The problem:
Let $K$ and $L$ be subfields of a field $\Omega$, and let $k\subset K\cap L$ be a common subfield.
(a) Show that there exists a unique ring homomorphism $f_{K,L}:K\otimes_k L\to \Omega$ such that $f(x\otimes y) = xy$ for all $x\in K, y\in L$.
(b) Prove that $f_{L,K}$ is injective if and only if $K$ and $L$ are linearly disjoint over $k$.
My problem: I've never really seen tensor products before. It's just being assumed known for this course and it's 3 weeks until our algebra course gets there, so I'm a little hazy, and was hoping I could pass my work over some delectable scrutiny :).
(a) We first prove the map $f_{K,L}$ is a homomorphism of rings. Let $k_1,k_2\in K$ and $\ell_1,\ell_2\in L$. Then we have \begin{align*} f_{K,L}(k_1\otimes \ell_1) + f_{K,L}(k_2\otimes \ell_2) &= k_1\ell_1 + k_2\ell_2\newline &= (k_1\ell_1\ell_2^{-1} + k_2)\ell_2\newline &= (k_1\ell + k_2)\ell_2 \end{align*} where $\ell = \ell_1\ell_2^{-1} \in L$. If $\ell\in K$ (moreover, $\ell \in k$), then we know it has inverse \begin{align*} k_1\ell + k_2 \otimes \ell_2 &= k_1\ell \otimes \ell_2 + k_2\otimes \ell_2\newline &= k_1 \otimes \ell\ell_2 + k_2\otimes \ell_2\newline &= k_1 \otimes \ell_1 + k_2\otimes \ell_2. \end{align*} Hence, $f_{L,K}(k_1\otimes \ell_1 + k_2\otimes\ell_2) = f_{L,K}(k_1\otimes\ell_1) + f_{L,K}(k_2\otimes\ell_2)$.
Now this relies on $\ell$ being in $K$, which I don't see why this should always be true, given the $\ell_i$'s are arbitrary. Another approach, we could look at $f_{K,L}(k_1\otimes \ell_1 + k_2\otimes \ell_2)$ instead, but there's no need for $k_1\otimes \ell_1 + k_2\otimes \ell_2$ to be expressible as a simple tensor, so this expression doesn't really make sense given the definition of the map..?
We also have $ f_{L,K}(k_1\otimes \ell_1)f_{L,K}(k_2\otimes\ell_2) = k_1k_2\ell_1\ell_2 = f_{L,K}(k_1k_2\otimes\ell_1\ell_2)$ and $ f_{L,K}(1\otimes 1) = 1.$ Thus $f_{L,K}$ is a ring homomorphism. Now as $K\otimes_k L$ is completely determined by the simple tensors on bases for $K$ and $L$ (which always exist, if necessary, using the axiom of choice), we have that this homomorphism must be unique by construction.
(b) We know that $K$ and $L$ are not linearly disjoint over $k$ if and only if there exists a finite set of elements of $K$, say $\{a_1,\ldots,a_n\}$ that are linearly independent over $k$, but not over $L$. In other words, there would be elmenets $\{b_1,\ldots,b_n\}$ of $L$ such that $ a_1b_1 + \ldots + a_nb_n = 0.$ But this is equivalent to having $ f_{L,K}(a_1\otimes b_1) + \ldots + f_{L,K}(a_n\otimes b_n) = f_{L,K}(a_1\otimes b_1 + \ldots + a_n\otimes b_n) = 0.$ So $f_{L,K}$ is injective when $K$ and $L$ are not linearly disjoint if and only if $ a_1\otimes b_1 = -a_2\otimes b_2 - \ldots - a_n\otimes b_n.$ By construction, the $a_i$ are linearly independent over $k$, so this dependence must comes from the $b_i$. But, since the left hand side is a simple tensor, and the $a_i$ are linearly independent over $k$, all the $b_i$ must be multiples of some element $b\in L$, and reducing the right hand side via bilinear relations shows $a_1$ is a $k$-linear combination of $a_2,\ldots,a_n$, which is not possible.
Thus $f_{L,K}$ is injective if and only if $L$ and $K$ are linearly disjoint over $k$.