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My attempt and thoughts:

I was able to show that $E\otimes_F E$ is not a field: Consider the multiplication map $\phi: E \otimes_F E \rightarrow E$ where $a\otimes b \mapsto ab$. Observe that $0 \neq \ker \phi \subsetneq E \otimes_F E$ since we can take an element $x\in E \setminus F$ and observe that $0\neq x \otimes 1 - 1 \otimes x \in \ker \phi$. Thus, it is not a field.

A comment on this post (Tensor products of fields) claims that $E \otimes_F E$ is a field if and only if it is a domain. But I am not sure how one would start this proof.

I have also seen another post that claims that the Tensor product $L \otimes_K L$ has no nilpotent elements iff $I/I^2=0$ where $I=\ker \phi$ from above. However, they rely on an argument using Kahler differentials $\Omega_L$ which I do not have. I tried studying $i \otimes 1 - 1 \otimes i \in \mathbb{Q}(i) \otimes_\mathbb{Q} \mathbb{Q}(i)$ but I can't see how it is nilpotent? I am currently in the belief that it is not actually nilpotent in this ring.

I also thought about possibly considering an element $\alpha \in E \setminus F$ and its minimal polynomial $f(x) \in F$. Then, $f(x) = (x-\alpha)g(x) \in F(\alpha)[x]$ and possibly cooking up a zero divisor using $g(\alpha)$ but I failed so far.

mathlover314
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    Did you look up this in the notes "My field theory notes" cited here? "You can find some material on when $L \otimes_F K$ is a field (and also when it is a domain) in $\S 12$ of my field theory notes". – Dietrich Burde Jan 01 '23 at 19:56
  • @DietrichBurde Yes, I have seen that his notes have information on sufficient conditions on when $K\otimes_F L$ is a domain, but there wasn't anything I could pick out that would help me show that $E\otimes_F E$ in my problem is not a domain. – mathlover314 Jan 01 '23 at 20:09
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    I wonder if you can do this in two parts. If $E$ is separable over $F$, then this is a sort of standard exercise using the primitive element theorem, using the observation that $E \cong F[X]/\langle f(X) \rangle$ for some appropriate $f(X) \in F[X]$. If $E$ is not separable over $F$, then let $F \subset E' \subset E$ be the maximal separable sub-extension. If I'm not mistaken, I think $E' \otimes_{F} E'$ embeds into $E \otimes_{F} E$ by faithful flatness, so at last I think we can reduce to the case where $E$ is purely inseparable over $F$. From here, I'm not so sure what to do, but... – Alex Wertheim Jan 01 '23 at 21:47
  • ...I expect that you might be able to make some kind of $p$-nilpotence argument where $p$ is the characteristic of $F$. I haven't really thought too hard about this though, so I warn you that this approach might not lead anywhere useful! – Alex Wertheim Jan 01 '23 at 21:48
  • Hm, coming back to this, maybe something like the following works: suppose that $F$ has characteristic $p$, and that $E/F$ is purely inseparable. Let $x \in E \setminus F$. Then I claim that $y := x \otimes 1 - 1 \otimes x$ is nilpotent. Indeed, there exists some natural number $n$ such that $x^{p^{n}} \in F$. Since $E \otimes_{F} E$ also has characteristic $p$, $y^{p^{n}} = x^{p^{n}} \otimes 1 - 1 \otimes x^{p^{n}}$ by Frobenius. But since $x^{p^{n}} \in F$, it commutes across the tensor product, so $y^{p^{n}} = 0$. I haven't carefully checked this for errors, though. – Alex Wertheim Jan 02 '23 at 04:04
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    Continuing the line of thought @AlexWertheim started. If $\alpha\in E\setminus F$ is algebraic over $F$, and $f(x)$ its minimal polynomial, then we have a short exact sequence $$0\to F[x]\to F[x]\to F(\alpha)\to0$$ with the injection coming from multiplication by $f(x)$, and the surjection by evaluating polynomials at $\alpha$. Tensoring this by $E$ gives another SES with the object at the right end being $$E\otimes_F F(\alpha)\simeq E[x]/\langle f(x)\rangle.$$ This ring thus has zero divisors, because $f(x)$ is divisible by $x-\alpha$. – Jyrki Lahtonen Jan 02 '23 at 04:21
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    (cont'd) And, as Alex said, $E\otimes_F F(\alpha)$ imbeds into $E\otimes_F E$, so this ring has zero divisors also. So I think that in/separability poses no problems. What about a purely transcendental extension? IIRC this is covered in Pete L. Clark's notes. Follow Dietrich's link. – Jyrki Lahtonen Jan 02 '23 at 04:23
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    @JyrkiLahtonen ah, very nice - cut right through to the heart of it! And the title says finite extension, so I don't think the OP is considering the subtle case of purely transcendental extensions. – Alex Wertheim Jan 02 '23 at 04:26
  • @JyrkiLahtonen & Alex Thank you!!! That proof makes sense. Two last small questions: Is $E \otimes_F F[x] \cong E[x]$? (I always get confused about this because I am worried about how much the choice of the field, in this case, $F$ in $\otimes_F$, matters) And we are getting the isomorphism you defined by applying the first isomorphism theorem on $$E \otimes_F F[x] \rightarrow E \otimes_F F(\alpha)$$ since the kernel of this map is exactly $\langle f(x) \rangle$? – mathlover314 Jan 02 '23 at 04:54
  • Also, I wish you posted your comment as a solution so I can mark this question as solved :) – mathlover314 Jan 02 '23 at 04:55

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Let's prove the following: If $E/F$ is an algebraic extension of fields such that $E \otimes_F E$ is a domain, then $E = F$.

So let $\alpha \in E$. Then $F(\alpha) \otimes_F F(\alpha)$ embeds into $E \otimes_F E$, which is therefore also a domain.

Let $f \in F[T]$ be the minimal polynomial of $\alpha$. Then we compute the tensor product as $$F(\alpha) \otimes_F F(\alpha) \cong F(\alpha) \otimes_F F[T]/ \langle f \rangle \cong F(\alpha)[T] / \langle f \rangle.$$ Now, this ring is a domain. This means that $f = 0$ (which is not the case) or that $f$ is irreducible over $F(\alpha)$. But $f$ has a root in $F(\alpha)$, namely $\alpha$. It follows $\deg(f)=1$, i.e. $\alpha \in F$.

Notice that the claim is not true for non-algebraic extensions. For example, $F(T) \otimes_F F(T)$ is a domain (not a field!), since it is a localization of $F[T] \otimes_F F(T) \cong F(T)[T']$, which is a domain.

Martin Brandenburg
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