prime ideals of $\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}}$ $\bar{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$.

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*Is every prime ideal of $\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}}$ maximal?


*Can one describe all prime ideals clearly?
I tried to work on the algebraic structure of the tensor product but
I got nothing.
This is the link of the same question Describe prime ideals and Krull dimension of $\overline{\mathbb{Q}} \otimes_{\mathbb{Q}} \overline{\mathbb{Q}}$ but
I don't really understand the answer.
Is there any other way to solve this problem?
 A: I’ll have a try, but I’m not sure that can be as good as the answer you linked – at least as long as you don’t explain what precisely upsets you there.
Let $K=\overline{\mathbb{Q}}$, $R=K \otimes_{\mathbb{Q}} K$.
For your first point:
Let $p \subset R$ be a prime ideal. Now, consider $R$ as a $K$-algebra where $a \cdot (b \otimes c)=(ab) \otimes c$. Then $R/p$ is an integral domain and a $K$-algebra.
Moreover, it’s easy to see that any element of $R$ generates a finite-dimensional $K$-algebra. Therefore, every element of $R/p$ generates a finite-dimensional $K$-algebra without zero divisors, so said algebra is a field; it follows that $R/p$ is a field and $p$ is maximal.
For your second question:
Maximal ideals of $R$ correspond to kernels of morphisms $R \rightarrow F$ onto a field.
As $R$ is a $K$-algebra, this makes $F$ a $K$-algebra. But as above, every element of $R$ is integral over $K$, so $F$ is algebraic over $K$. But $K$ is algebraically closed, so $F=K$. But this doesn’t mean that $K \rightarrow F$ is the identity: it could be any automorphism. So, composing with the inverse map, we find that the prime ideals of $R$ are the kernels $\kappa_{\mu}$ of the $K$-linear morphisms $\mu: R \rightarrow K$.
Let us show too that $\mu \longmapsto \kappa_{\mu}$ is injective, this will show the map to be a bijection. But if we have $\kappa_{\mu}=\kappa_{\nu}$, then clearly $\nu=\mu \circ s$, where $s : K \rightarrow K$ is a morphism. As $\nu$ is onto, so is $s$; $K$ is a field so $s$ is injective, $\mu$ and $\nu$ are $K$-linear, thus so is $s$ hence $s$ is the identity and $\mu=\nu$.
For such a $\mu$ (ie a $K$-linear ring homomorphism $R \rightarrow K$), define $\sigma_{\mu}(x \in K)=\mu(1 \otimes x) \in K$, which is a $\mathbb{Q}$-automorphism of $K$.
Clearly, $\mu \longmapsto \sigma_{\mu}$ is a bijection between the set of $K$-linear morphisms $R \rightarrow K$ and $Gal(K/\mathbb{Q})$ (the inverse map maps an automorphism $\sigma$ of $K$ to $a \otimes b \longmapsto a\sigma(b)$).
It follows that there is a bijective correspondance between the elements of $Gal(K/\mathbb{Q})$ and the prime ideals of $R$, given by $\sigma \longmapsto Z_{\sigma}=\mathrm{ker}\,(a \otimes b \longmapsto a\sigma(b))$. But it’s easy to see $Z_{\sigma}$ to be generated by the $\sigma(x) \otimes 1-1\otimes x$ for $x \in K$.
So, if, for $\sigma \in Gal(K/\mathbb{Q})$, $I_{\sigma}$ is the ideal generated by the $\sigma(x) \otimes 1-1\otimes x$ for $x \in K$, then $\sigma \longmapsto I_{\sigma}$ is a bijection with the prime ideals of $R$.
A: I did not look at the earlier answer or the linked post, so this will be written in completely independent terms, so hopefully avoid whatever the source of confusion was.  Apologies if I just end up regurgitating the previous explanations.
Given any $\alpha\in \bar{\mathbb{Q}}$, let $\beta_1,\cdots \beta_k$ be its conjugates (including itself).  In other words $$\prod_{i=1}^k (x-\beta_i)$$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.
Then $$\prod_{i=1}^k (\alpha\otimes 1-1\otimes\beta_i)=0,$$
as symmetric polynomials expressions in the $\beta_i$ are rational (they are the coefficients of the minimal polynomial of $\alpha)$ , so slip across the tensor product once the above expression is multiplied out.
Thus if $p\subseteq \bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}}$ is prime, then for each $\alpha\in \bar{\mathbb{Q}}$, the prime $p$ must contain $$\alpha\otimes 1-1\otimes\beta,$$ for some conjugate $\beta$ of $\alpha$.
Thus in the quotient $(\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}})/p$ we must have for any $\alpha,\gamma\in \bar{\mathbb{Q}}$:
\begin{eqnarray*}\alpha\otimes \gamma&=&(\alpha\otimes 1)(1\otimes \gamma)\\&=&(1\otimes \beta)(1\otimes\gamma)\\&=&(1\otimes \beta\gamma),\end{eqnarray*}
where $\beta$ is some conjugate of $\alpha$.
That is $(\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}})/p$ is a quotient of $\bar{\mathbb{Q}}$, hence isomorphic to $\bar{\mathbb{Q}}$.  In particular the answer to your first question is yes: $\bar{\mathbb{Q}}$ is a field, so $p$ must be maximal.
In answer to your second question, the prime ideals of $\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}}$ are precisely the kernels of ring homomorphisms $\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}}\to \bar{\mathbb{Q}}$, mapping $\alpha\otimes \gamma \mapsto \beta\gamma$ as above.
Note these restrict to the identity on the second factor.
In particular, if $f\colon \bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}}\to\bar{\mathbb{Q}}$ is such a ring homomorphism, then pre-composing with the inclusion: $$\bar{\mathbb{Q}}\to\bar{\mathbb{Q}}\otimes_{\mathbb{Q}}\bar{\mathbb{Q}},$$
mapping $\alpha\mapsto \alpha\otimes 1$, we get a field automorphism $g\colon\bar{\mathbb{Q}}\to \bar{\mathbb{Q}}$.
Given $g\in$Gal$(\bar{\mathbb{Q}} /\mathbb{Q})$, we have a prime ideal $p_g$ given by the kernel of the map $\bar{\mathbb{Q}}\otimes \bar{\mathbb{Q}}\to \bar{\mathbb{Q}}$, sending:
$$(\alpha\otimes\beta)\to g(\alpha)\beta,$$
and we have shown that every prime ideal has this form.
Finally note that given $g,g'\in$Gal$(\bar{\mathbb{Q}} /\mathbb{Q})$, if for some $\alpha\in \bar{\mathbb{Q}}$ we have $g(\alpha)\neq g'(\alpha)$, then $$
\alpha\otimes1-1\otimes g(\alpha)\in p_g,\qquad \alpha\otimes1-1\otimes g'(\alpha)\in p_{g'}.$$
As $1\otimes(g(\alpha)-g'(\alpha))$ is a unit, we may conclude that $p_g\neq p_{g'}$.
Thus we may conclude that the prime ideals of $\bar{\mathbb{Q}}\otimes \bar{\mathbb{Q}}$ are in one to one correspondence with
${\rm Gal}(\bar{\mathbb{Q}} /\mathbb{Q})$.
