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$(Y-a) g = 0 \in k(a)[Y]/(f)$
$(a'-a) g(a') = 0 \in k(a)(a')$
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

Mike Miller

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

Mike Miller

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…

Leaky Nun

7:11 AM

@MikeMiller i.e. $K[X,Y]/(X^2-2,Y^2-2) = K(\sqrt2)[X]/(X^2-2)$

5 hours later…

Mike Miller

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…

Mike Miller

1:05 PM

@LeakyNun did you like the second argument tho

6 hours later…

Leaky Nun

7:18 PM

@MikeMiller nice!

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