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
You should take a look at Theorem 3.4 (p 85) of Farb and Dennis' book Noncommutative Algebra. The statement is:
Let $L/k$ be a finite extension of fields. Then $K\otimes _k L$ is semisimple for every field $K\supseteq k$ if and only if $L/k$ is a separable extension.
That the tensor product is semisimple implies that its Jacobson radical vanishes. Conversely, any artinian ring with trivial Jacobson radical is semisimple. Therefore your equivalent formulation of separability is true.