This is the most complete treatment I could come up with. Let $k \subseteq K^{\operatorname{sep}} \subseteq K^{\operatorname{alg}} \subseteq K$ and $k \subseteq L^{\operatorname{sep}} \subseteq L^{\operatorname{alg}} \subseteq L$ be the separable algebraic and algebraic closures of $k$ in $K$ and $L$. The result is the following.
Theorem. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then
- $K \otimes_k L$ is irreducible if and only if $k \subseteq K^{\operatorname{sep}}$ and $k \subseteq L^{\operatorname{sep}}$ are linearly disjoint with respect to one (equivalently, every) choice of embedding into $\bar k$.
- $K \otimes_k L$ is reduced if and only if the intersection of $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $K \otimes_k \ker(\Omega_k \otimes_k L \to \Omega_L)$ inside $\Omega_k \otimes_k (K \otimes_k L)$ is generically trivial.
- $K \otimes_k L$ has dimension $0$ if and only if $k \subseteq K$ or $k \subseteq L$ is algebraic.
In particular, $K \otimes_k L$ is a field if and only if all three criteria hold.
Here we say a property holds generically if it holds for the localisation at all minimal primes. The second criterion is not very nice in general, but if $K \otimes_k L$ has dimension $0$ we can remove the word 'generically', so we get a rather clean criterion. This applies in particular in the case the OP is interested in, for $K \otimes_k L$ to be a field. On the other hand, in Example 3 we show that 'generically' in part 2 of the theorem cannot be removed, even in the case $K = L$.
In Proposition 3 below we give a more conceptual (but more technical) criterion equivalent to the one stated here.
Statements 1, 2, and 3 of the theorem generalise without difficulty to finite tensor products of fields. The statement is the 'obvious generalisation' of the version for binary tensor products. For example, in 1 we need that $K_1^{\operatorname{sep}} \otimes_k \ldots \otimes_k K_n^{\operatorname{sep}} \to \bar k$ is injective for one (equivalently, any) choice of embeddings into $\bar k$. In 2, we need that generically the sum of the subspaces is direct. In 3, we need that at most one of the extensions has a transcendental part.
Remark. We will use the algebro-geometric properties of field extensions: a field extension $k \subseteq \ell$ is
- algebraic if and only if it is integral (obvious);
- separable if and only if it has geometrically reduced fibres (Tag 030W);
- separable algebraic if and only if it is integral with geometrically reduced fibres (combine 1 and 2);
- purely inseparable (algebraic) if and only if it is radicial (by definition, see Tag 01S3);
- primary if and only if it has geometrically irreducible fibres (partially explained in Tag 037Q);
- regular if and only if it has geometrically integral fibres (combine 2 and 5).
This is useful because not all rings we encounter are fields, and the algebro-geometric properties are preserved by base change. In fact all these properties descend under faithfully flat morphisms (see [EGA IV$_2$, Prop 2.6.1(v)] for radicial, [EGA IV$_2$, Prop 2.7.1(xvii)] for integral, and obvious for all other properties), so $k \to K$ has one of the properties above if and only if $L \to K \otimes L$ does.
Below we will treat parts 1, 2, and 3 of the theorem separately; see Lemma 1, Proposition 3, and Lemma 4. Also have a look at examples 1, 2, 3, and 4, because they provide some insight into the types of behaviours you can expect.
1. Irreducibility
The precise version of part 1 of the theorem is as follows.
Lemma 1. Let $k \subseteq K$ and $k \subseteq L$ be field extensions, and let $\bar k$ be a separable algebraic closure of $k$. Then the following are equivalent:
- $K \otimes_k L$ is irreducible;
- $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is irreducible;
- $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is a field;
- for every pair of embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$, the subfields $i(K)$ and $j(L)$ of $\bar k$ are linearly disjoint;
- for one pair of embeddings $i \colon K^{\operatorname{sep}} \to \bar k$ and $j \colon L^{\operatorname{sep}} \to \bar k$, the subfields $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ of $\bar k$ are linearly disjoint.
Proof. Note that $K^{\operatorname{sep}} \to K$ and $L^{\operatorname{sep}} \to L$ are primary, hence flat with geometrically irreducible fibres. Then the same goes for $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}} \to K \otimes_k L$, so (1) $\Leftrightarrow$ (2) follows from [EGA IV$_2$, Prop. 4.5.7] (or modify the proof of Tag 038F). Since $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is reduced of dimension $0$, it is a field if and only if it is irreducible, proving (2) $\Leftrightarrow$ (3).
If $i(K^{\operatorname{sep}})$ and $j(L^{\operatorname{sep}})$ are linearly disjoint for one choice of $i$ and $j$, then
$$i \otimes j \colon K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}} \to \bar k$$
is injective, so $K^{\operatorname{sep}} \otimes_k L^{\operatorname{sep}}$ is a field. The converse is also clear, showing (3) $\Leftrightarrow$ (4) $\Leftrightarrow$ (5). $\square$
We will generalise properties 4 and 5 in Further remarks and examples to arbitrary field extensions. In the separable algebraic case, it has a reinterpretation in terms of Galois theory: if $i \colon K \to \bar k$ and $j \colon L \to \bar k$ is a choice of embeddings, then we get closed subgroups $H_K$ and $H_L$ of $G_k = \operatorname{Gal}(\bar k/k)$. The condition on linear disjointness then means that for any open subgroups $U_K \supseteq H_K$ and $U_L \supseteq H_L$, we have $[G_k : U_K \cap U_L] = [G_k : U_K] \cdot [G_k : U_L]$.
Remark. One is tempted to believe that $K$ and $L$ are linearly disjoint if and only if $i(K) \cap j(L) = k$ for every pair of embeddings $i \colon K \to \bar k$ and $j \colon L \to \bar k$, but this is not true:
Example 1. Let $k \subseteq M$ be an $S_5$-extension; for example $k = \mathbf C(\sigma_1,\sigma_2,\sigma_3,\sigma_4.\sigma_5)$ and $M = \mathbf C(x_1, x_2, x_3, x_4,x_5)$, where
$$\prod_{i=1}^5(X-x_i) = X^5 - \sigma_1 X^4 + \sigma_2 X^3 - \sigma_3 X^2 + \sigma_4 X - \sigma_5.$$
Let $K$ be the fixed field of $(12)$ and $L$ the fixed field of $(12345)$.
Then $K \cap L$ is fixed by $(12)$ and $(12345)$, hence equals $k$. What's more, any conjugate of $K$ is the fixed field of a transposition, and any conjugate of $L$ is the fixed field of a $5$-cycle, hence their intersection is still $k$ since any pair of a transposition and a $5$-cycle generate $S_5$. However, $K$ and $L$ cannot be linearly disjoint, because $[K:k] = 60$ and $[L:k] = 24$, but they're both contained in a degree $120$ extension of $k$.
2. Reducedness
We will use repeatedly that if $k \subseteq K$ and $k \subseteq L$ are separable, then they are geometrically reduced, hence $K \otimes_k L$ is reduced by Tag 034N. In particular, if $\operatorname{char} k = 0$ there is nothing to discuss, so assume $\operatorname{char} k = p > 0$.
For any field $k$ of characteristic $p$, write $\Omega_k = \Omega_{k/\mathbf Z} = \Omega_{k/\mathbf F_p}$ for the module of absolute differentials. Although my criterion was originally based on the idea of $p$-bases (Tag 07P0), I was ultimately unable to prove it in that language. The proof uses the (naive) cotangent complex $\mathbf L_{B/A}$ (see Tags 00S0 and 08R6) instead, where again we write $\mathbf L_A$ for $\mathbf L_{A/\mathbf F_p}$.
Lemma 2. Let $k \subseteq \ell$ be a field extension in characteristic $p > 0$, and let $A$ be a successive extension
$$\ell = A_0 \subseteq A_1 \subseteq \ldots \subseteq A_r = A$$
where $A_i = A_{i-1}[X_i]/(X_i^p - x_i)$ for some $x_i \in A_{i-1}$. Then the following are equivalent:
- $A$ is a field;
- $H_1\mathbf L_A = 0$;
- the boundary map $d \colon H_1\mathbf L_{A/k} \to \Omega_k \otimes_k A$ is injective.
- the boundary map $d \colon H_1\mathbf L_{A/\ell} \to \Omega_\ell \otimes_\ell A$ is injective.
Proof. For any field $F$ of characteristic $p$, the extension $\mathbf F_p \subseteq F$ is separable as $\mathbf F_p$ is perfect. Therefore, we have $H_1\mathbf L_F = 0$ (see for example Tag 07E5, but it also follows from Tag 08Q1, localisation, and the étale case of Tag 0D0M). Thus $H_1\mathbf L_A = 0$ if $A$ is a field, showing (1) $\Rightarrow$ (2).
If $A$ is not a field, then let $i$ be the largest index such that $A_i$ is a field, so by the above $H_1 \mathbf L_{A_i} = 0$. Then $A_{i+1}$ is not a field, meaning that $x_{i+1}$ is a $p$-th power in $A_i$. Then the long exact cotangent sequence (Tag 00S2) for $\mathbf F_p \subseteq A_i \subseteq A$ reads
$$0 \to H_1 \mathbf L_A \to H_1 \mathbf L_{A/A_i} \stackrel d\to \Omega_{A_i} \underset {A_i}\otimes A \to \ldots.$$
Hence $H_1 \mathbf L_A \neq 0$ since $X_{i+1}^p - x_{i+1}$ maps to $-dx_{i+1} = 0$ under $d$, showing (2) $\Rightarrow$ (1).
Finally, the long exact cotangent sequence for $\mathbf F_p \subseteq k \subseteq A$ reads
$$0 \to H_1\mathbf L_A \to H_1\mathbf L_{A/k} \stackrel d\to \Omega_k \underset k\otimes A \to \Omega_A \to \Omega_{A/k} \to 0.$$
This shows (2) $\Leftrightarrow$ (3), and replacing $k$ by $\ell$ gives (2) $\Leftrightarrow$ (4). $\square$
Proposition 3. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then the following are equivalent:
- $K \otimes_k L$ is reduced;
- $H_1 \mathbf L_{K \otimes_k L}$ is generically trivial;
- the intersection of $\ker(\Omega_k \otimes_k K \to \Omega_K) \otimes_k L$ and $K \otimes_k \ker(\Omega_k \otimes_k L \to \Omega_L)$ (in $\Omega_k \otimes_k (K \otimes_k L)$) is generically trivial.
Proof. We can reduce to the case that $k \subseteq K$ and $k \subseteq L$ are finitely generated, since tensor product, formation of $\Omega$ and $\mathbf L$, and computation of kernel commute with filtered colimits, and nilpotents are defined over a finitely generated subring.
By 'baby Serre's criterion' (Tag 031R), we need to check that $K \otimes_k L$ is $S_1$ and $R_0$. But $S_1$ is automatic, for example by Tag 0339, so it suffices to check $R_0$. Thus, we see that $K \otimes_k L$ is reduced if and only if every localisation $A$ of $K \otimes_k L$ at a minimal prime $\mathfrak p$ is a field.
We claim that every such $A$ is of the form described in Lemma 2. Indeed, let $k \subseteq K' \subseteq K$ be a subextension such that $k \subseteq K'$ is separable and $K' \subseteq K$ is purely inseparable; for example by taking $K'$ to be the elements separable over a given transcendence basis. Then $K' \otimes_k L$ is reduced and $K' \otimes_k L \to K \otimes_k L$ is radicial and faithfully flat (in particular a bijection on irreducible components by Tag 01S4).
If $\mathfrak q$ is the preimage of $\mathfrak p$ in $K' \otimes_k L$ and $\ell$ is the localisation of $K' \otimes_k L$ at $\mathfrak q$, then $\ell$ is a field since $\mathfrak q$ is minimal and $K' \otimes_k L$ reduced. The map $\ell \to A$ is a map as described in Lemma 2, since $K' \subseteq K$ is of that form. Thus from Lemma 2 we see that $A$ is a field if and only if $H_1 \mathbf L_A = 0$. Since formation of $\mathbf L_A$ commutes with localisation by Tag 00S7, we see that $K \otimes_k L$ is reduced if and only if $H_1\mathbf L_{K \otimes_k L}$ is generically trivial, proving (1) $\Leftrightarrow$ (2).
For (2) $\Leftrightarrow$ (3), the exact sequences
\begin{alignat*}{1}
0 & \to H_1\mathbf L_{K/k} & \stackrel d\to \Omega_k \underset k\otimes K & \to \Omega_K & \to \Omega_{K/k} & \to 0,\\
0 & \to H_1\mathbf L_{L/k} & \stackrel d\to \Omega_k \underset k\otimes L & \to \Omega_L & \to \Omega_{L/k} & \to 0.
\end{alignat*}
for $\mathbf F_p \subseteq k \subseteq K$ and $\mathbf F_p \subseteq k \subseteq L$ identify $H_1\mathbf L_{K/k}$ and $H_1\mathbf L_{L/k}$ as the kernels of the maps on differentials. Choosing a presentation of $K \otimes_k L$ by tensoring presentations for $K$ and $L$ over $k$, we see that
$$\mathbf L_{K \otimes_k L/k} \simeq \left( \mathbf L_{K/k} \underset k\otimes L \right) \oplus \left( \mathbf L_{L/k} \underset k\otimes K \right),$$
and the connecting map $H_1\mathbf L_{K \otimes_k L} \to \Omega_k \otimes_k (K \otimes_k L)$ is given by $(d,d)$ with respect to this decomposition. In particular, the connecting map for $K \otimes_k L$ is generically injective if and only if the images of the connecting maps for $K$ and $L$ have generically trivial intersection. $\square$
Remark. However, it is not true that $K \otimes_k L$ is a field if and only if there are no elements that become a $p$-th power in both; see Example 2 below (even in the finite purely inseparable case). In fact, $K \otimes_k K$ may be reduced even if $k \subseteq K$ is inseparable; see Example 3 below (a transcendental primary inseparable extension).
On the other hand, this last phenomenon does not happen for algebraic inseparable extensions $k \subseteq K$, since $k[x]/f \otimes_k k[x]/f \subseteq K \otimes_k K$ clearly picks up nilpotents if $f$ is inseparable. See also Example 4 at Further remarks and examples for an example of the interplay between the separable algebraic part and the purely inseparable part.
Example 2. Let $k = \mathbf F_p(x,y,z)$, let $K = k[u,v]/(u^{p^2}-z, v^p-xu^p-y)$ be the field adjoining a $p^2$-th root $u$ of $z$ and a $p^2$-th root $v$ of $x^pz+y^p$ (this example is due to Sweedler, as far as I can tell), and let $L = k(x^{1/p}, y^{1/p})$. Then
$$K \underset k\otimes L = \frac{\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})[v]}{(v^p-xz^{1/p}-y)} \cong \frac{\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})[X]}{(X^p)},$$
since $xz^{1/p}+y$ is a $p$-th power in $\mathbf F_p(x^{1/p}, y^{1/p}, z^{1/p^2})$. Thus, $K \otimes_k L$ is not reduced.
The elements in $k$ that become $p$-th powers in $K$ are given by
$$k \cap K^p = \mathbf F_p(x^p,y^p,z).\label{Eq become p-th power}\tag{1}$$
Indeed, the inclusion $\supseteq$ is obvious. For the converse, consider the chain of field extensions
$$K^{p^2} = k \cap K^{p^2} \subseteq k \cap K^p \subseteq k \cap K = k.$$
We have $[k:K^{p^2}] = [k:k^{p^2}]/[K^{p^2}:k^{p^2}] = (\operatorname{tr. deg.} k)^2/[K:k] = p^3$, and $x^p$ is in $k \cap K^p$ but not $k \cap K^{p^2}$ since $dx \neq 0 \in \Omega_K$ (see computation below). Thus, $[k:k \cap K^p] \leq p^2$, which gives equality in (\ref{Eq become p-th power}) by a dimension count.
By a similar (but easier) computation, we get
$$k \cap L^p = \mathbf F_p(x,y,z^p).$$
Thus, the only elements in $k$ that become $p$-th powers in both $K$ and $L$ are already $p$-th powers in $k$ to begin with.
What's going on is that $\ker(\Omega_k \otimes_k K \to \Omega_K)$ is not defined over $k$, and a fortiori cannot be spanned by elements of the form $dx_i$ for $x_i \in k$. Indeed, $\Omega_k = k \cdot dx \oplus k \cdot dy \oplus k \cdot dz$, and one computes
$$\Omega_K = \frac{K \cdot dx \oplus K \cdot dy \oplus K \cdot du \oplus K \cdot dv}{u^p\ dx + dy}.$$
The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is generated by $dz$ and $u^p\ dx + dy$, which is not defined over $k$.
Example 3. Let $k = \mathbf F_p(s,t)$, and let $K$ be the fraction field of $k[x,y]/(sx^p + ty^p - 1)$ (this example is due to Mac Lane, according to MO user nfdc23). Then $K$ is geometrically irreducible since
$$\bar k[x,y]/(sx^p + ty^p - 1) = \bar k[x,y]/(ux+vy-1)^p,$$
where $u^p = s$ and $v^p = t$, which is a thickening of $\bar k[x,y]/(ux+vy-1) = \bar k[x]$. In particular, $k \subseteq K$ is primary, i.e. $k$ is algebraically closed in $K$.
The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is generated by $df = x^p\ ds + y^p\ dt$, where $f = sx^p + ty^p - 1$. Again this is not defined over $k$, and in this case not even over an algebraic extension of $k$.
If we take two copies of $K$, then the elements
\begin{align*}
df_1 &= x_1^p\ ds + y_1^p\ dt\\
df_2 &= x_2^p\ ds + y_2^p\ dt
\end{align*}
agree at the locus $V((x_1y_2-y_1x_2)^p)$, which is a strict closed subscheme. Thus the intersection of the kernels is generically trivial (since $K$ is the function field of a regular but not geometrically regular curve over $k$, so the diagonal is not an irreducible component), so $K \otimes_k K$ is reduced.
(In a previous version of this answer, I erroneously stated a version of the criterion for reducedness that did not take this type of example into account.)
On the other hand, $K \otimes_k K \otimes_k K$ cannot be reduced, because we get $3$ relations $df_1$, $df_2$, and $df_3$ involving only $2$ variables $ds$ and $dt$.
3. Zero-dimensionality
Lemma 4. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is $0$-dimensional if and only if one of $k \subseteq K$ and $k \subseteq L$ is algebraic.
Proof. Immediate by Grothendieck–Sharp.
This can actually be seen directly as well: clearly the tensor product is $0$-dimensional if one of $k \subseteq K$ and $k \subseteq L$ is algebraic. Conversely, if $K$ and $L$ both contain an element transcendental over $k$, then this gives embeddings $k \subseteq k(x) \subseteq K$ and $k \subseteq k(x) \subseteq L$. View $K$ and $L$ as extensions of $k(x)$ and choose $k(x)$-linear embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into an algebraically closed extension $k(x) \subseteq \Omega$. Then $x \otimes 1 - 1 \otimes x$ is in the kernel $\mathfrak p$ of $i \otimes j \colon K \otimes_k L \to \Omega$.
The maps $k(x) \to K$ and $k(x) \to L$ are flat, hence the same goes for $k(x) \otimes_k k(x) \to K \otimes_k L$. The preimage of $\mathfrak p$ under this map contains $x \otimes 1 - 1 \otimes x$, hence is a nonzero ideal. But $k(x) \otimes_k k(x)$ is a domain, so going down for flat morphisms (see Tag 00HS) shows that $\mathfrak p$ cannot be minimal. Hence, $K \otimes_k L$ has positive dimension. $\square$
Further remarks and examples
Here is an interesting example of the interplay between the separable and purely inseparable parts, in light of statement 2 of the precise version of the theorem (note that already the separable part disqualifies this example from producing a field, but the point is only the reducedness statement now):
Example 4. Let $k = \mathbf F_p(\sigma_1,\sigma_2)$, and consider the extension $M = \mathbf F_p(x_1,x_2)$ given by $(X-x_1)(X-x_2) = X^2 - \sigma_1X + \sigma_2$. This is a $\mathbf Z/2$-Galois cover (even if $p = 2$, but feel free to assume $p > 2$). Let $K = \mathbf F_p(x_1^{1/p},x_2)$ and $L = \mathbf F_p(x_1,x_2^{1/p})$.
Then $K^{\operatorname{sep}} = L^{\operatorname{sep}} = M$, and $M \otimes_k M \cong M^{\mathbf Z/2}$ given by $m_1 \otimes m_2 \mapsto (m_1\sigma(m_2))_{\sigma \in \mathbf Z_2}$. That is, on the factor $M_0$ the inclusions are both given by $x_i \mapsto x_i$, and on the factor $M_1$ the inclusion from $L^{\operatorname{sep}}$ swaps $x_1$ and $x_2$.
Then $K \otimes_k L^{\operatorname{sep}} = M_0(x_1^{1/p}) \times M_1(x_1^{1/p})$, and $K^{\operatorname{sep}} \otimes_k L \cong M_0(x_2^{1/p}) \times M_1(x_1^{1/p})$ since the embedding $L^{\operatorname{sep}} \to M_1$ sends $x_2$ to $x_1$. We see that $K \otimes_k L$ is not reduced, because it contains $M_1(x_1^{1/p}) \otimes_{M_1} M_1(x_1^{1/p})$.
In terms of differentials, the relations $\sigma_1 = x_1+x_2$ and $\sigma_2 = x_1x_2$ give $d\sigma_1 = dx_1 + dx_2$ and $d\sigma_2 = x_1\ dx_2 + x_2\ dx_1$. The kernel of $\Omega_k \otimes_k K \to \Omega_K$ is spanned by
$$dx_1 = \frac{d\sigma_2 - x_1\ d\sigma_1}{x_2-x_1},$$
and the opposite happens for the kernel of $\Omega_k \otimes_k L \to \Omega_L$. These subspaces of $\Omega_k \otimes_k K$ and $\Omega_k \otimes_k L$ are not defined over $k$, so their intersection does not show up over $k$. In the factor $M_0$ of $K \otimes_k L$ they are linearly independent (note that indeed $M_0(x_1^{1/p}) \otimes_{M_0} M_0(x_2^{1/p})$ is a field), but in the other factor $M_1$ they give the same subspace because $x_2$ and $x_1$ are swapped in $L^{\operatorname{sep}} \to M_1$.
We end with the generalisation of Lemma 1 to arbitrary field extensions:
Lemma. Let $k \subseteq K$ and $k \subseteq L$ be field extensions. Then $K \otimes_k L$ is a field if and only if for every pair of embeddings $i \colon K \to \Omega$ and $j \colon L \to \Omega$ into an algebraically closed field $\Omega$, the images $i(K)$ and $j(L)$ are linearly disjoint (i.e. $i \otimes j \colon K \otimes_k L \to \Omega$ is injective). If $K$ and $L$ are algebraic over $k$, it suffices to take $\Omega$ an algebraic closure of $k$, and it suffices to take one pair $(i,j)$ of embeddings.
Proof. Since
$$\operatorname{Hom}_k(K,\Omega) \times \operatorname{Hom}_k(L,\Omega) = \operatorname{Hom}_k\left(K \underset k\otimes L, \Omega\right),$$
the result follows since a commutative ring $R$ is a field if and only if every map $R \to \Omega$ to a(n algebraically closed) field is injective.
This proves the first statement, and the final statement is Lemma 1. $\square$
Best Answer
Surely you mean "former to latter"?
I think the natural map is injective. Let $V$ and $W$ have bases $v_1,\ldots,v_r$ and $w_1,\ldots,w_s$ respectively. Let $g_1,\ldots,g_t$ be coset representatives for $H$ in $G$. Then a basis for $\mathrm{Ind}_H^G V\otimes \mathrm{Ind}_H^G W$ consists of the $(v_i g_k)\otimes(w_j g_l)$. The image of the natural injection from $\mathrm{Ind}_H^G(V\otimes W)$ is spanned by the $(v_i g_k)\otimes(w_j g_l)$ with $k=l$. There are exactly the right number of these.