Which vector $v\in \mathbb {R}^2 \otimes \mathbb {R}^2$ can be written as $p\otimes q$ for some $p,q\in\mathbb{R}^2$? Let $(e_1,e_2)$ be the canonical basis of $\mathbb {R}^2$. Look at the tensor product $\mathbb {R}^2 \otimes \mathbb {R}^2$ of
a. $v = e_1 \otimes e_1 + e_1 \otimes e_2\,.$
b. $v = e_1 \otimes e_1 + e_2 \otimes e_2\,.$
Do the vectors $p,q \in \mathbb {R}^2$ with $v = p \otimes q$ exist?

You can rewrite a with $e_1 \otimes (e_1 + e_2)$, you can't do that with b. I think that a is true and that b is false, is that correct?
 A: Here is a general result.  This might be useful for you if you have to deal with a large tensor element to check.
Suppose that $U$ and $V$ are finite-dimensional vector spaces over a field $\mathbb{K}$.  Write $m:=\dim_\mathbb{K}(U)$ and $n:=\dim_\mathbb{K}(V)$.  Let $t$ be an element of $U\otimes V$.  For an integer $k$ such that $0\leq k\leq\min\{m,n\}$, we say that $t$ is $k$-fold decomposable if $k$ is the smallest nonnegative integer such that $$t=u_1\otimes v_1+u_2\otimes v_2+\ldots +u_k\otimes v_k$$ for some $u_1,u_2,\ldots,u_k\in U$ and $v_1,v_2,\ldots,v_k\in V$. Otherwise, we say that $t$ is $k$-fold indecomposable.
Let $f\in U^*$ be an arbitrary linear functional.  We define $\phi_f:U\otimes V \to V$ to be the extension of $f$ by setting $\phi_f(u\otimes v):=f(u)\,v$ for all $u\in U$ and $v\in V$, and extend $\phi_f$ linearly.

Proposition.  Let $k$ be an integer such that $0\leq k\leq \min\{m,n\}$.  An element $t\in U\otimes V$  is $k$-fold decomposable if and only if the set $$X(t):=\big\{f\in U^*\,|\,\phi_f(t)=0\big\}$$
is a subspace of $U^*$ of codimension $k$.

First, suppose that $X(t)$ is a codimension-$k$ subspace of $U^*$ with $0\leq k\leq \min\{m,n\}$.  Pick $k$ elements $f_1,f_2,\ldots,f_k\in U^*$ such that the images of $f_1,f_2,\ldots,f_k$ under the canonical projection $U^*\to \big(U^*/X(t)\big)$ form a basis of $U^*/X(t)$.  Pick a basis $\{f_{k+1},f_{k+2},\ldots,f_{m}\}$ of $X(t)$.  Therefore, $\{f_1,f_2,\ldots,f_m\}$ is a basis of $U^*$.  Write $\{u_1,u_2,\ldots,u_m\}$ for the basis of $U$ dual to $\{f_1,f_2,\ldots,f_m\}$ (that is, $f_i(u_j)=\delta_{i,j}$ for $i,j=1,2,\ldots,m$, where $\delta$ is the Kronecker delta).  Now, let $$v_i:=\phi_{f_i}(t)\text{ for }i=1,2,\ldots,k\,.$$
We claim that $t=u_1\otimes v_1+u_2\otimes v_2+\ldots+u_k\otimes v_k$.
Observe that $v_1,v_2,\ldots,v_k$ are linearly independent elements of $V$.  To show this, suppose that
$$\sum_{j=1}^k\,\mu_j\,v_j=0$$
for some $\mu_1,\mu_2,\ldots,\mu_k\in\mathbb{K}$.  If $\mu_r\neq 0$ for some $r=1,2,\ldots,k$, we assume without loss of generality that $r=k$, and $\mu_k=-1$, so
$$v_k=\sum_{j=1}^{k-1}\,\mu_j\,v_j\,.$$
Consider the subspace $U'$ of $U$ spanned by $\{u_1+\mu_1 u_k,u_2+\mu_2u_k,\ldots,u_{k-1}+\mu_{k-1}u_{k}\}$.  This is a subspace of dimension $k-1$ of $U$, and therefore, the subspace $X'$ of $U^*$ consisting of all linear functional $f\in U^*$ such that $f(u)=0$ for all $u\in U'$ has codimension $k-1$ in $U^*$ with
$$\{f_k-\mu_1\,f_1-\mu_2\,f_2-\ldots-\mu_{k-1}\,f_{k-1},f_{k+1},f_{k+2},\ldots,f_m\}$$
as a basis.  However, it can be readily verified that $X'\subseteq X(t)$, which contradicts the assumption that $X(t)$ has codimension $k$ in $U^*$.
Consider a basis $\{v_1,v_2,\ldots,v_n\}$ of $V$ extending the linearly independt subset $\{v_1,v_2,\ldots,v_k\}$ of $V$.  Write $t=\sum\limits_{i=1}^m\,\sum\limits_{j=1}^n\,\lambda_{i,j}\,u_i\otimes v_j$ with $\lambda_{i,j}\in\mathbb{K}$ for all $i=1,2,\ldots,m$ and $j=1,2,\ldots,n$.  We have
$$0=\phi_{f_s}(t)=\sum_{j=1}^n\,\lambda_{s,j}\,v_j\,,$$
where $s=k+1,k+2,\ldots,m$.  Therefore, $\lambda_{s,j}=0$ for all $s=k+1,k+2,\ldots,m$ and $j=1,2,\ldots,n$.  This implies
$$t=\sum_{i=1}^k\,\sum_{j=1}^n\,\lambda_{i,j}\,u_i\otimes v_j\,.$$
That is, for $i=1,2,\ldots,k$, we have
$$v_i=\phi_{f_i}(t)=\sum_{j=1}^n\,\lambda_{i,j}\,v_j\,.$$
Ergo, for $i=1,2,\ldots,k$ and $j=1,2,\ldots,n$, we have $\lambda_{i,j}=0$ when $j\neq i$, and $\lambda_{i,i}=1$.  That is, $t=\sum\limits_{i=1}^k\,u_i\otimes v_i$, as asserted.
Now, suppose that $t$ is $k$-fold decomposable.  Then, write $t=\sum\limits_{i=1}^k\,u_i\otimes v_i$ for some $u_1,u_2,\ldots,u_k\in U$ and $v_1,v_2,\ldots,v_k\in V$.  Note that $u_1,u_2,\ldots,u_k$ are linearly independent elements of $U$, and $v_1,v_2,\ldots,v_k$ are linearly independent elements of $V$.  Therefore, $X(t)$ contains all $f\in U^*$ such that $f(u_i)=0$ for $i=1,2,\ldots,k$, and this is a subspace of $U^*$ of codimension $k$.
Remark.  Even if $U$ or $V$ is not finite-dimensional, the proposition can still be used.  Note that each tensor element $t$ can be written as a finite sum of elements of the form $u\otimes v$, where $u\in U$ and $v\in V$.  Take $\tilde{U}$ to be the subspace of $U$ spanned by such elements $u$, and $\tilde{V}$ the subspace of $V$ spanned by such elements $v$.  Then, $\tilde{U}$ and $\tilde{V}$ are finite-dimensional subspaces of $U$ and $V$.  You can then work with $\tilde{U}$ and $\tilde{V}$ instead.

For your specific question, let $\{e_1^*,e_2^*\}$ be the basis of $(\mathbb{R}^2)^*$ dual to $\{e_1,e_2\}$.  For Part a, observe that $X(v)$ is the codimension-$1$ subspace of $(\mathbb{R}^2)^*$ spanned by $e_2^*$.  To show this, let $f\in (\mathbb{R}^2)^*$.  Then, $f=\lambda_1e_1^*+\lambda_2 e_2^*$ for some $\lambda_1,\lambda_2\in\mathbb{R}$.  Now,
$$\phi_f(v)=\lambda_1e_1+\lambda_1e_2\,,$$
which is $0$ if and only if $\lambda_1=0$.  That is, $f\in X(v)$ if and only if $f=\lambda_2e_2^*$ for some $\lambda_2\in\mathbb{R}$.  Hence, $v$ is $1$-fold decomposable.
For Part b, we shall prove that $X(v)=0$, which has codimension $2$ in $(\mathbb{R}^2)^*$.    To show this, let $f\in (\mathbb{R}^2)^*$.  Then, $f=\lambda_1e_1^*+\lambda_2 e_2^*$ for some $\lambda_1,\lambda_2\in\mathbb{R}$.  Now,
$$\phi_f(v)=\lambda_1e_1+\lambda_2e_2\,.$$
Therefore, $f\in X(v)$ if and only if $\lambda_1=\lambda_2=0$.  Ergo, $f\in X(v)$ if and only if $f=0$.   Consequently, $v$ is $2$-fold decomposable (whence, $1$-fold indecomposable).
A: Yes, you are correct, since $\{e_i \otimes e_j  : 1\leq i,j \leq2 \}$ is a $\mathbb{R}$-basis for $\mathbb R^2 \otimes \mathbb R^2 $.
Put $p=ae_1+be_2$ and $q=ce_1+de_2$. $(a,b,c,d \in \mathbb R)$ Then $$p \otimes q =(ac)e_1\otimes e_1 +(ad)e_1\otimes e_2 + (bc)e_2\otimes e_1 + (bd)e_2 \otimes e_2$$
So $p \otimes q = e_1 \otimes e_1 + e_2 \otimes e_2$ yields $ac=1$, $ad=bc=0$, $bd=1$.
This is impossible since $(abcd)=(ac)(bd)=(ad)(bc)$.
A: I like Batominovski's result, but I think that it can be stated and proved more concisely, at least in the case where one of the spaces is finite dimensional. With that goal, define the map $\phi: U^* \otimes (U \otimes V)$ by
$$
\phi(f \otimes x \otimes y) = f(x)\,y.
$$
We have the following:

Claim:  Suppose that $U$ is finite dimensional. An element $t \in U \otimes V$ is $k$-fold decomposable if and only if the map $\phi_t:U^* \to V$ defined by $\phi_t(f) = \phi(f \otimes t)$ has rank $k$.

Proof: We note that if $\phi_t$ has rank $k$, then it admits a rank factorization $\phi_t = \psi_1 \circ \psi_2$ with $\psi_2:U^* \to K^k$ and $\psi_1: K^k \to V$.  Because $U^{**}\cong U$, there exist elements $u_1,\dots,u_k$ for which
$$
\psi_1(f) = (f(u_1),\dots,f(u_k)).
$$
If we let $v_j = \psi_2(e_j)$, where $e_1,\dots,e_k$ are the canonical basis elements for $K^k$, we have
$$
\psi_1 \circ \psi_2 = f(u_1) v_1 + \cdots + f(u_k) v_k.
$$
Because $\phi(f,t) = \phi(f,u_1 \otimes v_1 + \cdots + u_k \otimes v_k)$ holds for all $f \in U^*$, we can conclude that
$$
t = u_1 \otimes v_1 + \cdots + u_k \otimes v_k.
$$
(This last step requires some justification when $V$ is infinite dimensional). It follows that $t$ is $k$-fold decomposable.
Conversely, suppose that $t = u_1 \otimes v_1 + \cdots + u_k \otimes v_k$. We could then define the maps $\psi_2:U^* \to K^k$ and $\psi_1: K^k \to V$ as above, and we find that $\phi_t = \psi_1 \circ \psi_2$, so that $\phi_t$ is indeed a map with rank $k$.
