# Decomposition of tensor product into direct sum of fields

If I have tensor product of two fields $V_1\otimes V_2$, what is the general approach to decompose this product into a direct sum of fields? In particular, I have

$\bullet\;\Bbb Q(\sqrt 2) \otimes_\Bbb Q \Bbb Q(\sqrt 3)$ and

$\bullet\;\Bbb{F_{q^2}} \otimes_{\Bbb {F_q}} \Bbb F_{q^3}$

How do I proceed with these tensor products? And is there a general algorithm?

Edit: Can I simply take basis of $V_1$ and $V_2$ separetely (let it be $\{e_i\}$ and $\{f_j\}$) and write something like $V_1\otimes V_2 \cong \bigoplus_{i,j} F(e_i\otimes f_j)\cong \bigoplus_{i,j}F$?
Then $\Bbb Q(\sqrt 2) \otimes_\Bbb Q \Bbb Q(\sqrt 3)\cong \Bbb Q \oplus \Bbb Q \oplus \Bbb Q \oplus \Bbb Q$ since $dim V_1=2$ and $dim V_2=2$ and
$\Bbb{F_{q^2}} \otimes_{\Bbb {F_q}} \Bbb F_{q^3}\cong \Bbb F_q^5$ since $dim V_1=2$ and $dim V_2=3$?

• The tensor product of fields is not, in general, a direct product of fields! – Mariano Suárez-Álvarez Jun 16 '14 at 2:16
• @Mariano Suárez-Alvarez: Maybe S/he is thinking of some sort of coalgebra structure? – user99680 Jun 16 '14 at 3:44
• @Mariano Suárez-Alvarez: BTW: would you help me merge two accounts if I sent you the two e-mail addresses by stack exchange moderator mail? I followed some protocol I was advised to follow to this effect, but no merging happened. – user99680 Jun 16 '14 at 3:45
• @user99680, sure. But we mods cannot merge accounts anymore but we can point you in the right direction :-) – Mariano Suárez-Álvarez Jun 16 '14 at 4:09
• In the listed cases the fields are linearly disjoint over the common subfield they are being tensored over. Isn't it the case that when $L_1$ and $L_2$ are linearly disjoint fields over $K$ (inside some common extension field), the bilinear mapping $$f:L_1\otimes_KL_2\to L_1L_2,\ell\otimes\ell'\mapsto \ell\ell'$$ to the compositum is an isomorphism of fields? – Jyrki Lahtonen Jun 17 '14 at 19:48

If $A$ and $B$ are algebras over a feld $K$ then $A\otimes_KB$ is also a $K$-algebra. If we have bases $\cal A$ and $\cal B$ of the spaces $A,B$ over $K$ respectively then $(\bigoplus aK)\otimes(\bigoplus bK)\cong \bigoplus abK$ is a valid direct sum decomposition of vector spaces, but generally not of algebras. Why should $abK$ be closed under $\times$?
If $A$ is a finite separable field extension of $K$, then $A=K(a)\cong K[x]/(f(x))$ by the primitive element theorem, and subsequently $A\otimes_K B\cong B[x]/(f(x))$, which can be resolved into a direct product of fields by resolving $f(x)$ into irreducible factors over $B$ and then invoking the chinese remainder thm.