Here are two things that I have mistakenly believed at various points in my "adult mathematical life":

For a field $k$, we have an equality of formal Laurent series fields $k((x,y)) = k((x))((y))$.

Note that the first one is the fraction field of the formal power series ring $k[[x,y]]$. For instance, for a sequence $\{a_n\}$ of elements of $k$, $\sum_{n=1}^{\infty} a_n x^{-n} y^n$ lies in the second field but not necessarily in the first. [Originally I had $a_n = 1$ for all $n$; quite a while after my original post, AS pointed out that that this actually does lie in the smaller field!]

I think this is a plausible mistaken belief, since e.g. the analogous statements for polynomial rings, fields of rational functions and rings of formal power series are true and very frequently used. No one ever warned me that formal Laurent series behave differently!

[Added later: I just found the following passage on p. 149 of Lam's **Introduction to Quadratic Forms over Fields**: "...bigger field $\mathbb{R}((x))((y))$. (This is an iterated Laurent series field, not to be confused with $\mathbb{R}((x,y))$, which is usually taken to mean the quotient field of the power series ring $\mathbb{R}[[x,y]]$.)" If only all math books were written by T.-Y. Lam...]

Note that, even more than KConrad's example of $\mathbb{Q}_p^{\operatorname{unr}}$ versus the fraction field of the Witt vector ring $W(\overline{\mathbb{F}_p})$, conflating these two fields is very likely to screw you up, since they are in fact very different (and, in particular, *not* elementarily equivalent). For instance, the field $\mathbb{C}((x))((y))$ has absolute Galois group isomorphic to $\hat{\mathbb{Z}}^2$ -- hence every finite extension is abelian -- whereas the field $\mathbb{C}((x,y))$ is **Hilbertian** so has e.g. finite Galois extensions with Galois group $S_n$ for all $n$ (and ~~conjecturally~~ provably every finite group arises as a Galois group!). In my early work on the period-index problem I actually reached a contradiction via this mistake and remained there for several days until Cathy O'Neil set me straight.

Every finite index subgroup of a profinite group is open.

This I believed as a postdoc, even while explicitly contemplating what is probably the easiest counterexample, the "Bernoulli group" $\mathbb{B} = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. Indeed, note that there are uncountably many index $2$ subgroups -- because they correspond to elements of the dual space of $\mathbb{B}$ viewed as a $\mathbb{F}_2$-vector space, whereas an open subgroup has to project surjectively onto all but finitely many factors, so there are certainly only countably many such (of any and all indices). Thanks to Hugo Chapdelaine for setting me straight, patiently and persistently. It took me a while to get it.

Again, I blame the standard expositions for not being more explicit about this. If you are a serious student of profinite groups, you will know that the property that every finite index subgroup is open is a very important one, called **strongly complete** and that recently it was proven that each topologically finitely generated profinite group is strongly complete. (This also comes up as a distinction between the two different kinds of "profinite completion": in the category of groups, or in the category of topological groups.)

Moreover, this point is usually sloughed over in discussions of local class field theory, in which they make a point of the theorem that every finite index *open* subgroup of $K^{\times}$ is the image of the norm of a finite abelian extension, but the obvious question of whether this includes every finite index subgroup is typically not addressed. In fact the answer is "yes" in characteristic zero (indeed $p$-adic fields have topologically finitely generated absolute Galois groups) and "no" in positive characteristic (indeed Laurent series fields do not, not that they usually tell you that either). I want to single out J. Milne's class field theory notes for being very clear and informative on this point. It is certainly the exception here.

3Most examples are fantastic especially for those preparing for qualifying/comprehensive exams. – Unknown – 2010-11-22T19:07:53.953

1In addition to common false beliefs, I find something somewhat amusingly alleged to be a common false belief: Some time around 2003 or 2004, when Wikipedia was less developed than it later became, its article about the product rule asserted that the derivative of a product of two functions is different from what "most people think" it is. Then it said "Most people think that $(fg)' = f'g'$. – Michael Hardy – 2010-12-18T20:05:35.747

5It's almost surely time for this to be closed. Flagging for moderator attention. – Todd Trimble – 2011-10-06T12:47:27.400

4I would vote to close at this point if I didn't have superpowers. It is a great question, but perhaps 17 months is long enough. – S. Carnahan – 2011-10-06T15:48:13.770

23

Metacreated http://tea.mathoverflow.net/discussion/1165/examples-of-common-false-believes/ – None – 2011-10-08T14:27:20.867Sorry for being late. Two common false beliefs: 1. Any ring epimorphism is surjective. 2. Suppose given a short exact sequence X'->X->X'' in an abelian category A. If a full subcategory B of A contains X' and X, but not X'', then X'->X does not have a cokernel in B. (Wrong for A = Z-mod, B = Z-free, (X'->X->X'') = (Z -2-> Z -> Z/2).) – Matthias Künzer – 2011-10-08T14:38:58.933

15I vote not to close – Gil Kalai – 2011-10-08T20:28:25.620

3@Matthias: the epimorphism thing might stem not so much from a false belief as from unfortunate terminology. For many people, the

definitionof epimorphismissurjective homomorphism. Presumably this definition predates the category-theoretic one by many decades. – Thierry Zell – 2011-10-09T20:55:34.463@Thierry: As far as I know, "epimorphism" is Bourbaki terminology. I think Weil insisted on not mixing Greek and Latin at this point. So yes, you're right, since Bourbaki's point of view is "sets with structure", the definition via surjectivity is the original one. – Matthias Künzer – 2011-10-13T06:09:52.160

Dear @Matthias, what was the proposed mixture of Greek and Latin ? – Georges Elencwajg – 2013-10-15T14:46:13.803

@GeorgesElencwajg I think the point is that surjective homomorphism would be such a mix (the former being 'Latin' and the latter 'Greek', at least in an ethymological sense). – None – 2013-10-15T17:11:39.400

1@quid: yes, that's a possibility. I know that long ago some purists objected to

televisionfor the same reason. – Georges Elencwajg – 2013-10-15T18:33:20.3531@Georges Elencwajg: if I recall correctly, someone suggested "unimorphism" (Latin/Greek-mixture), but Weil insisted on "monomorphism". – Matthias Künzer – 2013-11-07T08:58:24.480

This is such a wonderfull question! – CSA – 2014-06-30T17:18:39.150

2Over $200$ false beliefs so far… maybe true beliefs are even more, but certainly not as popular! – Pietro Majer – 2015-02-17T02:18:49.780

90I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan – 2010-05-06T00:55:08.100

24The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – Unknown – 2010-05-22T09:04:07.617

one typical mistake in matrix algebras: positive matrices must have positive entries. (However, for example $\begin{pmatrix} 1 & -1 \ -1 & 1 \end{pmatrix}$ is positive as well, since this matrix is self-adjoint and has non-negative eigenvalues) – Ypsilon – 2016-10-31T08:51:58.777

Students thinking that the field $\mathbb{F}_4$ is the ring $\mathbb{Z}/4\mathbb{Z}$... – Sabrina Gemsa – 2017-04-28T15:03:18.317

People think that in a complete lattice $T$, if $M\subset T$, then $\operatorname{inf} M\leq \operatorname{sup} M$ – Max – 2017-07-13T13:39:34.353

2I'm voting to close this question as off-topic because enough false beliefs already – Mikhail Katz – 2017-09-20T08:37:42.163

wow, this will soon reach 666 votes... a nice score for a question about false beliefs – Pietro Majer – 2017-10-22T12:33:21.357

20wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. – Suvrit – 2010-09-20T12:39:27.377

19It's a thought -- I might consider it. – gowers – 2010-10-04T20:13:17.327