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On the Commutative Rings with At Most Two Proper Subrings
The commutative rings with exactly two proper (unital) subrings are characterized. An initial step involves the description of the commutative rings having only one proper subring.
This paper is a sequel to . All rings considered below are commutative with identity; all subrings, inclusions of rings, ring extensions, ring or algebra homomorphisms, modules, and submodules are unital. Our interest here is in characterizing rings that have at most two proper subrings. Observe that a ring has no proper subrings if and only if either , , or for some integer and that these three cases are mutually exclusive. We will answer the analogous questions for rings having exactly one proper subring (resp., for rings having exactly two proper subrings) in Theorems 5 and 13 (resp., in Theorems 6 and 15). In doing so, we may tacitly ignore the zero ring, as does not have any proper subrings (or, for that matter, any proper ring extensions). Thus, we may organize our answers with a focus on the characteristic of a ring of interest, since every nonzero ring is a ring extension of an isomorphic copy of exactly one of the so-called prime rings (namely, and for ).
Notice that a ring has exactly one proper subring if and only if is a minimal ring extension (in the sense of ) of its prime (sub)ring. As the minimal ring extensions of a (commutative integral) domain were classified in , much of the technical work leading up to our main theorems will be concerned with determining the minimal ring extensions of the prime rings of positive characteristic. Then, by determining the minimal ring extensions of either those minimal ring extensions or the minimal ring extensions of , we will be led to a list of candidates for the characterizations in Theorems 6 and 15. Determining which of those candidates survive will depend on the main result of  which, for convenience, is restated here as Theorem 2. Carrying out this program will require details about minimal ring extensions, most of which are summarized in the next paragraph.
Recall (cf. ) that a ring extension is a minimal ring extension if there does not exist a ring properly contained between and . A minimal ring extension is either integrally closed (in the sense that is integrally closed in ) or integral. If is a minimal ring extension, it follows from [2, Théorème 2.2(i) and Lemme 1.3] that there exists a unique maximal ideal of (called the crucial maximal ideal of ) such that the canonical injective ring homomorphism can be viewed as a minimal ring extension while the canonical ring homomorphism is an isomorphism for all prime ideals of except . A minimal ring extension is integrally closed if and only if is a flat epimorphism (in the category of commutative rings). If is an integral minimal ring extension with crucial maximal ideal , there are three possibilities: is said to be, respectively, inert, ramified, or decomposed if is isomorphic, as an algebra over the field , to a minimal field extension of , , or . (As usual, denote an indeterminate over the ambient base ring.)
If is a ring, then denotes the characteristic of ; (resp., ) denotes the set of prime (resp., maximal) ideals of ; and by the dimension of , we mean the Krull dimension of , denoted by . If are rings, then denotes the set of rings such that ; and, as in , we say that satisfies FIP (for the “finitely many intermediate rings” property) if the set is finite. Following [5, page 28], we let INC, LO, and GU, respectively, denote the incomparable, lying-over, and going-up properties of ring extensions. Given rings with , then . As usual, denotes proper inclusion; denotes the finite field of cardinality ; and denotes the cardinality of a set . The reader may find it useful to have copies of [1, 6] at hand. Any otherwise unexplained material is standard, as in .
Our work will be separated into individual approaches that depend on the characteristic of the ambient prime ring. Each of those approaches will make use of the next two results, which are the main results of [1, 6]. Prior to giving the main results for each of our approaches, we will provide most of the new technical results that will be needed to prove those main results. We will not need to make the cumbersome conditions from [6, Proposition 3.5] (ultimately, from [7, Theorem 5.18]) explicit which are mentioned in parts (xii) and (xiii) of Theorem 1.
Theorem 1 (see [6, Theorem 4.1]). Let and be minimal ring extensions, with crucial maximal ideals and , respectively. Then satisfies if and only if (exactly) one of the following conditions holds:(i)Both and are integrally closed.(ii) is integral and is integrally closed.(iii) is integrally closed, is integral, and .(iv)Both and are integral and .(v)Both and are inert, , and either is finite or there exists such that .(vi) is decomposed, is inert, and .(vii)Both and are decomposed and .(viii) is inert, is decomposed, .(ix) is ramified, is decomposed, and .(x) is decomposed, is ramified, and .(xi) is ramified, is inert, and .(xii) is inert, is ramified, , and the two conditions stated in [6, Proposition 3.5(a)] hold.(xiii)Both and are ramified, , and the two conditions stated in [6, Proposition 3.5(b)] hold.
Theorem 2 (see [1, Theorem 2.9]). Consider the 13 conditions, (i)–(xiii), in the statement of Theorem 1. Then, consider the following:(a)If data satisfy condition (vi) or condition (xi), then .(b)If data satisfy any of the seven conditions (iii), (iv), (vii), (viii), (ix), (x), and (xii), then (and ).(c)For each of the four conditions (i), (ii), (v), and (xiii), there exist data satisfying this condition for which and there exist other data satisfying this condition for which (and ).
We will often use the fact [8, Theorem 25.1(3)] that if is a ring and is an -module, then . As we now begin the approach that is specific to the context of characteristic , it is also convenient to record here another fact that will see frequent use below. To wit: the integrally closed minimal ring extensions of are, up to isomorphism, the minimal overrings of (inside ), namely, the rings , where runs over the set of prime numbers. This fact seems to be well known and can be recovered from [9, Lemma V.2] (cf. also [3, Remark 2.8(a)]). More generally, the minimal overrings of an arbitrary principal ideal domain have been explicitly identified earlier: see, for instance, [10, Proposition 4.11].
Proposition 3. Let be a prime number and . Then, consider the following:(a)Let be a prime number that is distinct from . Then, up to -algebra isomorphism, is the unique ring such that is an integrally closed minimal ring extension whose crucial maximal ideal is .(b)There does not exist a ring such that is an integrally closed minimal ring extension whose crucial maximal ideal is .
Proof. We will show first that if is a prime number that is distinct from and we put , then is an integrally closed minimal ring extension whose crucial maximal ideal is . The “minimal ring extension” assertion follows from the above comments because each member of takes the form for some (uniquely determined) . The remaining parts of the assertion follow via the lore of the idealization construction. Indeed, by [8, Theorem 25.5(1)], the total quotient ring of is (canonically isomorphic to) , that is, to . As is integrally closed (in and, hence, in ), it now follows from [8, Theorem 25.5(1)] that coincides with its integral closure (in its total quotient ring). In particular, is an integrally closed ring extension. To show that the crucial maximal ideal of this extension is , [2, Théorème 2.2(i)] reduces our task to showing that . Using [8, Corollary 25.5(2)], we find canonical identifications and, similarly, .
It now suffices to show that if a ring is such that is an integrally closed minimal ring extension, then up to -algebra isomorphism, for some prime number . By [2, Théorème 2.2(iii)], the inclusion map is a flat epimorphism (in the category of rings). We claim that, up to -algebra isomorphism, is an overring of (inside the total quotient ring of ); that is, we claim that we can view .
Since is a flat epimorphism, we can view , where is a certain universal object introduced by Lazard in . Hence, to prove the above claim, it suffices to prove that coincides with the total quotient ring of . Therefore, by the proof of [11, Proposition 4.1, page 116], it suffices to show that the generization (inside ) of the set of weak Bourbaki associated primes of is the same as the canonical image inside of the set of all prime ideals of the total quotient ring of , that is, the same as the two-element set whose members are and . (Some authors translate “le générisé de” as “the generalization of”; we prefer the translation “the generization of” because it is more reminiscent of the relevant notion of a generic point.) By definition, a weak Bourbaki associated prime of is a minimal prime (ideal of ) over the annihilator of some element of . A routine case analysis shows that the set of such annihilators consists of , , and . It follows that the set of weak Bourbaki associated primes of is . As this set is stable under generization, the above claim has been proved.
We have shown that . It follows easily that for some ring such that and is a minimal ring extension. By the above comments, for some prime number . In short, . Finally, since .
Proposition 4. Let be a prime number and . Then there does not exist a ring such that is an inert extension whose crucial maximal ideal is .
Proof. Put . Then we can identify (cf. [8, Lemma 25.4 and Theorem 25.5(2)]). Now, suppose the assertion fails. Then ; and by [12, Proposition 4.6], is an inert extension. In particular, is a minimal field extension of . Hence, by a harmless change of notation, we can now take , with being an inert extension, necessarily with crucial maximal ideal . As we can view as a minimal field extension, the standard Galois theory of finite fields provides a prime number (possibly equal to ) such that , whence for each element .
Pick . Then by the minimality of . Also, since the integral extension must satisfy INC, LO, and GU (cf. [5, Theorem 44]), we see that is quasi-local and, in fact, that as sets. Since is a (nonzero) non-zero-divisor in , it follows that is a non-zero-divisor in . Also, since , there exist such that . Thus, satisfies Taking th powers, we get As is a non-zero-divisor in , we now have . Applying the canonical surjection leads to that is, . Thus, , the desired contradiction.
Theorem 5. Up to isomorphism, the rings of characteristic zero that have exactly one proper subring can be classified as the rings satisfying (exactly) one of the following conditions:(a), where is a prime number (which is uniquely determined by );(b), where is a prime number (which is uniquely determined by );(c), where is a prime number (which is uniquely determined by ).
Proof. We must identify, up to isomorphism, the rings such that is a minimal ring extension which is either (a) integrally closed, (b) decomposed, (c) ramified, or (d) inert. Thanks to [3, Theorem 2.7] (or [13, Theorem 2.4]), we are led to the list of rings in the statement of this result, including the fact that there is no ring such that is inert. To obtain the above formulation in (a), recall that the integrally closed minimal ring extensions of are, up to isomorphism, the minimal overrings of (inside ), namely, the rings , where runs over the set of prime numbers.
The above-cited references show that conditions (a), (b), and (c) are mutually exclusive. It remains to establish the uniqueness assertions in those conditions. For (a), note that is determined as the prime number that has a multiplicative inverse in . The uniqueness assertions in (b) and (c) were established in [3, Theorem 2.7].
We next confront the much more arduous task of classifying the rings of characteristic zero that have exactly two proper subrings. Much of the work below will use the following well known fact. If is a nontrivial direct product of (nonzero) rings and , then, up to isomorphism, the minimal ring extensions of take one of the forms and where is a minimal ring extension of for .
Theorem 6. (1) Up to isomorphism, the rings of characteristic zero that have exactly two proper subrings can be characterized as the rings satisfying (exactly) one of the following two conditions:(a), where and are (possibly equal) prime numbers (which are uniquely determined by );(b)for some prime number , there is a ramified extension with crucial maximal ideal and, furthermore, for all .(2) For each prime number , there exist (nonisomorphic) rings and such that is a ramified extension with crucial maximal ideal for each , has exactly two proper subrings, and has more than two proper subrings.
Proof. (1) Thanks to Theorem 2, we need only address a ring that can result from a construction described in one of six of the 13 conditions in the statement of Theorem 1, namely, conditions (i), (ii), (v), (vi), (xi), and (xiii). We will deal with these six conditions in the stipulated order.
Suppose, first, that is a ring that results from condition (i) in the statement of Theorem 1. Then, as recalled prior to Proposition 3, there is a prime number such that is a minimal overring of the principal ideal domain . By [10, Proposition 4.11], is then (up to isomorphism) where is an irreducible element of . Such elements take the form , where and is a prime number that is distinct from . Then However, and are distinct proper subrings of this . In particular, this satisfies . (In fact, .) Thus, condition (i) does not produce any contributions to the list that we are building/verifying.
Next, suppose that results from condition (ii). Then, as noted in Theorem 5, there is a prime number such that is an integrally closed minimal ring extension of either or . We will first show that the first of these alternatives does not contribute to the list that we are building/verifying. As there is no integrally closed minimal ring extension of the form , it follows from the above comments that, up to isomorphism, , where is an integrally closed minimal ring extension of . As recalled above, this means that there is a prime number (possibly equal to ) such that we can take . In short, . It will suffice to show that does not satisfy FIP. In fact, we will show that if (e.g., ) and (e.g., ), then is such that does not satisfy FIP. This will follow from [14, Proposition 3.1] after we establish three facts (to show that the cited result is applicable here). The first of these facts is that is not an integral extension; this is clear since condition (ii) ensures that the integral closure of in is . The second required fact (actually, slightly more than is needed) is that, for all nonzero elements , there exist integers and , with , such that . (For a proof, pick a positive integer such that ; note that for some integer ; ; and so and satisfy .) The last required fact is that is a residually finite ring. This completes the proof concerning .
In the remaining subcase pertinent to condition (ii), is an integrally closed minimal ring extension of . Then by Proposition 3, there is a (uniquely determined) prime number such that, up to -algebra isomorphism, . With modest changes to the above argument that applied [14, Proposition 3.1], we can show that does not satisfy FIP. Thus, also fails to lead to a contribution to the list that we are building. This completes the discussion relative to condition (ii).
Next, note that condition (v) cannot lead to a contribution to our developing list because there is no ring such that is inert [3, Theorem 2.7]. We turn to condition (vi). This will lead to the first entry to our list. Indeed, up to isomorphism, the towers such that is decomposed with crucial maximal ideal and is inert with crucial maximal ideal such that can be characterized via and , where and are (possibly equal) prime numbers. (This can be seen by combining [3, Theorem 2.7], [2, Lemme 1.2], and the classical Galois theory of finite fields. To verify the required behavior of the crucial maximal ideals, observe that and .) As for the uniqueness conclusions in the assertion (a), note first that is the only prime number which, when viewed inside , is a zero-divisor in . It is slightly harder to establish the uniqueness of without mentioning and explicitly, but it can be done as follows (now that the uniqueness of has been shown). The factor ring is isomorphic to (a unique direct product of the form) , where is a field (actually, ) whose vector space dimension over is .
By Proposition 4, condition (xi) cannot lead to a contribution to our developing list. It remains only to describe the possible contributions from condition (xiii). Assume, then, that both and are ramified extensions whose crucial maximal ideals, denoted by and , respectively, satisfy . By [3, Theorem 2.7], there exists a uniquely determined prime number such that we can identify . Then, necessarily, and . The cumbersome conditions from [6, Proposition 3.5(b)] that were alluded to in the statement of condition (xiii) hold automatically in the present context. In other words, satisfies FIP. The easiest way to see this is to apply the noncumbersome part of [6, Proposition 3.5(b)], the point being that is finite. However, our relative lack of understanding of ramified extensions (cf. the third paragraph of [1, Remark 2.11(a)]) has made us settle here for the nonspecificity in assertion (b). One should note that the part of assertion (b) which follows the word “furthermore” is what allows us to conclude that has exactly two proper subrings. As we showed above that our data satisfy the conditions from [6, Proposition 3.5(b)], it is now clear that the earlier part of assertion (b) simply ensures that we are addressing data satisfying condition (xiii).
(2) We have , with . Noting that can be identified with , put By , is a minimal ring extension. Its crucial maximal ideal is ; ; and this minimal ring extension is ramified (cf. [16, Lemma 2.1]). Consider the ring . It is clear that , and so , as required. We turn next to the more difficult construction of a suitable .
Once again, we are working with and . The -algebra will be constructed so as to have essentially the same -module structure as . Specifically, as a set (and then as an additive group with the “” symbols being viewed as “”), let where and each are generators of (distinct) one-dimensional vector spaces over . Next, we define the multiplication in by setting for all and all . (Of course, expressions such as and are interpreted by using the additive structure of .)
It is straightforward (albeit somewhat tedious) to verify that is a ring (having as its multiplicative identity element). Moreover, the function , given by for all and , is an injective (unital) ring homomorphism, thus allowing us to view as a subring of . A more intuitive understanding of the structure of as a -algebra is that we required the multiplication to satisfy the axiomatic restrictions and the additional relations , , and . (As has been identified with , we see that the last two relations differ from their analogues in .)
Recall that is a ramified extension with crucial maximal ideal . We claim that is a ramified extension with crucial maximal ideal (such that ). Consider the set (described additively as) . It is straightforward to verify that is a prime ideal of such that . As is integral over , the fact that is a maximal ideal of ensures that is a maximal ideal of . It is also straightforward to verify that and that the canonical map is an isomorphism. In addition, one can verify that is an ideal of (since and ) and is a two-dimensional vector space over . Therefore, according to [12, Theorem 4.2(c)], the claim has now been proved.
It remains to show that has only two proper subrings. It is enough to show that if , then . Without loss of generality, for some . Note that since . Thus, the above rules for multiplication lead to . In particular, contains the element for each integer . In other words, . But . Now, since , we get that . Since , we can see, by taking integer multiples as above, that . It is now evident that , which completes the proof.
One can fairly state that the formulation of assertion (b) in part (1) of the preceding theorem means that we have given a characterization but not a classification of the rings in question. As noted in the proof, this lack of a classification is due solely to our current inability to classify certain ramified extensions (up to isomorphism). This feature of having a characterization, but not a classification, will recur when we treat the corresponding problems for base rings of positive characteristic; there, too, the sole reason for a lack of a classification will be our incomplete information concerning certain ramified extensions.
It was shown in [9, Proposition V.1] that if a nonzero ring has only finitely many subrings and , then is a finite ring. Having dispatched the case of characteristic (and the trivial case of the zero ring), we henceforth devote our attention to nonzero finite rings. The first technical lemma leading to our main results on finite rings is given next. It will be applicable because any (nonzero) finite ring is zero-dimensional.
Proposition 7. Let be a (nonzero) zero-dimensional ring. Then there does not exist a ring such that is an integrally closed minimal ring extension.
Proof. Suppose the assertion fails. Let denote the crucial maximal ideal of an integrally closed minimal ring extension . As is a minimal ring extension which is not integral, it follows from [2, Théorème 2.2(ii)] that there does not exist a prime ideal of such that . However, is a minimal prime ideal of (since ), and so by an application of Zorn’s Lemma [5, Exercise 1, page 41], such a prime ideal does exist, the desired contradiction.
Thanks to Proposition 7 and the comments that preceded it, we may now focus on integral minimal ring extensions of finite rings. The next result is in the same spirit as Proposition 3. To further motivate it, first recall from [2, Lemme 1.2] that if is a prime number, the inert extensions of (i.e., of ) are the minimal field extensions of . The next result shows that if (and is a prime number), then the base ring behaves differently. It will also be useful to note that the hypothesis of the next result accommodates base rings such as . Indeed, it accommodates base rings that are arbitrary special principal ideal rings (other than fields).
Proposition 8. Let be a quasi-local zero-dimensional principal ideal ring which is not a field. Then there does not exist a ring such that is an inert extension.
Proof. Observe that is the only prime ideal of . As is a principal ideal ring, there exists such that . As is not a field, , and so . Suppose the assertion fails. Then is the crucial maximal ideal of some inert extension . The “inert” condition ensures that is quasi-local. Next, choose any element . Then , by the minimality of . As , there exists such that , and so . Since , cannot be a unit of ; that is, . Hence, , the desired contradiction.
Remark 9. Although Proposition 8 has been stated in a form that will be useful below, it may be of interest to record the following generalization of it. If is a (nonzero) zero-dimensional ring such that is a principal ideal ring and for each , then there does not exist a ring such that is an inert extension. For a proof by contradiction, take to be the crucial maximal ideal of an inert extension . Next, apply [12, Proposition 4.6] to replace with . Then repeat the proof of Proposition 8 to get the desired contradiction.
In contrast to Propositions 7 and 8, Proposition 10 shows that there does exist a ring with the relevant property (and that it is unique up to isomorphism). As an application, it also shows a way in which the base ring behaves regardless of whether the positive integer exceeds . In view of Proposition 8, this highlights an important difference in the behavior of the “inert” and “decomposed” concepts.
Proposition 10. Let be a quasi-local zero-dimensional principal ideal ring. Then, up to -algebra isomorphism, the unique ring such that is a decomposed extension is .
Proof. By [3, Lemma 2.2] (or the first paragraph of the proof of [13, Corollary 2.5]), is a minimal ring extension. As this minimal ring extension is integral, it follows from [2, Théorème 2.2(ii)] that its crucial maximal ideal is its conductor, namely, (viewed as ). Since , we see that is a decomposed extension.
Conversely, suppose that is a decomposed extension. We will prove that is -algebra isomorphic to . By [2, Lemme 1.2], we may assume that is not a field. In particular, . Since is a principal ideal ring, there exists such that . As , we get . As is the only prime ideal of , it consists of all the nilpotent elements of . Let be the index of nilpotence of . Clearly, is a nilpotent ideal whose index of nilpotence is . Next, notice that is an Artinian ring, since it is a zero-dimensional Noetherian ring. Since is finitely generated as an -module, it follows that is also an Artinian ring. By a fundamental structure theorem [17, Theorem 3, page 205], is (essentially uniquely expressible as) the direct product of finitely many (nonzero Artinian quasi-) local rings, say, . In fact, since the “decomposed” hypothesis ensures that has exactly two maximal ideals, say and . The “decomposed” hypothesis also ensures that ; and the canonical ring (field) homomorphism is an isomorphism for . With denoting the unique maximal ideal of , there is no harm in taking and . It will be useful later to notice that for .
Next, note that is a special principal ideal ring (in short, an SPIR), in the sense of [17, page 245]. (The literature contains a variety of definitions of an SPIR, but those definitions are all equivalent.) Consequently, each nonzero proper ideal of is of the form for some integer such that [17, page 245]. Then, for each , the First Isomorphism Theorem shows that the composition of the inclusion map with the canonical projection map has an image which is a subring of that is isomorphic as an -algebra to for some nonnegative integer . There is no harm in viewing as a subring of for . Observe that the canonical map is an injection (since the composition of with the inclusion map is the inclusion map ). For each , we have . Put . Then . Since is an injection, . Thus, either or is . Without loss of generality . Hence, we can view as a subring of .
We claim that is not a proper subring of . To see this, suppose, on the contrary, that . Using to identify with its image in , we then get , and so the minimality of yields that . However, this gives a contradiction, since has only one prime ideal while has two distinct prime ideals. This contradiction proves the claim. Thus, . It remains to show that as algebras over .
We have and . It follows that Of course, when viewed inside , each element of is identified with , where is the image of under the canonical projection map . Thus, viewing matters in , we have Each occurs as the first coordinate of only one element in the last-displayed set, and so must be a singleton set. Hence, ; that is, is a field. Therefore, , as desired.
The next result classifies certain ramified extensions. Its proof served us as motivation for the second construction that was used in the proof of Theorem 6(2).
Proposition 11. Let , where is a prime number. As usual, write , where . Then, up to -algebra isomorphism, there are exactly two rings such that is a ramified extension. One of these rings, say , can be constructed as the -algebra whose additive structure as a vector space over is given by for some nonzero element and whose multiplication is determined by the relations and (and ). The other ring can be taken as . Moreover, has exactly two proper subrings, while has more than two proper subrings (and so is not isomorphic to ).
Proof. To shape the constructions of the rings , consider any ramified extension of . The crucial maximal ideal of is necessarily then the unique prime ideal of ; namely, . As recalled in the proof of Proposition 10, the fact that is an integral minimal ring extension gives that . In particular, is an ideal of . Moreover, since is ramified and is quasi-local, it follows that must be quasi-local, say with maximal ideal , such that , (), and . Hence , and so . Therefore, taking to be any element of , we have . Observe that .
We claim that . In fact, and so for some . Thus . Since , it follows that is not invertible in . As , we see that is nilpotent. Hence, is not invertible (in ). Since is a member of the field , we get , whence , thus proving the above claim. Note also that since .
Since and , it follows that the linear structure of must be given by . Given this explicit linear structure and the specific facts that we have found about multiplication in , there are now two general ways to describe a set of necessary conditions for the -algebra . One would impose the following set of relations: , , and . This set of conditions determines the -algebra . It is known that is ramified (see  and [16, Lemma 2.1]), and so this given set of relations should not be augmented in a possible search for other possible . To show that has more than two proper subrings, one need only tweak an argument that was given in the proof of Theorem 6(2): consider .
However, there is one other general way to prescribe the multiplication of , namely, as follows: , , and , where is some nonzero element of . We next explain that any such prescription does give a -algebra and that any two such prescriptions (using different values of the nonzero element ) give isomorphic -algebras. The first of these tasks can be handled classically, as we have merely stipulated how to multiply the basis elements of a vector space. (Lacking that vectorial context led us to the more complicated argumentation given for the somewhat similar construction in the proof of Theorem 6(2).) Let denote the -algebra that is constructed when we use . The second task is now handled by noticing that any other nonzero element , say , gives a -algebra that is isomorphic to . What it does give is a multiplication that is determined on the elements of the -basis of the -algebra by , , and . Consider the induced multiplication on the basis consisting of , , and . One easily finds that , , and . Thus, the construction using has produced an algebra that is isomorphic (via the change of bases from to ) to the algebra that was constructed using .
Only two tasks remain: to prove that is ramified and that has only two proper subrings. To accomplish the first of these, we will invoke [12, Theorem 4.2]. For that, it suffices (since ) to show the following: is a maximal ideal of such that , , and the canonical map is an isomorphism. All of the preceding assertions are easily verified and so we omit those details. Finally, to show that has only two proper subrings, it suffices to show that if , then . Without loss of generality, for some . To complete this proof, it suffices to repeat the argument in the final paragraph of the proof of Theorem 6, with that proof’s coefficient ring now being modified to for the purposes of this proof.
Note that the idealization used in Proposition 11 can also be described as . In the same spirit, one could describe the ring from that result as . This description of is admittedly more compact than the earlier one, but that earlier description seemed more suitable in presenting the proof of Proposition 11.
Part (a) of the next result was developed in correspondence with Jay Shapiro and is included here with his kind permission.
Proposition 12. Let , where is a prime number and is an integer. Let , the maximal ideal of . Then, consider the following:(a)Let be a ring such that is a ramified extension. Let denote the maximal ideal of . Then there exists such that , , and . In fact, . Moreover, there exist units and of and uniquely determined integers and such that , , and . Also, . If , then is even and (so ). If , then .(b)Let be a ring such that is a ramified extension. Let be as in (a). Let in . Then is a subset (not necessarily an additive subgroup) of of cardinality . Moreover, the assignment gives a bijection (which does not necessarily preserve any algebraic structure).(c)If is a ring such that is a ramified extension, then .(d) is a ramified extension.(e)Up to isomorphism, has exactly two ramified extensions, namely, the ring given by (d) and the ring .(f)Suppose that is an odd prime number. Then, up to isomorphism, has exactly three ramified extensions, namely, the ring given by (d), the ring , and the ring , where is a(ny) quadratic nonresidue of .
Proof. (a) By [13, Proposition 2.12], there exists such that , , , and . In particular, . As is an SPIR whose unique prime ideal is generated by , it follows from [17, page 245] that there exist uniquely determined integers and and units and of such that , , and . Thus If , then the right-hand side of the displayed equation is nonzero (again, by [17, page 245]), and thus so is the left-hand side, whence (again, by [17, page 245]), so that and . In particular, if , then is even, , and . On the other hand, if , then the right-hand side of the displayed equation is , hence so is its left-hand side, whence and .
It remains only to prove that . First, note that is nilpotent since is a member of the only prime ideal of . (In general, .) Recall that . As and , it suffices to show that . Suppose, on the contrary, that . Then . As the nilpotence of implies that is a unit of , it follows that , contradicting .
(b) Let with . By the division algorithm, there exist nonnegative integers and such that and . Thus, can be expressed as the sum of an element of with an element of . We will show that this expression is unique.
We claim that there cannot exist nonnegative integers such that . For a proof, suppose, on the contrary, that satisfies . Since , and are relatively prime, and so there exist integers , such that . Hence, . As and are elements of , it follows that , the desired contradiction. This proves the above claim.
Now suppose that , where and . As , it follows from the above claim that (hence ) and . This proves the above uniqueness assertion. It also shows that and establishes the asserted bijection.
(c) One proof of (c) is via (b), for . We next give an alternate proof. Let denote the maximal ideal of . Since is ramified, the canonical map is an isomorphism and . Therefore, by Lagrange’s Theorem, .
(d) By , is a minimal ring extension of ; and this extension is ramified (i.e., subintegral) by [16, Lemma 2.1].
(e), (f) Let be a ramified extension of and let denote the maximal ideal of . We adopt the other data from (a) (now with ). Recall from (a) that . As here, . Next, since , it follows as above via [17, page 245] that for some unit of and some (uniquely determined) positive integer . It cannot be the case that . (Otherwise, and , with and suitable units of as above, and so , a contradiction since .) On the other hand, if , then (the point being that if , then ). Thus, is either or . In any event, , and so . In summary, we have , , and is either or of the form (for some unit of ).
We first consider (e), where . Then, since the set of units of is , we have that is either or . By (b) and (c), has cardinality . Since the algebraic structure of is determined by the values of and , it follows that (up to isomorphism) there are at most two ramified extensions of . The first of these candidates is , which arises from the identities and . Note that is isomorphic to , which we know, by (d), is a ramified extension of (where can be taken to be ). Hence, is a ramified extension of .
The remaining candidate is , which arises from the identities and . While it was rather obvious that , we next include a proof that . To view , we will show that the canonical -algebra homomorphism is an injection; in other words, we claim that . Suppose, on the contrary, that for some and some . Equating constant terms shows that . As , it follows that the constant term of must be a unit of . Hence, the coefficient of in is a unit of . This is a contradiction (since ). This proves the above claim. We leave to the reader a similar argument (focusing on the coefficient of ) which reveals that . Thus, satisfies , with and . Next, note that the set of nilpotent elements of , namely, its unique prime ideal, is . (The same could be said of .) By verifying the conditions in [12, Theorem 4.2(c)], it is straightforward to verify that is a ramified extension of . (In detail, , so ; since ; and .)
It remains only to verify that and are not isomorphic. For each of these rings , the only elements of that can play the role of in (a) are and ; and . In (resp., ), each element of has index of nilpotence (resp., ). This completes the proof of (e).
We turn to (f), where is an odd prime and . The first candidate for a ramified extension of is , which arises from the identities and . As above, we see via (d) that is a ramified extension of since . Any other (nonisomorphic) candidate for a ramified extension of must arise from the identities and for some unit of . This candidate can be viewed as . Ostensibly, could be the coset of in that is represented by any positive integer which is not divisible by . However, it suffices to further restrict , for if the division algorithm gives , then leads to since . As the above illustrates, it will be convenient (and harmless) to blur the distinction between and its coset representative , so we will view . As is well known (cf. [18, Theorem 9.1]), exactly half of the elements in the set are quadratic nonresidues of (and the other half are quadratic residues of ). To finish the proof of (f), it suffices to establish the following five facts (where ): each is a ramified extension of ; is not isomorphic to ; if and each are quadratic residues of , then ; if and each are quadratic nonresidues of , then ; and if is a quadratic residue of and is a quadratic nonresidue of , then and are not isomorphic.
Let . By reworking the above analysis of in the proof of (e), we can view and see that satisfies , with and . Next, by reworking the proof of (b), one can show that each element of can be uniquely expressed as the sum of an element of and an element of the form where . It follows easily that . Moreover, the set of nilpotent elements of is the ideal of cardinality . It follows that is the unique maximal ideal of ; ; ; and the canonical ring homomorphism is an isomorphism (since it is an injective map of fields both of which have cardinality ). In addition, since and , we see that is an ideal of . Then must be a two-dimensional vector space over since . We have verified all the criteria in [12, Theorem 4.2(c)], and so is a ramified extension.
Next, we show that there cannot exist an -algebra isomorphism . Suppose, on the contrary, that such exists. View with . As , we see by applying that must satisfy and . Consequently, . Therefore, as noted in the preceding paragraph (and using its notation), for some and some . (Note that since .) Then (since is not divisible by ), the desired contradiction. Hence, is not isomorphic to .
Next, to show that whenever and are quadratic residues of , it is enough to show that if and is a quadratic residue of , then . It suffices to prove that there exists such that , , and (for then ). By hypothesis, there exists such that . By the definition of , there exists such that , , and . Put . Since is a unit of , we have , and so . It follows that . It is easy to check that . Moreover, since for some “integer” and , we get , as required.
Next, we will show that whenever and each are quadratic nonresidues of . By the well known multiplicative property of the Legendre symbols [18, Theorem 9.3(ii)], is a quadratic residue of . Pick such that . Let be the element of such that . Then, working in , we have and so . Given such that , , and , put . As is a unit in , we see as above that , and so . It is easy to check that . Moreover, for some “integer” , and so , as required.
Finally, to complete the proof, it suffices to show that if is a quadratic nonresidue of , then is not isomorphic to . Suppose, on the contrary, that there exists an isomorphism . By hypothesis, there exists such that , , and ; and there exists such that , , and . Put . Since , we see by applying the isomorphism that . Therefore, by (b), for some (uniquely determined) and . Since is nilpotent, so is . But is also nilpotent. Thus, is nilpotent; that is, . Hence . In addition, ; that is, . Also,