Not especially famous, long-open problems which higher mathematics beginners can understand This is a pair to
Not especially famous, long-open problems which anyone can understand
So this time I'm asking for open questions so easy to state for students of subjects such as undergraduate abstract algebra, linear algebra, real analysis, topology (etc.?) that a teacher could mention the questions almost immediately after stating very basic definitions.  
If they weren't either too famous or already solved, appropriate answers might include the Koethe Nil Conjecture, the existence of odd order finite simple groups or the continuum hypothesis.
Thanks
 A: Does $\lim_{n\to\infty}R(n,n)^{1/n}$ exist? (Where $R(n,n)$ is the classical Ramsey number.)
A: I think the following were already mentioned in answers to 
Not especially famous, long-open problems which anyone can understand
Are there circulant Hadamard matrices of degree > 4?
What is the actual value of R(5,5)?
The integer brick problem.
A: Is there a Borel set in the plane which meets every straight line in exactly two points?
A: The Dixmier conjecture in rank $n$ asserts that any endomorphism of the $n$-th Weyl algebra (the algebra of polynomial differential operators in $n$ variables) is invertible.
See "The Jacobian Conjecture is stably equivalent to the Dixmier Conjecture"
Alexei Belov-Kanel, Maxim Kontsevich

A conjecture (Kontsevich???) which says that the automorphism group
of the Weyl algebra in characteristic zero is canonically isomorphic to the
automorphism group of the corresponding Poisson algebra of classical polynomial symbols.
See "Automorphisms of the Weyl algebra"
Alexei Belov-Kanel, Maxim Kontsevich

Kaplansky's 6th conjecture: dim(Irrep) | dim(algebra) - for semi-simple Hopf algebras
See Kaplansky's 6th conjecture: dim(Irrep) | dim(algebra) - for semi-simple Hopf algebras 

Kaplansky zero-divisor conjecture
Let K be a field and G a group. The so called zero-divisor conjecture for group rings asserts that the group ring K[G] is a domain if and only if G is a torsion-free group.
See
What is the current status of the Kaplansky zero-divisor conjecture for group rings?
Zero divisor conjecture and idempotent conjecture
A: Given $4$ involutions $\alpha_1,\alpha_2,\alpha_3,\alpha_4\in\text{Sym}(\mathbb N)$, do there exist $3$ involutions $\beta_1,\beta_2,\beta_3\in\text{Sym}(\mathbb N)$ such that $\{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}\subseteq\langle\beta_1,\beta_2,\beta_3\rangle$?
This is a slightly restated form of the following question from p. 231 of F. Galvin, Generating countable sets of permutations, JLMS 51, 1995:

Question 1.4. Let $E$ be an infinite set. Is every countable subset of $\operatorname{Sym}(E)$ contained in a group generated by three involutions?

A: Given a unit square, can you find any point in the same plane, either inside or outside the square, that is a rational distance from all four corners? 
Or, put another way, given a square $ABCD$ of any size, can you find a point $P$ in the same plane such that the distances $AB$, $PA$, $PB$, $PC$, and $PD$ are all integers?
 
References:


*

*Guy, Richard K. Unsolved Problems in Number Theory, Vol. 1, Springer-Verlag,        2nd ed. 1991, 181-185.

*Barbara, Roy. "The rational distance problem", Mathematical Gazette 95, March              2011, 59-61.
A: Does there exist a completely separable MAD family in ZFC? That is, an infinite family $\mathcal S$ of infinite subsets of $\mathbb N$ such that (1) the intersection of any two members of $\mathcal S$ is finite, and (2) every subset of $\mathbb N$ either contains a member of $\mathcal S$ or else is covered by finitely many members of $\mathcal S$.
It's an easy transfinite construction assuming the continuum hypothesis; that's why only a construction in ZFC counts.
A: The Hot spot conjecture The conjecture seems quite amazing and simple to formulate,
it  can be even understood by persons "from the street"  seems its prediction can be tested experimentally. It is a subject of "polymath project 7".
Let me quote:

The hotspots conjecture can be expressed in simple English as:
Suppose a flat piece of metal, represented by a two-dimensional bounded connected domain, is given an initial heat distribution which then flows throughout the metal. Assuming the metal is insulated (i.e. no heat escapes from the piece of metal), then given enough time, the hottest point on the metal will lie on its boundary.
In mathematical terms, we consider a two-dimensional bounded connected domain D and let u(x,t) (the heat at point x at time t) satisfy the heat equation with Neumann boundary conditions. We then conjecture that
For sufficiently large t > 0, u(x,t) achieves its maximum on the boundary of D
This conjecture has been proven for some domains and proven to be false for others. In particular it has been proven to be true for obtuse and right triangles, but the case of an acute triangle remains open. The proposal is that we prove the Hot Spots conjecture for acute triangles!
  Note: strictly speaking, the conjecture is only believed to hold for generic solutions to the heat equation. As such, the conjecture is then equivalent to the assertion that the generic eigenvectors of the second eigenvalue of the Laplacian attain their maximum on the boundary.
  A stronger version of the conjecture asserts that
For all non-equilateral acute triangles, the second Neumann eigenvalue is simple; 
  and
  The second Neumann eigenfunction attains its extrema only at the boundary of the triangle.
(In fact, it appears numerically that for acute triangles, the second eigenfunction only attains its maximum on the vertices of the longest side.)

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May be this problem can be mentioned when teaching determinants and  in particular: 
$\det(AB)= \det(A)\det(B).$
There  are so-called Capelli identities which generalize this formula for specific matrices with non-commutative entries.
In the paper Noncommutative determinants, Cauchy-Binet formulae, and Capelli-type identities by Sergio Caracciolo, Andrea Sportiello, Alan D. Sokal they formulate 
certain conjectures of the type $$\det(A)\det(B)=\det(AB+\text{correction})$$
on the page 36 (bottom),  conjectures 5.1, 5.2.
I think these are quite non-trivial, but probably some smart young mathematician may solve them,
given some amount of time (some months may be).
I spent some amount of time thinking on them without success, and moreover
let me mention that  D. Zeileberger and D. Foata also 
failed to find a combinatorial proof of the Capelli identity of very similar type -- 
the one proved by Kostant-Sahi and Howe-Umeda -- 
see their comments in Combinatorial Proofs of Capelli's and Turnbull's Identities from Classical Invariant Theory page 9 bottom: "Although we are unable to prove the above identity combinatorially ... ".
So words above are some idications of non-triviality of the conjectures.
Personally I am quite interested in a proof, probably it can give clue for further generalizations.
A: These two are probably my favourite.
We all know that cardinals are ordered by injections, i.e. $|A|\leq|B|$ if there is an injection from $A$ into $B$. But it's clear that we can also order the cardinals by surjections, $|A|\leq^\ast|B|$ if $A$ is empty, or $B$ can be mapped onto $A$.
Assuming the axiom of choice these two notions are obviously equivalent. But what happens without the axiom of choice?
Well, it is easy enough to give counterexamples that $|A|\leq^\ast|B|$ and $|A|\nleq|B|$ (e.g. Dedekind-finite sets that can be mapped onto $\omega$; $\Bbb R$ can be mapped onto $\aleph_1$ in Solovay's model).

The Partition Principle. If $|A|\leq^\ast|B|$ then $|A|\leq|B|$.

For over a century now it is open whether or not this principle implies the axiom of choice in ZF. Russell claimed to have a proof, but it was never published.

The second open problem is also related to cardinals and their order:
Assuming the axiom of choice all the cardinals are ordinals, and since there are no decreasing sequences of ordinals, the $\leq$ relation is well-founded.
But what happens without the axiom of choice? Well we know that every partial order can be realized as cardinals of some model of ZF (in fact we can assume dependent choice to hold as high as we want to). An immediate consequence is that it is consistent that the cardinals are not well-founded. Even the existence of a Dedekind-finite set implies this easily, because removing one element decreases the cardinality and so we can find a decreasing sequence of cardinals (but we can't find a decreasing chain of subsets!).
The question whether or not the assumption that there is no decreasing sequence of cardinals implies the axiom of choice in ZF is open, for about a century as well.
Unlike the previous problem where pretty much everyone feels that the answer is positive, this problem had people arguing for both sides. Some people suggested that it will imply choice, others suggested that it will not imply the axiom of choice. But no one has an answer yet.
A: Showing that if product of $ n $ Toeplitz operators is again a Toeplitz Operator.this is open and quite elementary to state. http://www.mathnet.or.kr/mathnet/thesis_file/WYLee.pdf
