I am dealing with this question in An Introduction to Galois Theory course:
Which of the following algebras are fields? Products of fields? Describe these fields.
- $\mathbb{Q}(\sqrt[3]{2}) \space \otimes_{\mathbb{Q}} \space \mathbb{Q}(\sqrt{2}) $
- $\mathbb{Q}(\sqrt[4]{2}) \space \otimes_{\mathbb{Q}} \space \mathbb{Q}(\sqrt{2}) $
- $\mathbb{F}_2(\sqrt{T}) \space \otimes_{\mathbb{F}_2(T)} \mathbb{F}_2(\sqrt{T}) $
- $\mathbb{F}_4(\sqrt[3]{T}) \space \otimes_{\mathbb{F}_4(T)} \mathbb{F}_4(\sqrt[3]{T}) $
Clarifying the question I might say :
$.\otimes_{A}.$ is a notation used for the Tensor product of two algebras or modules over the ring A.
Tensor product is defined with the universal property in my course.
$\mathbb{F}_2$ and $\mathbb{F_4}$ denote $\mathbb{Z}_2$ and $\mathbb{Z}_4$ respectively.
My Progress:
I know that any finite algebra have finitely many maximal ideals.
Say $m_1,…,m_k$ be the maximal ideals of our finite algebra A.
Then $A \cong \frac{A}{m_1^{n_1}}\times …\times \frac{A}{m_r^{n_r}}$ for some $n_i\in\mathbb{N}$.
Hence if $\space $$\forall i \space ;n_i=1$ , then A is a product of fields.
Also there are some helpful theorems that I've referred , in my answers document, which is linked bellow in my answer to this problem.
I've written all of my detailed answers in the following document, but I'm not quite sure about them (Specially about parts 3 and 4).
Click here for reaching the google document link.
After you saw my answers I'd like to add the following:
In part 3, I've shown in my answers that:
$\mathbb{F}_2(\sqrt{T}) \space \otimes_{\mathbb{F}_2(T)} \mathbb{F}_2(\sqrt{T}) \cong \frac{\mathbb{F}_2(\sqrt{T})[U]}{ (U-\sqrt{T})^2}$ where $U$ is a variable.
So this is not a field because of the presence of nilpotent elements like $U-\sqrt{T}$.
But I can not show that whether it can be a product of fields or not?
Also In part 4, I've shown in my answers that:
$\mathbb{F}_4(\sqrt[3]{T}) \space \otimes_{\mathbb{F}_4(T)} \mathbb{F}_4(\sqrt[3]{T}) \cong \frac{\mathbb{F}_4(\sqrt[3]{T})[U] }{U^3-T}$
But now I've got stucked and can not say anymore about the product of fields.
Any help leading to a progression would be greatly appreciated.
Best Answer
For 3, as you say there is a nilpotent element in the tensor product. This cannot happen in a product of fields. If we have $R=F_1\times\cdots\times F_n$, a product of fields, and $(a_1,\ldots,a_n)^2$ is zero in $R$ then $a_i^2=0$ in each $F_i$ so that $a_i=0$ (as $F_i$ is a field). In this example, then the tensor product is not a product of fields.
For field extensions $F_1/F$, $F_2/F$ the tensor product $F_1\otimes_F F_2$ can only fail to be a product of fields when both $F_1/F$ and $F_2/F$ are inseparable extensions, and that is precisely the case here.
But in case 4, you have separable extensions. Indeed $F_1/F$ is a Kummer extension here as $F=\Bbb F_4(T)$ has three cube roots of unity: $1$, $\omega$ and $\omega^2$. Then $$F_1\otimes _F F_1\cong F_1\times F_1\times F_1$$ via $$a\times b\mapsto (ab,a\sigma(b),a\sigma^2(b))$$ where $\sigma:F_1\to F_1$ is the automorphism taking $\sqrt[3]T$ to $\omega\sqrt[3]T$.