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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 $k$.

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Alright, well I spoke to my Algebra prof about it, and it seems quite trivial with the right point of view! (I.e., universal property.)

Namely, consider the map $f:K\times L \to \Omega$ given by $f((k_0,\ell)) = k_0\ell$. Firstly, we show this map is bilinear. Notice $$f((k_1 + k_2,\ell)) = (k_1+k_2)\ell = k_1\ell + k_2\ell = f((k_1,\ell)) + f((k_2,\ell)).$$ Similarly, $$ f((k_0,\ell_1+\ell_2)) = k_0(\ell_1+\ell_2) = k_0\ell_1 + k_0\ell_2 = f((k_0,\ell_1)) + f((k_0,\ell_2)),$$ and lastly, for any $r\in k$, we have $$ f((rk_0,\ell)) = k_0r\ell = f((k_0,r\ell)) = r(k_0\ell) = rf((k_0,\ell)).$$

Now that $f$ is bilinear, we know there must be a unique linear map, $g:K\otimes_k L\to \Omega$ such that $$g(k_0\otimes \ell) = f((k_0,\ell)) = k_0\ell,$$ by the universal property of tensor products. But then we are almost done. By construction, we see $g = f_{L,K}$, and all that's left is to show that $f_{L,K}$ satisfies $$ f_{L,K}(k_1,\ell_1)f_{L,K}(k_2,\ell_2) = f_{L,K}(k_1k_2,\ell_1\ell_2)$$ and $$ f_{L,K}(1_K\otimes 1_L) = 1_\Omega.$$ But these are clear. We 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 the latter follows by the definition of $f_{L,K}$, since $1_K = 1_L = 1_\Omega$.

Hence, there is a unique ring homomorphism $f_{L,K}:K\otimes_k L\to \Omega$ such that $f_{L,K}(x\otimes y) = xy$ for all $x\in K$ and $y\in L$. Definitely interesting to see that the hands on approach is so far from the nicest way, given it seemed so easy at first...

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