I'm working through the last exercise in Tignol's Galois' theory of algebraic equations, which is about an unusual proof of the Galois correspondence. The setup and the part of the question I'm struggling with are excerpted below.
Let $ K/F $ be a Galois extension of fields with Galois group $G$ and let $ \mathcal{F}(G,K) $ be the $K$-vector space of all maps from $G$ to $K$.
(a) Show that there is a well-defined $F$-linear map
\begin{equation*}
\varphi: K \otimes_{F} K \to \mathcal{F}(G,K)
\end{equation*}
such that $ \varphi( a \otimes b )(\sigma) = \sigma(a)b $ for $a,b \in K $.
As far as I can see, this is a consequence $\sigma \in G$ fixing $F$, so $\psi(a,b,\sigma)=\sigma(a)b$ is $F$-bilinear, and of the universal property of the tensor product.
(b) Consider $ K \otimes_{F} K $ as a vector space over $K$ by $ ( a \otimes b ) \cdot k = a \otimes bk $. Show that $ \varphi $ is $K$-linear and bijective.
$K$-linear is easy enough, given this definition of the scalar multiplication. Bijective follows by results about the rank of the matrix of the map with respect to the bases $(e_j \otimes 1)_{1 \leq j \leq n}$ and $f_k:\sigma_i \mapsto \delta_{ik}$, which are proved in the text and the previous question, but I don't think this part is relevant beyond the definitions.
(d) Let $G$ act on $K \otimes K $ by $\sigma(a \otimes b) = a \otimes \sigma(b) $. [...]
(e) Show that a $K$-subalgebra of $K \otimes K $ is globally invariant under the action of $G$ defined in (d) if and only if it has the form $ L \otimes_{F} K $ for some subfield $L$ of $K$ containing $F$.
[Hint: for any $K$-subalgebra $A$ of $K \otimes K$, define $ L = \{ x \in K \mid x \otimes 1 \in A \} $. Show that every $ \sum_{i=1}^{r} a_{i} \otimes b_{i} \in A $ is in $ L \otimes K $ by induction on the number $r$ of terms.]
This is the part I'm stuck on. I seem to be able to do this, but I don't seem to have used the global invariance under $G$, which is odd, so maybe I have misunderstood the question. Facts that I think I know:
We need to equip $K \otimes K$ (presumably this is still $K \otimes_F K $) with a bilinear product to turn it from a $K$-vector space into a $K$-algebra. The obvious way to do this is to set $(a \otimes b) \cdot (c \otimes d) = (ac) \otimes (bd)$ and extend linearly. Since the book says nothing about this, presumably it's simple, and this seems the simplest way to me.
$K \otimes K$ the $K$-algebra is composed of elements of the form
$$ \sum_{i=1}^r a_i \otimes b_i, \qquad r \geq 1, a_i,b_i \in K. $$
We have $F$-linearity in both arguments of $a \otimes b$ and $K$-linearity in the second by the definition of $K \otimes K$ as a $K$-vector space in (b), and the product mentioned in the previous point that turns $K \otimes K$ into a $K$-algebra.Now let $A$ be a $K$-subalgebra of $K \otimes K$. This means that $K$-linear combinations of elements of $A$ are in $A$, and the product of two elements of $A$ is in $A$. If in addition $A$ is globally invariant, $G(A) \subseteq A$.
We want to show that ($A$ is a $K$-subalgebra that is is globally invariant under the action of $G$) $\iff$ ($A = L \otimes_F K$ where $L$ is a subfield of $K$ and contains $F$)
So suppose first that we have $L\otimes_F K$ where $L$ is a subfield of $K$ and contains $F$. We want to show that this is a subalgebra of $K \otimes K$ and that it is globally invariant under the action of $G$.
Check it's a subalgebra: the elements of $L \otimes K$ are all of the form $\sum_{i=1}^r c_i \otimes b_i$ with $r \geq 1$, $c_i \in L$ and $b_i \in K$. Obviously a $K$-linear combination of these is still in $L \otimes K $ since it is of the same form. Equally, $(c \otimes b) \cdot (c' \otimes b') = (cc') \otimes (bb')$ is in $L \otimes K$, so bilinearity and induction give that $L \otimes F$ is a $K$-subalgebra.
Check it's invariant under the action of $G$: for $\sigma \in G$, the definition in (d) gives $\sigma \left( \sum_{i=1}^r c_i \otimes b_i \right) = \sum_{i=1}^r c_i \otimes \sigma(b_i) $ (assuming the action extends linearly to elements of higher rank—what else could it do?), and $\sigma(b_i) \in K$, so we get $\impliedby$. Fine.
Now the other direction. Suppose that $A$ is a subalgebra. Define $L$ as in the Hint.
First we show that $L$ is an $F$-vector space (and hence an abelian group). Suppose that $x,y \in L$. Then $x \otimes 1, y \otimes 1 \in A$ by definition. Then if $a,b \in F$, $ (ax+by) \otimes 1 = a(x \otimes 1) + b(y \otimes 1) \in A $ since elements of $K \otimes K$ are $F$-linear in the first argument and $A$ is an $F$-vector space. Hence $ax+by \in L$ and so $L$ is an $F$-vector space.
Next we show $L$ is closed under products. With $x,y$ as before, $ (xy) \otimes 1 = (x \otimes 1) \cdot (y \otimes 1) \in A$ by the definition of the product and $A$'s being a subalgebra, so $L$ is indeed closed under products.
Now we show $F \subseteq L$. Let $x \in L$. Then $x^k \in L$ for $k \geq 0$, and the usual linear algebra implies that there is a monic irreducible polynomial $p(X) = X^m + p_{m-1}X^{m-1} + \dotsc + p_0 \in F[X]$ of degree $m \leq [K:F]$ so that $p(x)=0$. Since it is irreducible, $p(0) = p_0 \neq 0$, and since $p(x) \neq 0$, $p_0 = -x^m-p_{m-1}x^{m-1}- \dotsb -p_1 x $. $L$ is an $F$-vector space, and the right-hand side is an $F$-linear combination of elements of $L$, so $p_0 \in F \cap L$. But for any $q \in F$, $qp_0^{-1} \in F$, so $qp_0^{-1}p_0 = q \in L$ for any $q \in F$. Hence $F \subseteq L$.
Lastly we need to show $x^{-1} \in L$. Suppose again that $x \in L$ with minimal polynomial $p$, then we have $x^{-1} = -p_0^{-1} (x^{m-1} + p_{m-1}x^{m-2} + \dotsb + p_1) \in L $ by a similar argument to before.
Therefore $L$ is a subfield of $K$ that contains $F$.
We now need to show that $A = L \otimes K $. Proceed as the Hint suggests: let us start with elements of the form $a \otimes b \in A$. Then $A$ is a $K$-vector space, so $ A \ni (a \otimes b) \cdot (b^{-1}) = a \otimes 1$, so $a \in L$ and hence $a \otimes b \in L \otimes K$.
Now suppose that every member of $A$ of the form $\sum_{i=1}^r a_i \otimes b_i$ is in $L \otimes K$, and consider $ \sum_{i=1}^{r+1} a_i \otimes b_i $. But this is the sum of two terms: $ \sum_{i=1}^{r} a_i \otimes b_i \in L \otimes K $ by the induction hypothesis, and $a_{r+1} \otimes b_{r+1} \in A$ since it is the difference of two elements of $A$, which lies in $L \otimes K$ by the basis case. Hence $\sum_{i=1}^{r+1} a_i \otimes b_i$ is the sum of two elements of $L \otimes K$ and so itself lies in $L \otimes K$.
Hence every element of $A$ lies in $L \otimes K$, so $A \subseteq L \otimes K$. Lastly, we want to show that $L \otimes K \subseteq A$, which follows because every element of $L \otimes K$ can be written as a $K$-linear combination of elements of the form $a \otimes 1 \in L \otimes K$.
So that would seem to be it. But I haven't used anything about $G$ at all in the $\implies$ proof! So this seems to imply that all $K$-subalgebras of $K \otimes_F K $ are globally invariant under this $G$-action. But apparently this is not the case, given that the next part of the question involves showing that subalgebras of $\mathcal{F}(G,K)$ (which are presumably in bijection with those of $K \otimes_F K$ by (a) and (b)) are in bijection with partitions of $G$, and are globally $G$-invariant precisely when they correspond to a partition of $G$ into cosets of a subgroup, and obviously not all partitions of $G$ are such.
Can someone point out my error?
Answer
Your induction step does not work. By the induction hypothesis, you only know that every element of the form $\sum_{i=1}^r a_i \otimes b_i$ which is in $A$ is also in $L\otimes K$. In general, you can't conclude from $\sum_{i=1}^{r+1} a_i \otimes b_i \in A$ that $\sum_{i=1}^r a_i \otimes b_i \in A$, so the induction hypothesis does not apply.
I think it's instructive to give a concrete example to show where it goes wrong:
Let $F=\Bbb Q$ and $K=\Bbb Q(i,\sqrt{2})$, then a $K$-basis of $K\otimes_F K$ (with respect to the action of $K$ as in the OP) is given by $(1\otimes1,i\otimes1,\sqrt{2}\otimes 1, i\sqrt{2}\otimes 1)$.
Consider the element $$\alpha=2 i\otimes i - \sqrt{2} \otimes \sqrt{2} - i\sqrt{2}\otimes i\sqrt{2}\in K\otimes_F K$$ A simple computation shows that $\alpha^2=12\otimes 1 +4\alpha$, so the $K$-subvector space spanned by $1\otimes 1$ and $\alpha$ is closed under multiplication, thus it is a subalgebra, which I will call $A$. It's not difficult to compute that $L=\Bbb Q$ in this case. If we try to apply the induction step to $\alpha$, we need to have that $2i\otimes i -\sqrt{2}\otimes\sqrt{2} \in A$, but this element is not a linear combination of $\alpha$ and $1\otimes 1$.
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