# Results from abstract algebra which look wrong (but are true)

There are many statements in abstract algebra, often asked by beginners, which are just too good to be true. For example, if $$N$$ is a normal subgroup of a group $$G$$, is $$G/N$$ isomorphic to a subgroup of $$G$$? As an experienced mathematician, we see immediately that there is no reason for this to be true — even without thinking about this in detail. Often we can quickly come up with counterexamples. Sometimes, it is hard to find counterexamples.

Many questions fall into this category, for example:

• If $$f : R \to S$$ is a ring homomorphism and $$I \subseteq R$$ is an ideal, is then $$f(I) \subseteq S$$ an ideal? (SE/2200335)
• $$\DeclareMathOperator\Aut{Aut}$$If $$G,H$$ are groups, do we have $$\Aut(G \times H) \cong \Aut(G) \times \Aut(H)$$? (SE/1236571)
• Is every submodule of a finitely generated module also finitely generated? (SE/83078)
• If $$A$$ is an abelian group with $$A^3 \cong A$$, does this imply $$A^2 \cong A$$? (MO/10128)
• If $$A$$ is an abelian group with $$A \oplus \mathbb{Z}^2 \cong A$$, does this imply $$A \oplus \mathbb{Z} \cong A$$? (MO/218113)
• If $$G$$, $$H$$ are groups whose group algebras $$\mathbb{Q}[G]$$, $$\mathbb{Q}[H]$$ are isomorphic, are then $$G$$, $$H$$ isomorphic? (SE/1342851)

But my question is actually about situations where, for some strange reason, our first gut feeling is not correct and a wrong-looking statement turns out to be true. Examples will be abundant, which is why I want to restrict this question to examples coming from abstract algebra (you are welcome to open similar questions for other branches and flavors of mathematics, and please let me know if there are already questions of this type).

Here are some examples which come to my mind:

• Every group homomorphism $$\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}$$ is a finite linear combination of projections. In fact, $$(\mathbb{Z}^{ \mathbb{N}})^* \cong \mathbb{Z}^{\oplus \mathbb{N}}$$. (Specker 1950)
• If $$A$$, $$B$$ are finitely generated abelian groups (more generally, finitely generated modules over a commutative Noetherian ring) and $$f : A \to A \oplus B$$, $$g : A \oplus B \to B$$ are homomorphisms such that $$0 \to A \xrightarrow{f} A \oplus B \xrightarrow{g} B \to 0$$ is exact, then it is split exact.
• If $$A$$, $$B$$, $$C$$ are finite groups such that $$A \times B \cong A \times C$$, then $$B \cong C$$. (SE/3579745)
• every negation of the examples mentioned above, for example: There is an abelian group $$A$$ with $$A \cong A^3$$ and $$A \not\cong A^2$$. (However, I am more interested in "positive" results.)

I am looking for statements in abstract algebra where this is your reaction when you learn that they are actually true.

Please try to include a reference for the statement and proof.

• How about every simple group is generated by at most 2 elements. And if it is non-abelian every element is a commutator Jan 19 at 23:33
• Yes, I made the WTF face! ;-) You are welcome to post your examples as answers :-) Jan 19 at 23:34
• This might be borderline but two finite fields are isomorphic iff they have the same size. This fails for finite groups or rings or pretty much any other standard algebraic structure except Boolean algebras. Jan 19 at 23:49
• @ycor: I am trying to imagine why you find that depressing. Jan 20 at 0:09
• The fact that $(\mathbb R, +)$, $(\mathbb C, +)$ and $(\mathbb{C}[X],+)$ are isomorphic as groups looked wrong to me first time I have seen it. Jan 20 at 17:06

The free group with infinitely many generators is a subgroup of the free group with two generators.

• I'm still not over this. Jan 20 at 9:17
• The statement is kind of obvious for monoids. Digital technology depends on it.
Jan 20 at 11:43
• @wlad: Those generators work, but so do (uncountably!) many others. Perhaps the most obvious are $\{b^nab^{-n}\mid n\in\mathbb{Z}\}$.
– HJRW
Jan 20 at 12:09
• @Sam: The question is "results which look wrong". Naively, one would perhaps expect that the free functor $\bf Set\to Grp$ would somehow an isomorphism. So the fact that "a set of two elements" has an "infinite subset", in this sense, seems very wrong. Jan 20 at 15:07
• Topologically, I don't really find this too surprising. Something a lot more non-trivial and general is that for any (countable) group with $k$ defining relations, it embeds in a $2$-generated group with $k$ defining relations (the free group case being, of course, $k=0$). Jan 20 at 15:20

Let $$G$$ be a finite group and $$n \mid |G|$$.

If $$S = \{x \in G : x^n = 1\}$$ contains exactly $$n$$ elements, then $$S$$ is a subgroup of $$G$$.

There seems no a priori reason to expect $$S$$ to be a subgroup if $$G$$ is non-abelian. The proof (see Iiyori and Yamaki - On a conjecture of Frobenius (MSN)) uses the classification of finite simple groups.

• I think (surprisingly in my view) this was reduced to the case that $n$ and $\frac{G|}{n}$ are coprime by fairly elementary means by Kate Fenchel. It is conceivable that the general might be provable by character theoretic methods, though many tried ( probably including Frobenius, who first proved that the number of solutions of $x^{n} = 1$ is a multiple of $n$ when $n$ divides $|G|$, and Brauer, who proved this last fact using his characterization of characters), and no-one has succeeded to date. Jan 20 at 14:22
• Some more discussion in answers to an older question. Any proof of the result would probably need character theory at some level, since the fact that Frobenius kernels are subgroups follows as a corollary (see comment here). Jan 21 at 1:32

Every finite simple group can be generated by at most $$2$$ elements. This is another famous consequence of the classification.

• @GeoffRobinson, my understanding is that the question is asking for examples where a property is too good to be true and there is no reason to believe expect it from the definitions but it turns out to be true. What in the definition of simple group seems to imply it should be generated by two elements? Infinite simple groups can require more than 2 generators. Jan 20 at 13:43
• @GeoffRobinson, I guess I interpreted the question as statements where your first inclination would be to look for a coubterexample even if it is hard to find Jan 20 at 16:43
• This answer actually includes two WTF moments, one for each sentence. +1 Jan 20 at 18:48
• @GeoffRobinson, the way I would look at a question "is a finite simple group generated by 2 elements" would be like this: "take the first element, generate a cyclic subgroup with it, and act on it with commutators from the cyclic group generated by the other element; would the resulting set sufficiently dense, for the lack of a better word, in the given simple group to even suspect that it covers entire group?" And the intuitive answer would be "heck no". It is kind of amazing that you can always find two elements that defy that intuitive answer. Jan 21 at 0:09

Every finite index subgroup of a finitely generated profinite group is open. The converse is obvious, but this direction was quite surprising to me. This is a result of Nikolov and Segal and uses the classification of finite simple groups.

• +1 It is amazing how the classification of finite simple groups pops up in so many different, unexepected situations! Several answers here rely on it. Jan 20 at 19:06
• @MartinBrandenburg, the classification leads to many too good to be true statements because it is a real miracle. Jan 21 at 1:23

Every element of a finite simple non-abelian group is a commutator. This is the positive solution to the Ore conjecture (see Liebeck, O’Brien, Shalev, and Tiep - The Ore conjecture) and uses the classification.

• As well as making heavy use of the properties of the finite simple groups, this proof involved about 150 hours (i.e. nearly three years) of computer CPU time to handle base cases. I sometimes wonder whether results like this are somehow true by accident rather than for any genuine reason. Jan 20 at 14:02
• @AndreiSmolensky Thanks for the observation about the analogy with semisimple Lie groups. In my previous comment I meant 150 weeks of CPU time rather than 150 hours. Jan 20 at 16:43
• @DerekHolt Well, some hours before coffee do feel like weeks. Jan 20 at 21:19
• @AndreiSmolensky For something to "not look coincidental," either for finite simple groups or semisimple Lie groups, it might suffice to have a proof that does not rely explicitly on the classification. Jan 21 at 1:48
• H. Blau gave an example of a quasisimple group in which some element is not a commutator. Jan 22 at 11:08

The Nielsen-Schreier theorem that subgroups of free groups are free might have seemed surprising from am algebraic view given the analogue for many other algebraic structures is false. While this is easy to prove topologically, the original algebraic proof is in my view just an algebraic translation of the topological proof that for some strange reason, preceded the topological one.

• As some shameless self-promotion, if one considers varieties of algebras over a field of characteristic zero with one binary operation, a recent result of myself and Umirbaev is that there are infinitely many such varieties that are Nielsen-Schreier (before that, only six such varieties were known - so this was very surprising for us when we proved it). Jan 20 at 12:24
• @VladimirDotsenko, is that Dotsenko and Umirbaev - An effective criterion for Nielsen–Schreier varieties? Jan 20 at 18:17
• @LSpice it is indeed! Jan 21 at 8:04
• @VladimirDotsenko the varieties of groups with this property are known to be the varieties of all groups, abelian groups, and abelian exponent $p$ groups, as proved by P. Neumann and J. Wiegold in "Schreier varieties of groups", Math. Z. (1964). Jan 24 at 8:53
• @GilesGardam yes exactly. And it is equally restrictive for varieties of Lie algebras, for instance. So I think that a lot of people expected a list of possible varieties of nonassociative k-linear algebras with this property to be finite or at least discrete - and we even found continuous families... Jan 24 at 9:00

For a group with finitely many elements of finite order, the set of elements of finite order is a subgroup.

• Interesting! Can you add a reference? Jan 20 at 18:10
• This follows from Schur's Theorem: $[G:Z(G)]<\infty\Rightarrow |G'|<\infty$. WLOG, we can replace $G$ with the subgroup generated by finite order elements. Hence $\cap_{o(x)<\infty}C_G(x)\subseteq Z(G)$. But each of these centralizers $C_G(x)$ is of finite index because the conjugacy class of $x$ is finite (its elements are of finite order). Hence $[G:Z(G)]<\infty$ which shows that $G'$ is finite. Now if $x,y$ are of finite order, then so is the element $xyG'$ of the abelian group $G/G'$. But if $(xy)^n\in G'$ for some finite $n$, $xy$ itself should be of finite order because $|G'|<\infty$. Jan 20 at 19:17
• Here is a completely explicit and elementary proof (in French). Jan 20 at 20:22

As fields, the algebraic closures of the fields $${\bf Q}_p$$ are isomorphic, and are isomorphic to the complex numbers.

• This is a funny one, in that I think it depends on how much you know. When you're first learning field theory, this sounds impossible! (Surely there's some sort of "generalised characteristic" that can algebraically tell $\mathbb Q_p$ and its extensions apart from $\mathbb C$?) But, once you know some more field theory, you get the informal sense (that can of course be made precise via uncountably categorical-ness) that there just aren't that many (big, characteristic-$0$) algebraically closed fields. Jan 20 at 23:47
• It's worth pointing out that some people don't even believe this is true. Jan 21 at 1:42

In combinatorial group theory, loosely speaking almost any problem one can imagine, in full generality, turns out to be undecidable. This includes the word problem, the isomorphism problem, the triviality problem, etc. Here's an example of a very general problem which nevertheless is decidable.

In the mid 1960s and early 1970s, the "equation problem" or "Diophantine problem" for free semigroups (and groups) was studied. Given a fixed free group $$F$$ (for simplicity, say of rank $$2$$ on the generators $$a$$ and $$b$$), this asks: let $$w_1(a, b, X,Y,Z,\dots)$$ and $$w_2(a,b, X,Y,Z,\dots)$$ be two words written in the generators $$a,b$$ and their inverses, together with some "variables" $$X, Y, Z, \dots$$ and their formal inverses. Are there words $$w_X, w_Y, \dots, w_Z \in F$$ such that the equation $$w_1(a, b, w_X, w_Y, w_Z, \dots) = w_2(a,b, w_X, w_Y, w_Z, \dots)$$ holds true in $$F$$? For example, is there a solution to $$XbXY^{-1} = Zb$$ in the free group on $$a$$ and $$b$$? (Yes, e.g. take $$X = Y = Z = a$$). The same problem can be asked for free semigroups instead (in which case one just omits the inverses -- these are just called "word equations").

The "real" Diophantine problem, i.e. the problem of determining whether a polynomial over $$\mathbf{Z}$$ have integer roots, or Hilbert's Tenth Problem, was proved undecidable in general by Matiyasevich in 1970. For a long time it was believed that Hilbert's Tenth Problem should be reducible to the Diophantine problem in free (semi)groups, and thereby prove this latter problem undecidable too.

But this was not to be. In 1977, Makanin proved that free semigroups have decidable Diophantine problem, and a few years later, he also proved that free groups have decidable Diophantine problem. His solutions are incredibly intricate, but have been generalised to all hyperbolic groups by Dahmani & Guirardel (this latter solution is based on Razborov diagrams; Razborov was, like Makanin, also a student of Adian's, and worked to make a geometric version of Makanin's combinatorial arguments).

This is one of few instances of a very general problem which is decidable in combinatorial group theory. The first time I saw it, I was certain I had misread "undecidable" as "decidable", given the general gloom of undecidability in this area (cf. the Adian-Rabin theorem). I think this qualifies!

• There is a remarkable fact underlying this work, which is possibly worthy of its own answer: Every system of equations over a free monoid/free group having a finite number of variables is equivalent to a finite subsystem of it. This was the Ehrenfeucht conjecture, proven by Albert-Lawrence and Guva in the 1980s.
Jan 20 at 16:02
• @ADL While I agree that the work by Albert-Lawrence and Guba is remarkable, Makanin’s work predates them, so I’m not sure what the “underlying” part refers to. Jan 20 at 16:09
• Sorry, wasn't quite thinking correctly - the Guba/Albert-Lawrence result means that the nice descriptions of solutions sets, via Makanin-Razborov diagrams and as EDT0L languages, hold for all systems of equations, rather than just finite systems.
Jan 20 at 16:23
• @ADL I see what you mean now, and yes, I definitely agree! (And: hello! I didn't realise it was you until I clicked your profile!) Jan 20 at 19:55
• @HJRW AL’s and Guba’s proofs in the monoid case are both very short and also go via the Hilbert basis theorem. Jan 22 at 11:59

There exists a finitely-generated infinite group with only two conjugacy classes, a difficult result of Osin.

• I’m not sure if this is more or less surprising than the many other kinds of finitely generated monsters that exist. The ur-example is surely the negative solution to the Burnside problem.
– HJRW
Jan 22 at 15:09

In a finite Frobenius group, the set of all fixed point free elements together with the identity forms a subgroup.

This might not have such a shocking effect to us, since usually when we first hear about Frobenius groups, it's precisely with the goal of proving this statement, but then again, usually in group theory, when some subsets in a large class of groups don't have a "natural" reason for being subgroups, sooner or later they won't be (and also, the above becomes false in infinite Frobenius groups).

• To see how surprising this is, it’s worth considering how dramatically this fails in the case of infinite groups. In terminology usually used by infinite group theorists, Frobenius’ theorem says that every malnormal subgroup of a finite group is a retract.
– HJRW
Jan 20 at 8:56

It seems to me appropriate to name the following totally unexpected result, which is too good to be true, yet is true.

Are there only finitely many finite groups with $$m$$ generators of exponent $$n$$, up to isomorphism (Restricted Burnside problem)?

In the case of the prime exponent $$p$$, this problem was extensively studied by A. I. Kostrikin during the 1950s, prior to the negative solution of the general Burnside problem. The case of arbitrary exponent has been completely settled in the affirmative by Efim Zelmanov, who was awarded the Fields Medal in 1994 for his work.

• This is not a bad example, but I feel it has a minor aesthetic flaw, in the following sense. To me, it seems too good to be true only after learning (the highly nontrivial fact) that the original Burnside problem is too good to be true. Without that background, I would have no intuition about whether the restricted Burnside problem is too good to be true. Jan 20 at 23:26
• @TimothyChow You are absolutely right. In posting this example I was guided only by my own impressions. After the Adyan-Novikov and Tarski monsters, a positive solution to the restricted Burnside problem seemed completely implausible. Jan 21 at 7:19

The Auslander–Buchsbaum theorem that every regular local ring is a unique factorization domain.

I should say that the first time I saw this theorem stated, I was not immediately surprised, but that was because I did not yet have enough experience with commutative algebra to have a well-developed intuition either way. Somehow, the Auslander–Buchsbaum theorem became more amazing to me the more I learned.

I suppose there is a case for saying that Jordan's theorem on finite complex linear groups might be such a result: there is a function $$f: \mathbb{N} \to \mathbb{N}$$ such that for every $$n \in \mathbb{N}$$, every finite subgroup $$G$$ of $${\rm GL}(n,\mathbb{C})$$ has an Abelian normal subgroup $$A$$ with $$[G:A] \leq f(n).$$ This is well known to fail if we try to replace $$\mathbb{C}$$ by an algebraically closed field of characteristic $$p > 0$$.

• And yet there is a true variant in positive characteristic, and its method of proof is astounding (dare I say?, you won't believe this one weird trick!): Larsen and Pink - Finite subgroups of algebraic groups. Jan 20 at 18:08
• @LSpice: Yes, thanks,, I knew about the work of Larsen and Pink, if you "throw in" Lie type characteristic $p$ components ( subnormal quasisimple subgroups). I agree it is an amazing result. Jan 20 at 20:41

Quite a few things in the Hopf algebra world are surprising:

• Takeuchi's theorem: Every connected graded bialgebra is a Hopf algebra. (No finiteness assumptions!) Takeuchi was actually more general: If $$H$$ is an $$\mathbb N$$-graded bialgebra and its $$0$$-th graded component $$H_0$$ is Hopf, then $$H$$ is Hopf.

• The Cartier-Milnor-Moore theorem: Every cocommutative graded (= $$\mathbb N$$-graded) bialgebra in characteristic $$0$$ is isomorphic to the universal enveloping algebra of its primitive space (= space of primitive elements). This means that understanding cocommutative graded bialgebras in characteristic $$0$$ is essentially equivalent to understanding graded Lie algebras. Commutative graded bialgebras can be understood likewise by duality. You can replace "graded" by "filtered", and the theorem still holds with appropriate changes. To see proper wild behavior, you need to go to the neither-commutative-nor-cocommutative case, or to positive characteristics (or work over rings). I consider the proof of Cartier-Milnor-Moore (by studying Eulerian idempotents) to be full of surprises as well, but maybe more combinatorial than algebraic ones.

• The Nichols-Zoeller theorem: If $$H$$ is a finite-dimensional Hopf algebra over a field, and $$A$$ is a Hopf subalgebra of $$H$$, then $$H$$ is a free left $$A$$-module and a free right $$A$$-module. (Perhaps this is somewhat less surprising if you think of it as generalizing Lagrange's theorem for finite groups, but there are so many more Hopf algebras than groups!) Generalized even further by Skryabin (2006).

• Zelevinsky's theory of PSH algebras, showing that every $$\mathbb Z$$-Hopf algebra that satisfies certain positivity properties is isomorphic to a tensor product of degree-stretched copies of the ring of symmetric functions.

Once you get to symmetric functions, the surprises start multiplying: I find the "Schur polynomial = alternant divided by Vandermonde determinant" identity surprising no matter how many different proofs I see; the Littlewood-Richardson rule in its many forms; the semistandard tableaux forming a section of the plactic monoids while also indexing a basis of irreducible $$\operatorname{GL}_n$$-modules; ... But maybe surprises are somewhat less surprising when they come from combinatorics, as we are used to think of algebra as formal manipulations and of combinatorics as a jungle full of life?

And then, back in algebra, there is of course the Fundamental Theorem of Galois theory.

• The symmetric function examples don't quite fit the spirit of the question, IMO. It's not enough for something to be a surprise; the result should look "obviously false" and yet be true. For example, let's compare Littlewood-Richardson coefficients and Kronecker coefficients. A priori, do we expect there to be a nice combinatorial interpretation (in which case Kronecker coefficients are surprisingly nasty) or do we expect there not to be a nice combinatorial interpretation (in which case LR coefficients are surprisingly nice)? I don't think either one is the "obvious default assumption." Jan 20 at 23:37
• Yeah, if you come to the LR coefficients from the Kronecker coefficients, the rule looks too good to be true, but if you go the other way round, it's a disappointment. Jan 21 at 0:01
• Out of curiosity, what exactly surprises you in "Schur polynomial = alternant divided by Vandermonde determinant" ? Jan 21 at 8:09
• @VladimirDotsenko: The fact that the LHS is a weighted sum of tableaux and the RHS is something completely different (the positivity of the coefficients is far from obvious!). Jan 21 at 15:24
• @darijgrinberg ah ok, that makes sense. I tend to think of the formula "Schur polynomial = alternant divided by Vandermonde determinant" as almost the definition of the Schur polynomial (and a particular case of Weyl character formula), and the formula as the weighted sum of tableaux as some surprising addendum, hence the question! Jan 21 at 17:51

Let $$F$$ be a non-abelian free group and let $$G=\prod_\omega F$$ be the direct product of infinitely many copies of $$F$$. Then the abelianisation of $$G$$ has torsion (of order $$2$$), by a theorem of Kharlampovich and Myasnikov ["Implicit function theorem over free groups and genus problem", Knots, braids, and mapping class groups—papers dedicated to Joan S. Birman].

Granted, this is an example of an unreasonably bad phenomenon, not an unreasonably good phenomenon, but I still couldn't believe my ears when I was told it (by Lars Louder).

• That's incredible (especially given how easy it is to show torsion-freeness in the case of a finite direct product). Since the article is 200+ pages long, what theorem of theirs does it follow from? Feb 1 at 16:30
• @Carl-FredrikNybergBrodda: It follows from the following claim: there is a sequence of elements $w_n\in [F,F]$ such that the commutator length $\mathrm{cl}(w_n)$ tends to infinity, but $\mathrm{cl}(w_n^2)=2$. It's then clear that $(w_n)$ is a torsion element of $G$. The existence of such a sequence can be deduced from KM's "Implicit Function Theorem" (Sela has an equivalent result he calls "Merzlyakov's theorem"), but Lars tells me that no explicit construction of $w_n$ is known! He also referred me to this paper: arxiv.org/abs/1504.04261 .
– HJRW
Feb 1 at 17:59
1. There is some theory about maximal valuation rings (a special type of ring with linearly ordered ideals, not necessarily a domain) and then there are almost-maximal valuation rings which is of course a relaxation. But if you look for an almost-maximal valuation ring which is not maximal, it has to be a domain for some reason (Gill - Almost Maximal Valuation Rings).

2. The descending chain condition on right ideals of a ring implies the ascending chain condition on right ideals. The descending chain condition on principal right ideals does not imply the ascending chain condition on principal right ideals: instead it implies the ascending chain condition on principal left ideals (Jonah - Rings with the minimum condition for principal right ideals have the maximum condition for principal left ideals).

The projective dimension of $${\mathbb C}(x,y,z)$$ as a module over $${\mathbb C}[x,y,z]$$ is two if the continuum hypothesis holds, and three otherwise.

If $$k$$ is an algebraic number field then for every positive integer $$n$$ there exist infinitely many field extensions of $$k$$ of degree $$n$$ having no proper subfields over $$k$$.

An example might be that the category of abelian groups is hereditary. That is, every complex of abelian groups is quasi-isomorphic to the (graded) direct sum of its cohomologies. Although this is very straightforward, I know of at least one person (me) who was surprised by this statement when learning homological algebra

That there exist finitely presentable non-Hopfian groups. [I still remember my shock when I was first learned this result!]

A group G is Hopfian if every surjective homomorphism $$\phi:G\to G$$ is in fact bijective. Most common-or-garden finitely generated groups are Hopfian, e.g. linear groups, finitely generated residually finite groups, and hyperbolic groups are all Hopfian. For non-finitely generated groups we can mimic Hilbert's hotel, so for example the infinite product $$\mathbb{Z}\times\mathbb{Z}\cdots$$ is non-Hopfian.

The Baumslag-Solitar group $$\operatorname{BS}(2, 3)=\langle a, b\mid b^{-1}a^2b=a^3\rangle$$ is non-Hopfian, with the relevant map being $$a\mapsto a^2, b\mapsto b$$.

• This seems me like something too good to be true and it isn't Jan 21 at 0:04
• Note that the first known finitely presented non-Hopfian group was due to Higman (1951, so 10 years before the Baumslag-Solitar groups), and is two-related. Interestingly, Higman claims, indirectly, in that same paper that all one-relator groups are Hopfian -- oops! He attributes this to an unpublished result of H & B.H. Neumann. Jan 21 at 11:46
• @BenjaminSteinberg I don't disagree - I focussed too much on the title of the tread. But I did make the reaction when I learned this result!
Jan 23 at 11:16

A subring of a Noetherian ring need not be Noetherian.

Given all the stability properties that Noetherian rings enjoy, this may sound surprising at first, but it becomes much more obvious if you think about the fact that any domain can be embedded in a Noetherian ring - its field of fractions.

Perhaps more surprising though is that this can also fail for finitely generated rings (or finitely generated algebras over a field): the subring $$\mathbb Z[2X,2X^2,2X^3,\dots]$$ of $$\mathbb Z[X]$$ is not Noetherian, and neither is the subring $$k[XY,XY^2,XY^3,\dots]$$ of $$k[X,Y]$$ for any field $$k$$.

• Another example: the ring of integer-valued polynomials (a subring of $\mathbb{Q}[x]$) is not Noetherian. Jan 21 at 2:33
• @SamHopkins, is that easy to prove? Feb 20 at 18:21
• I think it’s pretty easy. See e.g. math.stackexchange.com/questions/408219/… Feb 20 at 20:55

A theorem of Bass: For a ring $$R$$, every left $$R$$-module has a projective cover if and only if $$R$$ satisfies the descending chain condition on principal right ideals.

In the same vein of the statement the OP included:$$G\times H \cong G\times K \Longrightarrow H\cong K$$ for product of finite groups, which can be rephrased as "product in finite groups is cancellative", a similar property holds true for product "powers".

$$\underbrace{G\times \cdots \times G}_{n\text{ times} } \cong \underbrace{H\times \cdots \times H}_{n\text{ times}} \Longrightarrow G\cong H$$

In other words to take powers of finite groups is cancellative.

• I think this follows from the Krull-Schmidt theorem.
A simple module $$S$$ over a finite dimensional algebra $$A$$ over an algebraically closed field $$K$$ such that there exists a non-split short exact sequence $$0\rightarrow S \rightarrow X \rightarrow S \rightarrow 0$$ has infinite projective dimension. This is the strong no loops conjecture and it is not known whether the assumption on the field can be removed, see for example Igusa, Liu, and Paquette - A proof of the strong no loop conjecture.
In ZFC, the complex number field has $$2^{2^{\aleph_0}}$$ automorphisms, whereas the real number field has just one, the identity.
• Since you were careful to specify the axiom system in your other answer—the proof of this that I know uses a transcendence basis of $\mathbb C$ over $\mathbb Q$, and hence, at least superficially, the axiom of choice. Is this still known to be true without AC? (I'm sure the answer is already elsewhere on MO ….) Feb 20 at 18:19
• @LSpice It is consistent with ZF that the only automorphisms of $\mathbb C$ are the identity and complex conjugation. Feb 21 at 3:38