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The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

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    $\begingroup$ I have to say this is proving to be one of the more useful CW big-list questions on the site... $\endgroup$ May 6, 2010 at 0:55
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    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$
    – Unknown
    May 22, 2010 at 9:04
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    $\begingroup$ wouldn'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. $\endgroup$
    – Suvrit
    Sep 20, 2010 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

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I used to think that the subset of even norm vectors in an integral lattice is a sub-lattice. This is true for the "classically integral" lattice defined by $<u,v> \in \mathbb{Z}$ for $u,v$ in the lattice because the even vectors is the kernel of the group homomorphism $v \rightarrow <v,v>$ mod 2. However this fails for the more general notion of "integer norm" lattice where we only require the quadratic form is integer valued (ie. the coefficients are all integral or that the off diagonal entries in the Gram matrix may be half integral eg $x^2+xy+2y^2$). For the hexagonal lattice $x^2+xy+y^2$ which is not classically integral, the even vectors is a sub-lattice but for a different reason that it is the lattice scaled by 2.

If $t = 3$ mod 4, the lattice $L_t$ with quadratic form $x^2+ty^2$ is classically integral. Its even sublattice $L_{t0}$ has quadratic form $4(x^2+xy+(t+1)y^2/4)$ which clearly equals 2$W_t$ where $W_t$ is the lattice with quadratic form $(x^2+xy+(t+1)y^2/4)$. If $t=7$ mod 8, the coefficients of the form is [odd,odd,even], the even vectors is not a subgroup since for example $[0,1]$ and $[1,1]$ has even norm but $[0,1]+[1,1]=[1,2]$ has odd norm. If $t$ is 3 mod 8, the form $(x^2+xy+(t+1)y^2/4)$ can only be even only if both $x,y$ are even since all coefficients are odd. So the even vectors in $W_t$ turn out to be $2W_t$. It is a sub-lattice and it is the subset of even vectors but it is index 4 in $W_t$. It is the 2-scaled sub-lattice.

If $v \in L_t$, $2v \in 2L_t \subset L_{t0}=2W_t$, so $L_t \subset W_t$. So the picture is $2W_t=L_{t0} \subset L_t \subset W_t$ with each containment is index 2.

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The GNS construction

Let $\phi$ be a state on a $C^*$-algebra $\mathcal{A}$, and put $N_{\phi}:=\{a\in\mathcal{A}\;|\;\phi(a^*a)=0\}$. Then $N_{\phi}$ is a norm-closed left ideal in $\mathcal{A}$. The sesquilinear form $\left<\cdot,\cdot\right>:\mathcal{A}/N_{\phi}\times\mathcal{A}/N_{\phi}\to\mathbb{C}$ defined by $\left<a+N_{\phi},b+N_{\phi}\right>:=\phi(b^*a)$ is a well-defined inner product on $\mathcal{A}/N_{\phi}$. The completion of $\mathcal{A}/N_{\phi}$ establishes a Hilbert space.

False belief: The completion is in the quotient norm.

Surprisingly, Wikipedia (as of April 27, 2018) presents a false statement "The quotient space of the A by the vector subspace I is an inner product space. The Cauchy completion of A/I in the quotient norm is a Hilbert space, which we label H."(https://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction#The_GNS_construction) First of all, the quotient of a Banach space by its closed subspace is again a Banach space in the quotient norm, which is a very elementary fact in functional analysis. Thus A/I is already complete in the quotient norm, and hence there is no need to complete it in the quotient norm!

The correct completion is, of course, in the norm induced by the inner product, and this norm is not equivalent to the quotient norm in general. In fact, let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space and $\{\xi_n\}_{n=1}^{\infty}$ be an orthonormal basis for $\mathcal{H}$. The linear functional $\phi:\mathbb{B}(\mathcal{H})\to\mathbb{C}$ defined by $\phi(a):=\sum_{n=1}^{\infty}\frac{1}{2^n}\left<a\xi_n,\xi_n\right>$ is a state on $\mathbb{B}(\mathcal{H})$, and $N_{\phi}=\{0\}$. Let $\xi_k\otimes\xi_k$ be the canonical rank-one operator, and put $p_n:=\sum_{k=1}^n\xi_k\otimes\xi_k$. Then $(p_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{B}(\mathcal{H})/N_{\phi}$ in the norm induced by the inner product defined at the beginning, but it is NOT a Cauchy sequence in the quotient norm.

I have never seen a remark which clearly states the distinction between the norm induced by the inner product and the quotient norm in the literature on $C^*$-algebras. Since a quotient space is involved, students are easily tempted to think that the completion is in the quotient norm. (Even the Wikipedia editor was confused!) Or, they may thoughtlessly assume that these two norms are the same. So it will be instructive to clearly state the distinction between these two norms when one teaches this subject to undergraduate students.

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    $\begingroup$ I think it is surprising to be surprised by a wrong statement on Wikipedia. Fortunately it need not remain wrong! $\endgroup$
    – LSpice
    Apr 27, 2018 at 22:31
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    $\begingroup$ @LSpice: Well, the surprising thing is that this false statement appeared on May 6, 2004 and has remained since then, and nobody has corrected the error for 14 years! See the editing history. $\endgroup$ Apr 27, 2018 at 22:55
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    $\begingroup$ I read the "quotient norm" statement as saying that it's the quotient of the seminorm induced by the inner product. Then it's correct, no? $\endgroup$ Apr 28, 2018 at 12:24
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    $\begingroup$ @MasayoshiKaneda: I agree that the Wikipedia entry has been ambiguous at that point, and I've clarified it. $\endgroup$ Apr 29, 2018 at 7:14
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    $\begingroup$ @Tobias Fritz: Good job! I also modified the sentence before the one you clarified. $\endgroup$ May 1, 2018 at 10:51
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Given a finite dimensional vector space $V$ over $\mathbb{R}$ it cannot be written as a countable union of proper subspaces. (This can be proved by algebraic arguments or by the Baire category theorem.)

This may lead one to believe that the same is true if $V$ is infinite dimensional. However, that is false!

The vector space $P$ of polynomials with real coefficients is the union of the subspaces $P_n$ of polynomials of degree $n$.

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    $\begingroup$ I like the example, but "This may lead one to believe" is not the same as "this is a common false belief". $\endgroup$ Jan 26, 2022 at 11:16
  • $\begingroup$ True enough! However, I have often seen people think that throwing in some additional conditions will make it true. For example, I have heard the assertion, "In a normed space a countable union of closed subspaces cannot be the whole space." This is also false with the same counter-example and a norm like the sup norm. $\endgroup$
    – Kapil
    Jan 26, 2022 at 11:29
  • $\begingroup$ Of course, throwing in enough additional conditions does make it true. No infinite-dimensional Bach space is the countable union of closed proper subspaces. $\endgroup$ Apr 13, 2022 at 20:41
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From Kleiman's article "Misconceptions about $\mathcal{K}_X$", the sheaf of meromorphic functions:

Denote by $A_\mathrm{tot}$ the total fraction ring of a ring $A$.

(1) $\mathcal{K}_X$ can be defined as the sheaf associated to the presheaf of total fraction rings $$U \mapsto \Gamma(U, \mathcal{O}_X)_\mathrm{tot}$$

(2) The stalks of the meromorphic functions are the total fraction rings of the stalks: $$\mathcal{K}_{X,x} = (\mathcal{O}_{X,x})_\mathrm{tot}$$

(3) If $U = \mathrm{Spec}(A) \subset X$ is an affine open, then $$\Gamma(U,\mathcal{K}_X) = A_\mathrm{tot}$$

The first two misconceptions apply to any ringed space $X$, and the third applies to a scheme. Please see his nice, three-page article for discussion and examples.

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If $R$ is a commutative ring with $char(R)=p$ ($p$ is prime), then

$$\varphi:R \to R$$ $$x\mapsto x^p$$

is an automorphism.

Which is false of course.

(If $R$ is a field then see comments).

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  • $\begingroup$ @SamHopkins Pay attention, I changed the statement because accidentally I wrote field instead of ring $\endgroup$
    – Or Shahar
    Apr 13, 2022 at 19:45
  • $\begingroup$ But even with field it is still false? $\endgroup$ Apr 13, 2022 at 19:47
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    $\begingroup$ @SamHopkins Well, if $R=\Bbb{F}_p$ then $x^p$ is an automorphism because it’s a 1 to 1 on a finite set and endomorphism. If the field is infinite, then take $\Bbb{F}_p(x)$, there isn’t exist $f(x)\in \Bbb{F}_p(x)$ such that $f(x)^p=x$ $\endgroup$
    – Or Shahar
    Apr 13, 2022 at 20:02
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    $\begingroup$ Yes. I guess I just meant that since the (false) assertion is "weaker" with field instead of ring, as a false belief it might be more common. $\endgroup$ Apr 13, 2022 at 20:03
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Cambridge mathematicians (with the notable exception of the exceptional Dirac) have often misunderstood the Euler-Heaviside fractional integro-derivative calculus, disadvantaging their students.

False belief: A erroneous claim that has been often repeated is that the fractional calculus (FC) as envisioned by Euler and Heaviside (Hadamard, Pincherle, and others) doesn't obey the law of exponents

$D^{\alpha}D^{\beta} =D^{\alpha+\beta},$

and, specifically, differentiation $D$ and integration $D^{-1}$ do not commute, and consequently, neither do they obey the law of exponents.

One example of this apparent lack of commutativity is given on the webpage Fractional Calculus III (circa 2008) by Beardon:

$$D^{-1}D \; e^x = \int_0^x e^t dt = e^x-1 \neq DD^{-1} e^x = D(e^x-1) = e^x = D^0e^x = D^{-1+1}e^x.$$

In fact, however, commutativity applies once the Heaviside step function $H(x)$ is correctly introduced since

$$D^{-1}D \; H(x) e^x =H(x) \int_0^x (\delta(t) +e^t) dt =H(x)( 1+ e^x-1) =H(x) e^x = D^0H(x) e^x .$$

The mistaken claim of lack of commutativity is repeated in another example by Beardon illustrated below (and similar examples by others).

More generally, the Euler-Heaviside FC can be framed a number of ways to ensure the fundamental operator action is interpreted as

$$D^{\alpha}D^{\beta} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\beta} D^{\alpha} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\alpha+\beta}H(x) \frac{x^{\gamma}}{\gamma!} = H(x) \frac{x^{\gamma-\alpha-\beta}}{(\gamma-\alpha-\beta)!}$$

for $\alpha,\beta,$ and $\gamma$ any real numbers.

In one interpretation (e.g., see Gelfand and Shilov's Generalized Functions Vol. I, p. 57),

$$H(x) \frac{x^{-n-1}}{(-n-1)!}= D^n \delta(x)= \delta^{(n)}(x)$$

such that, under a finite part construction or other analytic continuation,

$$D^{n} H(x)\frac{x^{\alpha}}{\alpha!} =H(x) \int_0^x \frac{(x-t)^{-n-1}}{(-n-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{n!}{(z-x)^{n+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-n}}{(\alpha-n)!}$$

and, more generally,

$$D^{\beta} H(x)\frac{x^{\alpha}}{\alpha!} = H(x)\int_{-\infty}^\infty H(x-t) \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} H(t)\frac{t^{\alpha}}{\alpha!}dt$$

$$= H(x)\int_0^x \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{\beta!}{(z-x)^{\beta+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-\beta}}{(\alpha-\beta)!}.$$

Then the Euler-Heaviside FC gives

$$D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!} = H(x)\frac{x^{-1}}{(-1)!} = \delta(x)$$

and

$$D^{\frac{1}{2}}D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!}$$$$ = D^{\frac{1}{2}} H(x)\frac{x^{-1}}{(-1)!} = H(x)\frac{x^{\frac{-3}{2}}}{(\frac{-3}{2})!} = D H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!},$$

so, in this case,

$$D^{\frac{1}{2}}D^{\frac{1}{2}} = D$$

whereas Beardon concludes that

$$D^{\frac{1}{2}}x^{\frac{-1}{2}} = 0,$$

implying that the law of exponents is violated since then

$$D^{\frac{1}{2}} D^{\frac{1}{2}}x^{\frac{-1}{2}} = D^{\frac{1}{2}}0 = 0 \neq D x^{\frac{-1}{2}} =\frac{-1}{2}x^{\frac{-3}{2}} .$$

H. and B. Jeffreys on p. 229 of their book Methods of Mathematical Physics (Second Ed., 1950), often referenced in discussions of operational calculus, assert unqualifiedly that integration and differentiation do not commute. And, Lavoie, Osler, and Trembley in "Fractional derivatives and special functions" (1976) repeated the false argument of the unqualified violation of the law of exponents. A lack of appreciation of the roles of the Heaviside step and delta functions in the Euler-Heaviside FC seems to be at the core of this oversight with a concomitant tendency (issuing from some blend of laziness, carelessness, and quasi-authoritarianism) to reprint the claims of previous researchers who have omitted discussions of the core constructs--the step and delta functions--in their analyses.

If the FC is constructed using an infinitesimal generator, the analytic continuations can be dodged, or a Pochhammer contour integral can be invoked for generalizing the beta function integral. These constructions are consistent with Laplace- and Mellin-transform approaches over the domains of common convergence and analytic continuation of the reps, with Pincherle's axiomatic treatment of a canonical FC, and with the calculus of Appell Sheffer polynomial sequences.

Mikusinski provided an analogous algebraic convolutional approach, and Sato, a hyperfunction approach, reflecting earlier presentations by Niels Nielsen and then much later than Nielsen by Feynman and others.

(An equally erroneous claim is made to the opposite extreme in an answer to this MO-Q that differentiation and integration operators commute unqualifiedly.)

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  • $\begingroup$ When contemporary mathematicians refer to all the families of fractional calculi as semi-groups, I believe they are propagating the same error. $\endgroup$ Dec 5, 2022 at 22:27
  • $\begingroup$ Could you indicate the specific answer to the question that you reference? $\endgroup$
    – LSpice
    Dec 6, 2022 at 1:04
  • $\begingroup$ Re, sorry, I was unclear. You say: "An equally erroneous claim is made to the opposite extreme in an answer to this MO-Q that differentiation and integration operators commute unqualifiedly." I was asking if you could indicate which answer makes that claim. $\endgroup$
    – LSpice
    Dec 6, 2022 at 1:35
  • $\begingroup$ @LSpice, Terry Tao's answer. I wrote a dissenting viewpoint with a link in my comment to that answer, demonstrating when commutation holds and when it doesn't, revolving around exact specification of the integration operator. Specification of an operator naturally involves what it acts upon. $\endgroup$ Dec 6, 2022 at 1:49
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Just today, I realised that I had been mis-interpreting the FTFGAG. One can speak of the $2^\infty$-torsion subgroup or the $2'$-torsion subgroup, the $3^\infty$-torsion or the $3'$-torsion subgroup, or even just the torsion subgroup of a FGAG or the maximal torsion-free quotient … so surely one can speak of the torsion-free subgroup, right? In fact, when I was corrected on this, my first thought was to reply: "just take the subgroup consisting of all infinite-order elements", and it only occurred to me as I was saying it to wonder how the identity element would squeeze its way into this so-called 'subgroup'.

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    $\begingroup$ A good way to think of this is as a natural, exact sequence of abelian groups $1\to tor(G)\to G\to G/tor(G)\to 1$ (and similar sequences). If $G$ is f.g., the sequence happens to split, but not in a natural way. Thus there is not such thing as "the" complement of the torsion subgroup. $\endgroup$ Mar 16, 2018 at 23:08
  • $\begingroup$ Consider the unit group $\mathbf Z[\sqrt{2}]^\times = \{\pm 1\} \times (1+\sqrt{2})^{\mathbf Z}$. Applying conjugation turns $(1+\sqrt{2})^{\mathbf Z}$ into $(1-\sqrt{2})^{\mathbf Z}$ and these are not the same subgroup complementing the torsion subgroup. $\endgroup$
    – KConrad
    Mar 16, 2018 at 23:39
  • $\begingroup$ To be clear, I am aware that this is false! $\endgroup$
    – LSpice
    Mar 17, 2018 at 0:46
  • $\begingroup$ @JohannesHahn, what was misleading me was the fact that, if $\pi$ is any set of primes and $G$ is torsion (as well as finitely generated Abelian), then $1 \to G[\pi^{\mathbb Z}] \to G \to G/G[\pi^{\mathbb Z}] \to 1$ does split canonically; so it seems so tempting to think that a similar result should hold without the torsion hypothesis …. $\endgroup$
    – LSpice
    Mar 17, 2018 at 1:52
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I’m not sure if this counts as “reasonably advanced”, but given the wide-ranging nature of the answers here, I feel like this is worth a mention, given that it plagued my mind for so long when I was in high school.

$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ for all real numbers $a, b$.

Of course, what exactly does $\sqrt{a}$ even mean when $a < 0$ is not completely clear. I’ll just take $\sqrt{a} = i\sqrt{|a|}$ for negative $a$, which seems perfectly reasonable given that we usually teach students $i = \sqrt{-1}$. Under this definition, the above statement is false. Indeed, $\frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i} = -i$ but $\sqrt{\frac{1}{-1}} = \sqrt{-1} = i$. When I first encountered complex numbers, I obtained a “proof” that $1 = -1$ by multiplying both sides of $i = \sqrt{-1} = \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i}$ by $i$ and I had been wondering all throughout my high school years where the error was in the proof until I worked out why exactly I thought $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ and figured out it doesn’t work for negative $a$ or $b$.

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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

https://en.wikipedia.org/wiki/Matrix_exponential#Evaluation_by_Laurent_series (current revision)

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:


There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.


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    $\begingroup$ Dear Johannes, please reread my post. $\endgroup$ May 12, 2010 at 15:18
  • $\begingroup$ Even a cursory examination of Nick Higham's book amazon.co.uk/Functions-Matrices-Computation-Nicholas-Higham/dp/… will show that both these opinions on the evaluation of matrix exponentials are hopelessly naive. $\endgroup$ May 15, 2010 at 9:17
  • $\begingroup$ Dear Robin, Thanks for your answer. I don't have Nick Higham's book. I was wondering if you could be more precise. Your comment is very surprising to me: I thought I was stating a triviality. Here are two references en.wikipedia.org/wiki/Matrix_exponential#Alternative iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients Looking forward to hearing from you. $\endgroup$ May 15, 2010 at 10:51
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    $\begingroup$ Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. $\endgroup$ May 15, 2010 at 14:07
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    $\begingroup$ I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. $\endgroup$ May 17, 2010 at 6:19
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I don't think I've seen it in here:

Every vector space has a non-trivial dual space ($L^p$ for $0 < p < 1$ was a counter-example only mentioned during one of the classes in measure theory)

And of course there's the common false belief of people outside of mathematics that "mathematicians work with numbers and formulae all day long" :)

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    $\begingroup$ Well, it is true that every vector space has a dual space, even $L^{1/2}$... and it is even true that every topological vector space has a continuous dual space... What you mean is that it is not true that every topological vector space has a non-trivial continuous dual space (or, that the continuous dual of a topological vector space does not necessarily separate points) $\endgroup$ Jul 7, 2010 at 18:54
  • $\begingroup$ You are indeed correct. I'll do better not to dismiss the trivial case the next time. $\endgroup$
    – Asaf Karagila
    Jul 7, 2010 at 19:31
  • $\begingroup$ The existence of a nonzero functional on a locally convex space is guarenteed by the Hahn-Banach theorem. $p$-Banach spaces are in general not locally convex if $p<1$. $\endgroup$
    – user20948
    Feb 8, 2021 at 22:45
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False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

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  • $\begingroup$ Yay! 100th answer! $\endgroup$ Oct 4, 2010 at 21:33
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    $\begingroup$ I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. $\endgroup$ Oct 5, 2010 at 0:35
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    $\begingroup$ True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. $\endgroup$ Oct 5, 2010 at 10:52
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The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$ (x + y) (x^2 + y^2) = z^2 $

On defining $ u = x/y $ and $ v = z/y $ one obtains $ y (u + 1) (u^2 + 1) = v^2 $, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $ x y (x y + 1) (x + y) = z^2 $ rational? I'm almost sure it isn't; but how can one be sure?

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  • $\begingroup$ Algebraic genus in software: Seems to be available in Singular and Maple and may be available in Mathematica by now (from mathematica.stackexchange.com/a/5453/16237 ). (Aside: I don't know anything about the software mentioned.) $\endgroup$ Aug 13, 2017 at 2:06
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If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

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    $\begingroup$ You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. $\endgroup$ Jan 28, 2011 at 16:46
  • $\begingroup$ But that's not a bilinear form. And it has no generalization to other fields (what is it on $\overline{\mathbb{F}_p}$?). How can it be standard? :) I certainly agree that people should know that matrices which are self-adjoint with respect to the standard sesquilinear form are diagonalizable. $\endgroup$ Jan 28, 2011 at 17:51
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    $\begingroup$ Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) $\endgroup$ Jan 28, 2011 at 21:24
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If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

  1. Any expression inside a radical evaluates to a real number
  2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

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Something I was sure about until earlier today:

Suppose $\kappa$ is an $\aleph$ number, then $AC_\kappa$ is equivalent to $W_\kappa$, namely the universe holds that the product of $\kappa$ many sets is non-empty if and only if every cardinality is either of size less than $\kappa$ or has a subset of cardinality $\kappa$.

In fact this is only true if you assume full $AC$, and $(\forall \kappa) AC_\kappa$ doesn't even imply $W_{\aleph_1}$, I was truly shocked.

Furthermore, $W_\kappa$ doesn't even imply $AC_\kappa$ in most cases.

The strongest psychological implication is that most people actually think of the well-ordering principle as a the "correct form" of choice, when it is actually Dependent Choice (limited to $\kappa$, or unbounded) which is the "proper" form, that is $DC_\kappa$ implies both $AC_\kappa$ and $W_\kappa$.

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    $\begingroup$ How common is this misconception? $\endgroup$ Apr 17, 2011 at 3:08
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    $\begingroup$ @Thierry: For the past couple of weeks I spent a lot time considering models without choice, not only I held that misconception but not once anyone corrected me about it - grad students and professors alike. $\endgroup$
    – Asaf Karagila
    Apr 17, 2011 at 6:09
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Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

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    $\begingroup$ Is this really a common(!) false belief? $\endgroup$ Oct 3, 2011 at 7:23
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Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.

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Here is a false belief I had. Let $f:X \to Y$ be a map of topological spaces having the property that for every finite CW complex $K$, the induced map $f_{\ast}:[K,X] \to [K,Y]$, on unpointed homotopy classes of maps, is a bijection. Then $f$ is a weak homotopy equivalence (that is, it induces isomorphisms on all homotopy groups relative to all basepoints). A counterexample is given by the stabilization map $B \Sigma_{\infty}\xrightarrow{+1} B \Sigma_{\infty}$, which is not an isomorphism on $\pi_1$.

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    $\begingroup$ Although the original intent of this question seems to have long since evaporated, I can't help asking: is this really a "common false belief"? $\endgroup$
    – Yemon Choi
    Feb 17, 2015 at 1:24
  • $\begingroup$ how about: if two CW complexes have all homotopy groups isomorphic, then they are homotopy equivalent? as i recall, you need those isomorphisms to be induced by a single continuous map. $\endgroup$
    – roy smith
    Apr 22, 2017 at 0:01
  • $\begingroup$ @roysmith Yes. You can even have two non weakly equivalent spaces having all Postnikov stages weakly equivalent $\endgroup$ May 8, 2017 at 10:47
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Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

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    $\begingroup$ Definitely integrals are numbers and $f(x)g(x)$ is a function of variable $x$. Formula as written is something very strange. $\endgroup$ Apr 20, 2016 at 18:58
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    $\begingroup$ @FedorPetrov More strange is that most students don't see such a very strange something :) $\endgroup$ Apr 20, 2016 at 19:50
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This might not be common, but I once believed the following.

Let $ A, B $ be integers, and define a sequence by the linear recurrence $ s_n = A s_{n-1} + B s_{n-2} $ with the base case $ s_0 = 0 $, $ s_1 = 1 $. Two important special cases are the Fibonacci sequence ($ A = B = 1 $) and the sequence $ s_n = 2^n - 1 $ (where $ A = 3 $, $ B = -2 $). Then, for any integers $ n $ and $ k $, $ \gcd(s_n, s_k) = s_{\gcd(n,k)} $.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $ A = B = k = 2 $, $ n = 3 $.

Update: corrected the powers of two minus one example from $B = 2$ to $B = -2$. Thanks to Harry Altman.

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  • $\begingroup$ Quick correction, that should be A=3, B=-2 for 2^n-1. $\endgroup$ Apr 7, 2011 at 21:23
  • $\begingroup$ Hmm, this raises an obvious question of whether it is true whenever (A,B)=1. $\endgroup$
    – JoshuaZ
    Apr 22, 2020 at 23:09
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The derived subgroup of a finite group equals to the set of all its commutators

or equivalently

A product of two commutators in a finite group is always a commutator

This mistake is very widespread, probably because counterexamples to it tend to be quite large. The smallest group, for which it is not true has order $96$.

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    $\begingroup$ What is your evidence that this is a commonly held belief? $\endgroup$ Aug 28, 2020 at 12:06
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Let $M_1$ be a finitely generated module over a PID and let $M_2$ be a submodule.

We may pick $L_i$ and $T_i$ submodules of $M_i$ such that $L_i$ is free, $T_i$ is torsion, $M_i = L_i \oplus T_i$, $L_2\subseteq L_1$ and $T_2\subseteq T_1$.

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If we regard a ring $R$ (with identity) as a right module ($R_{R}$), then there is a ring isomorphism $\text{End}(R_{R}) \simeq R$, however the same does not happen if we regard $R$ as a left module!

The correct is $\text{End}(_{R}R) \simeq R^{\text{op}}$.

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  • $\begingroup$ Here is a discussion about the condition for Morita equivalence between rings, which is related to this subtle detail: math.stackexchange.com/questions/3566579/… $\endgroup$
    – user144185
    Mar 27, 2020 at 12:19
  • $\begingroup$ But this is just a notational quirk. In some Russian algebra texts for example, the composition "f then g" in $\operatorname{End}_R({}_RR)$ is written $fg$, which leads to $R$ being isomorphic, as a ring, to its (left $R$-linear) endormorphism ring. $\endgroup$
    – Jo Mo
    Jan 27, 2022 at 16:11
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I don’t know how common is the following false belief, but I had it for several years, so maybe some other people also have it. I apologize to those to whom I shared this false belief. I hope this post will help.

Kaplansky’s 6th conjecture (here, 1975) states that if $H$ is a finite dimensional semisimple Hopf algebra and $V$ an irreducible representation of $H$, then $\dim (V)$ divides $\dim (H)$. This conjecture is open over the complex field $\mathbb{C}$, but false in positive characteristic. So we assume to be over $\mathbb{C}$.

For the group case, this property was proved by Frobenius, that is why a finite dimensional semisimple Hopf algebra (over $\mathbb{C}$) with this property is called of Frobenius type.

A finite dimensional Hopf algebra (over $\mathbb{C}$) is called a finite quantum group (or Kac algebra) if it has a $*$-structure. And then it is also semisimple. It is an open problem whether such a $*$-structure always exists.

False belief: George Kac proved Kaplansky’s 6th conjecture for the finite quantum groups.

This false belief was pointed out to me by Pavel Etingof after this talk I gave for Harvard University, and where I mentioned it. Fortunately, that does not affect the content of the talk.

What I had in mind is Theorem 2 in the following paper:
G. I. Kac, Certain arithmetic properties of ring groups., Funct. Anal. Appl., 6 (1972), pp. 158–160.

In modern language, Theorem 2 proves the following: let $H$ be a finite quantum group, and let $\mathcal{C} = Corep(H)$ be the fusion category of complex corepresentations of $H$. For every simple object $X$ of the Drinfeld center $Z(\mathcal{C})$ which contains the trivial object of $\mathcal{C}$ under the forgetful functor, $FPdim(X)$ divides $FPdim(\mathcal{C}) = \dim(H)$ (the quotients are called the formal codegrees).

Note that these $X$ correspond to the irreducible representations of the Grothendieck ring $K(\mathcal{C})$ of $\mathcal{C}$ (see Theorem 2.13 here). In particular, for $G$ a finite group, $\mathcal{C} = Corep(G) = Vec(G)$, and $Irr(K(\mathcal{C})) = Irr(G)$. That is why Theorem 2 implies Kaplansky’s 6th conjecture in the group case (i.e. covers Frobenius theorem). But it is not clear for a finite quantum group in general. It could be relevant to search in this direction, in particular to check whether for every object $Y$ of $Irr(H)$ there exists an $X$ as above such that $\dim(Y)$ divides $FPdim(X)$, because this would prove that $H$ is a Frobenius type.

Note that Theorem 2 (as stated above) holds more generally for every (complex) fusion category $\mathcal{C}$. The case $\mathcal{C} = Rep(G)$, with $G$ a finite group, recovers the fact that the size of each conjugacy class of $G$ divides $|G|$. Finally, according to Pavel, the theorem holds more generally without the assumption ‘which contains the trivial object’ (I don’t have the exact reference for that, so if you know it, please put it in comment).

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Here's one that I think will surprise some number theorists:

False belief. Let $E$ be an elliptic curve over an algebraically closed field $k$ of characteristic $p > 0$. Then $\operatorname{End}^\circ(E)$ is strictly larger than $\mathbb Q$.

While this is true for all elliptic curves defined over finite fields, most elliptic curves whose field of definition is transcendental over $\mathbb F_p$ have $\operatorname{End}^\circ(E) \cong \mathbb Q$. The extra automorphism on elliptic curves over a finite field comes from the geometric Frobenius. For varieties over larger fields, this is not a thing.

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“A reversible computer can factor integers efficiently in polynomial-time”: Since a reversible computer can multiply two integers efficiently in polynomial-time, it can also factor an integer efficiently in polynomial-time by just putting the computer in reverse.

I wish this were true.

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    $\begingroup$ As someone who knows hardly anything about the mathematical underpinnings of reversible computing, is the flaw that a reversible computer only guarantees that computations can be reversed, not that the inverse computation is as efficient as the original? (I also don't quite understand what it means to reverse a multiplication, since the factors are not uniquely determined. Does one insist on multiplying only distinct prime powers?) $\endgroup$
    – LSpice
    Jan 15, 2023 at 17:21
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    $\begingroup$ @LSpice The flaw is what is the output of multiplication between two integers on a reversible computer? It has to include more than just the product of the two integers in order for the computation to be reversible. For instance, the output might include one of the integers that one started with. $\endgroup$ Jan 15, 2023 at 17:55
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Fans: (related to the one of polytopes written above) all convex cones are rational, i.e. one would expect that a line would eventually hit a point in the lattice. It is obviously not true, just take the one-dimensional cone generated by $(1,\sqrt{2})$. A similar one was thinking that if I rotate the cone a bit, I can always make it rational.

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    $\begingroup$ reminds me of the curious fact that some circles in the plane, too, have no points in $\mathbb Q^2$. (proven simply by cardinality!) $\endgroup$ Oct 4, 2010 at 19:21
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Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

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  • $\begingroup$ I think, it is closely related to the following false "deduction": because invertible element cannot be at the same time zero divisor, therefore sum of any unit and zero divisor is not invertible. Ok, maybe it isn't popular, but I've got this belief at my first algebra course, until I discovered counterexample $1+X$ in $R[X]/X^2$. This is almost exactly the thing you mentioned, just put $B:=X, A:=X+1$. $\endgroup$
    – Przemek
    Nov 5, 2020 at 20:09
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Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

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    $\begingroup$ This does not seem like a common false belief. $\endgroup$
    – Yemon Choi
    May 15, 2016 at 11:47
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Let $M \subset B(H)$ be a von Neumann algebra, $p \in B(H)$ a projection and $q=I-p$.

False belief: If $pM=Mp$ then $M=pMp \oplus qMq$.
(I think it is a quite common careless mistake)

Counter-example: diagonal embedding of $\mathbb{C}$ into $M_2(\mathbb{C})$.

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