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Let $[K : k]$ be a Galois extension. Show that there is a bijection between set of all $K$-linear homomorphisms $K\otimes_k K \to K$ and set of all irreducible idempotents in $K\otimes_k K$.

Here we consider $K\otimes_k K$ as $K$-algebra that acts as $a(v_1 \otimes v_2) = av_1 \otimes v_2$.

Element $e$ is called an idempotent if $e^2=e$.

There is a hint: Prove that any such homomorphism sends all but one irreducible idempotents to zero.

I can't come up with how to use it, but here is my attempt: Consider action with left multiplication by $1-e$ (if $e$ is an idempotent so does $1-e$), then it maps to zero all elements of the form $e\otimes v_2$. If $v_2 = w$ (let $w$ be another idempotent in $K$), then $e \otimes w$ is an idempotent in $K\otimes_k K$. The problems are that, firstly, I'm not sure that any arbitrarily idempotent in $K \otimes_k K$ is a tensor product of $K$ idempotents and, secondly, this map doesn't take to zero everything except chosen one. Moreover, arbitrarily homomorphism $K \otimes_k K \to K$ may not be a multiplication with some idempotent element.

Can anyone explain me how to prove this statement? Any help will be very appreciative.

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There is a different description of $K \otimes_k K$ as a $K$-algebra. By the primitive element theorem, you can write $K = k(x)$ for some $x \in K$. Let $\mu$ be the minimal polynomial of $x$, so $K \cong k[T]/(\mu)$ as $k$-algebras. Then as $K$-algebras.

$$K \otimes_k K \cong k[T]/(\mu) \otimes_k K \cong K[T]/(\mu)$$

where $K[T]/(\mu)$ is the ideal generated by $\mu$ in $K[T]$, not $k[T]$. Since $K/k$ is normal and separable, and $\mu$ has a root $x \in K$, its remaining roots $x = x_1, ... , x_n$ also lie in $K$, and they are all distinct. Therefore $\mu = (T - x_1) \cdots (T - x_n)$, and by Chinese remainder theorem,

$$K[T]/(\mu) \cong \prod\limits_{i=1}^n k$$

as rings, where the isomorphism is given by $g + (\mu) \mapsto (g(x_1), ... , g(x_n))$.

If $(a_1, ... , a_n)$ is an idempotent element in $R = \prod\limits_{i=1}^n k$, then $a_i^2 = a_i$ in $k$, hence $a_i$ is zero or one. So we see that there are $n$ irreducible idempotent elements.

On the other hand, the kernel of a $K$-algebra homomorphism of $K[T]/(\mu)$ into $K$ must be a maximal ideal of $K[T]$ containing $\mu$, and the choices are $(T - a_1), ... , (T - a_n)$. So there are the same number of $K$-algebra homomorphisms $K \otimes_k K$ into $K$.

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  • $\begingroup$ That all assumes $[K:k]$ is finite, no? $\endgroup$ – sharding4 Jun 11 '17 at 18:28
  • $\begingroup$ Yes. $\space$ $\space$ $\endgroup$ – D_S Jun 11 '17 at 18:30
  • $\begingroup$ @ D_S 's answer can be generalized to the case of 2 extensions $K_i$ of $k$, one of them being finite separable, see e.g. math.stackexchange.com/a/1912067/300700 $\endgroup$ – nguyen quang do Jun 13 '17 at 7:35

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