q(R)^m such that f₁ ★ g₁+· · ·+f_m★g_m = 1; so f₁(0)g₁(0)+· · ·+f_m(0)g_m(0) = 1. Therefore f₁(0)R+· · ·+ f_m(0)R = R. Since R is an L-Hermite ring, there exists an m×m invertible matrix P=(f1(0)f2(0)⋯fm(0)r21r22⋯r2m⋮⋮⋱⋮rm1rm2⋯rmm) over R. Let Q=(f1f2⋯fmr21r22⋯r2m⋮⋮⋱⋮rm1rm2⋯rmm). Then Q is an m × m matrix over G_q(R). Note that the constant term of the determinant of Q is precisely the same as the determinant of P; so by Proposition 2.6, the determinant of Q is a unit in G_q(R). Hence Q is invertible. Thus G_q(R) is an L-Hermite ring.

(2) ⇒ (1) Let (a₁, …, a_m) ∈ R^m be such that a₁R + · · · + a_mR = R. Then a₁ ★ G_q(R) + · · · + a_m ★ G_q(R) = G_q(R). Since G_q(R) is an L-Hermite ring, there exists an m×m invertible matrix M=(a1a2⋯amf21f22⋯f2m⋮⋮⋱⋮fm1fm2⋯fmm) over G_q(R). Let N=(a1a2⋯amf21(0)f22(0)⋯f2m(0)⋮⋮⋱⋮fm1(0)fm2(0)⋯fmm(0)). Since the determinant of M is a unit in G_q(R), Proposition 2.6 shows that the determinant of N is a unit in R. Hence N is an m × m invertible matrix over R. Thus R is an L-Hermite ring.

Let R be a commutative ring with identity and M a unitary R-module. Then M is free if M has a basis; and M is stably free if there exist positive integers m and n such that M ⊕ R^m = Rⁿ. Clearly, free modules are stably free.

Note that if T is a commutative ring with identity, then T is an L-Hermite ring if and only if every stably free T-module is free [7, Chapter I, Corollary 4.5]. Thus the equivalence follows from Theorem 4.3.

(⇒) Let C = {c_i | i ∈ ℕ₀} be a countable subset of I, and let f=∑i=0∞ciXi∈Gq(I). Since G_q(I) = I ★ G_q(R), we can find b₁, …, b_n ∈ I and g₁, …, g_n ∈ G_q(R) such that f = b₁ ★ g₁ +· · ·+b_n ★ g_n. Let F = (b₁, …, b_n). Then F is a finitely generated ideal of R such that C ⊆ F ⊆ I.

(⇐) Clearly, I ★ G_q(R) ⊆ G_q(I), because I ⊆ G_q(I). For the reverse containment, let f=∑i=0∞aiXi∈Gq(I). Then there exist suitable elements b₁, …, b_n ∈ I such that {a_i | i ∈ ℕ₀} ⊆ (b₁, …, b_n); so for each i ∈ ℕ₀, ai=∑j=1nbjcij for some c_ij ∈ R. Hence we obtain

Thus G_q(I) = I ★ G_q(R).

As an immediate consequence of Lemma 5.1, we obtain

(1) ⇒ (2) This implication follows from Proposition 5.2, because every ideal of a Noetherian ring is finitely generated.

(2) ⇒ (1) Suppose to the contrary that R is not a Noetherian ring. Then there exists a strictly ascending chain of ideals (I_n)_n≥0 of R; so for each n ≥ 1, we can choose an element a_n ∈ I_nI_n−1. Let I=∪n=0∞In. Then I is an ideal of R. Since G_q(I) = I ★G_q(R), Lemma 5.1 indicates that there exists a finitely generated ideal F of R such that {a_n | n ∈ ℕ} ⊆ F ⊆ I. Since F is finitely generated, F ⊆ I_k for some positive integer k. Hence a_k+1 ∈ I_k, which is a contradiction. Thus R is a Noetherian ring.

Let R be a commutative ring with identity and I an ideal of R. Then I denotes the radical of I. We end this article with the radical property of Gaussian series rings.

Deny the conclusion. Then for each m ≥ 1, there exist b_m1, …, b_mm ∈ J such that b_m1 · · · b_mm ∉ I. Let C be the ideal of R generated by {b_mi |m ∈ ℕ and 1 ≤ i ≤ m}. Then C is a countably generated subideal of J such that C_m ⊈ I for all m ∈ ℕ. Since G_q(J) = J ★ G_q(R), by Lemma 5.1, there exists a finitely generated ideal F of R such that C ⊆ F ⊆ J. Since F is finitely generated and J⊆I, F^k ⊆ I for some k ∈ ℕ. Hence C^k ⊆ I, which is absurd. Thus Jⁿ ⊆ I for some n ≥ 1.