12:17 AM
in Mathematics, 11 mins ago, by Mike Miller
@Fargle Prove that if $K/k$ is a finite field extension, the ring $K \otimes_k K$ can never be an integral domain.
(later clarified : $K/k$ is non-trivial)
let then $a \in K \setminus k$ and $f = \operatorname{irr}(a,k)$
Consider $k[X]/(f) \to K$ injective
then $k[X]/(f) \otimes_k k[X]/(f) \to K \otimes_k K$ injective (there's no problem here, everything are vector spaces)
the LHS is $k[X,Y]/(f,f)$
$k(a)[X]/(f)$
$(f) \subseteq k(a)[X]$ is not a prime ideal though
letting $f=(x-a)g \in k(a)$ we see what I claimed
sloppy argument ^
trying to follow the steps to find two explicits elements in $K \otimes_k K$

I'll respond in 1/2 hour
Have to focus on something else for a bit

$(Y-a) g = 0 \in k(a)[Y]/(f)$
$(a'-a) g(a') = 0 \in k(a)(a')$
This is hard to convert to $k(a) \otimes k(a)$ explicitly because $g$ has both $a$ and $a'$
So let's work with an example: $K = \Bbb Q(\sqrt2)$, $k = \Bbb Q$
then $f = X^2 - 2 = (X-\sqrt2)(X+\sqrt2)$
replace $\sqrt2$ with $\sqrt 2 \otimes 1$ and $X$ with $1 \otimes \sqrt2$
we get $(1 \otimes \sqrt 2 - \sqrt 2 \otimes 1) (1 \otimes \sqrt 2 + \sqrt 2 \otimes 1) = 1 \otimes 2 - 2 \otimes 1 = 0$ as required
let's work with a harder example
$f = X^5 + X + 1$, $k = \Bbb Q$, $K = \Bbb Q[X]/(f)$
Let $\alpha \in K$ be such that $f(\alpha) = 0$
then $f = (X-\alpha)(X^4+\alpha X^3 + \alpha^2 X^2 + \alpha^3 X + \alpha^4 + 1)$
replace $\alpha$ with $\alpha \otimes 1$ and $X$ with $1 \otimes \alpha$
then we get $(1 \otimes \alpha - \alpha \otimes 1) (1 \otimes \alpha^4 + \alpha \otimes \alpha^3 + \alpha^2 \otimes \alpha^2 + \alpha^3 \otimes \alpha + \alpha^4 \otimes 1 + 1 \otimes 1)$
$= 1 \otimes \alpha^5 + 1 \otimes \alpha - \alpha^5 \otimes 1 - \alpha \otimes 1$
$= 1 \otimes (\alpha^5 + \alpha) - (\alpha^5 + \alpha) \otimes 1$
$= 1 \otimes (-1) - (-1) \otimes 1$
$= 0$
so we're done
so I believe this works

1:10 AM
@LeakyNun oh cool, i remember when i was working on this i tried to find an answer by finding explicit elements but couldn't figure it out
should I look over your argument first or tell you mine?
oh, u gone

1:31 AM
I don't understand why k[X,Y]/(f,f) is the same as k(a)[X]/(f)
But if that's true I understand the rest of the argument
Here's my argument
1) First our ring is not a field, because the product $K \otimes_k K \to K$ is a ring map; maps of fields are injective; this contradicts the dimension count.
2) Artinian domains are fields. For consider the decreasing sequence $(a^n)$ for nonzero $a$. This must stabilize, so for some $m$ we have $(a^m) = (a^{m+1})$. So $a^m = a^{m+1} y$ for some $y$.
Because the ring is a domain, $a^m$ is not zero. Therefore $y$ is not zero. So we have $a^m(ay-1) = 0$. We have constructed an inverse to an arbitrary nonzero $a$.
Finally $K \otimes_k K$ is Artinian, as ideals are $k$-subspaces and this is a finite dimensional $k$-vector space.

6 hours later…
7:11 AM
@MikeMiller i.e. $K[X,Y]/(X^2-2,Y^2-2) = K(\sqrt2)[X]/(X^2-2)$

5 hours later…
11:45 AM
@LeakyNun yeah clearly, thanks. i guess i got worried that we might forcibly add new roots in this extension but it's not a splitting field for that polynomial

1 hour later…
1:05 PM
@LeakyNun did you like the second argument tho

6 hours later…
7:18 PM
@MikeMiller nice!