Is it true that primitive idempotents of $kS_n \otimes_k kS_m$ are just tensor of idempotents of $kS_n$ and $kS_m$? I was working over a general field $k$ of characteristic $0$. For now, just think $k$ as the field of complex number.
Let $n$ and $m$ be some natural numbers. Let $kS_n$ be the group algebra of $S_n$, the symmetric group over $n$ objects. Let $A := kS_n \otimes_k kS_m$ be the tensor product of $k$-space equipped with a pointwise multiplication, i.e.
$$(a \otimes b)( c \otimes d) = ac \otimes bd.$$
Now, $A$ is a $k$-algebra.
My question is: is it true that any primitive idempotent of $kS_n \otimes_k kS_m$ is of the form $e_1 \otimes e_2$ where  $e_1$ and $e_2$ are primitive idempotents of $kS_n$ and $kS_m$ respectively?
Some work I have done: for the case when $n = 2 $ and $m=2$, my computation shows that it is true. Let $S_2 = \{1, u\}$ and
$$v = a(1 \otimes 1) + b(1 \otimes u) + c(u \otimes 1) +d(u \otimes u)$$
where $a,b,c,d \in k$. You get all the four primitive idempotents of $A$ by setting $(a,b,c,d)$ to $(x,x,x,x),(x,x,-x,-x),(x,-x,x,-x)$ and $(x,-x,-x,x)$ where $x = \frac{1}{4}$. I proved the completeness by computing out all the idempotents in $A$ which is a long process involving solving a system of equations. Note that all the four idempotents are precisely the tensor product of primitive idempotents of $kS_2$, which are either $\frac{1+u}{2}$ or $\frac{1-u}{2}$. So it is true for this case. I did not try larger case because it turns extreamly hard to determing all idempotents of $A$ as $n$ and $m$ grows larger.
 A: This is false for all $n, m \ge 3$.
As a general fact, the irreducible representations of $S_n$ over any field of characteristic $0$ are in fact already defined over $\mathbb{Q}$. This implies that $k[S_n]$ is a finite direct product of matrix algebras $M_{d_i}(k)$ over $k$ (rather than over division rings over $k$). Since $M_{d_1}(k) \otimes M_{d_2}(k) \cong M_{d_1 d_2}(k)$ the same is true for $k[S_n] \otimes k[S_m]$.
Hence an idempotent in $k[S_n] \otimes k[S_m]$ is exactly a choice of idempotent in each simple factor $M_{d_1 d_2}(k)$, and in particular is primitive iff it is a primitive idempotent in exactly one such simple factor. So the problem reduces to the corresponding problem for matrix algebras, namely:

Is a primitive idempotent in $M_{d_1 d_2}(k)$ the tensor product of a primitive idempotent in $M_{d_1}(k)$ and a primitive idempotent in $M_{d_2}(k)$?

The answer to this question is no. An idempotent $e \in M_d(k)$ is primitive iff it has rank $1$, meaning that its image $\text{im}(e) \subseteq k^d$ is a line (a $1$-dimensional subspace). An idempotent $e \in M_{d_1 d_2}(k)$ which is the tensor product of an idempotent $e_1 \in M_{d_1}(k)$ and an idempotent $e_2 \in M_{d_2}(k)$ has the property that its image $\text{im}(e)$ is a tensor product
$$\text{im}(e_1) \otimes \text{im}(e_2) \subseteq k^{d_1} \otimes k^{d_2} \cong k^{d_1 d_2}$$
of the images $\text{im}(e_1)$ and $\text{im}(e_2)$. If $d_1, d_2 \ge 2$ then there are lines in $k^{d_1 d_2}$ which are not tensor products of lines in $k^{d_1}$ and $k^{d_2}$ (in fact "almost all" lines, or the "generic" line, have this property), so a primitive idempotent whose image is such a line cannot be a tensor product. We can be quite explicit about this for $d_1 = d_2 = 2$: a tensor product line in $k^4$ is spanned by a vector of the form
$$(x_1, y_1) \otimes (x_2, y_2) = (x_1 x_2, y_1 x_2, x_1 y_2, y_1 y_2)$$
and the four components $(a, b, c, d)$ of such a vector must satisfy $ad - bc = 0$. So any vector $(a, b, c, d) \in k^4$ such that $ad - bc \neq 0$ spans a line which is not a tensor product. There's a more general version of this criterion involving vanishing of the $2 \times 2$ minors of an element of $k^{d_1 d_2}$ regarded as a matrix in $M_{d_1 \times d_2}(k)$.
Since $S_n$ is nonabelian for $n \ge 3$, $k[S_n]$ contains a simple factor $M_d(k)$ where $d \ge 2$ and so this argument applies.
However, the answer is yes if instead of primitive idempotents we consider centrally primitive idempotents, since $M_{d_1 d_2}(k)$ has a unique centrally primitive idempotent, namely the identity, which is the tensor product of the identities of $M_{d_1}(k)$ and $M_{d_2}(k)$. This corresponds to the statement that every irreducible representation of $S_n \times S_m$ is the tensor product of an irreducible representation of $S_n$ and an irreducible representation of $S_m$.
