I should make it clear that I don't understand tensor products. I generally understand things from abstract algebra when I can relate them to number theory, so this might be where I finally get something. I'm looking at a paper, and it contains the sentence:

A degree $n$ number field $K$ can be embedded into its Minkowski space as $j_{R}:K\hookrightarrow K \otimes_Q R \cong R^n$ yielding a rank $n$ lattice $j_R(\mathcal{O}_K)\subseteq R^n$

So, I'm familiar with the embedding they're describing, I think, in which the integer ring of a number field becomes a lattice in $R^n$. How is that the same thing as a tensor product of $K$ with $R$ over $Q$? It seems it's supposed to have something to do with products of elements in $K$ with real numbers, but not the usual products? What does it mean that it's over $Q$? Is that just because both $K$ and $R$ are already vector spaces over $Q$? From what I'm reading, the basis of this tensor product should consist of products of basis elements of $K$ and $R$, but a basis for $R$ over $Q$ is.... a mess? How does this all work to produce the embedding I'm familiar with?

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    $\begingroup$ If $F\subset F'$ are fields and $K$ is a vector space over $F$, then $K\otimes_{F}F'$ is a vector space over $F'$, and a basis of $K$ is still a basis in the tensor product. $\endgroup$ Commented Mar 14, 2018 at 2:28
  • $\begingroup$ I believe you, but I don't see how that's consistent with the definitions I've seen, especially the part about the basis... $\endgroup$ Commented Mar 14, 2018 at 2:49
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    $\begingroup$ The basis is not part of the definition. In any case, it's true if you're considering a basis over $\mathbb Q$, which is horrific, monstrous, and useless. What matters is the basis over $\mathbb R$. $\endgroup$ Commented Mar 14, 2018 at 2:51
  • $\begingroup$ I guess I hear what you're saying, but not understanding any more than before. :/ I'll think about it, and see what other people have to say. $\endgroup$ Commented Mar 14, 2018 at 3:03
  • $\begingroup$ Tony, what may be holding you back is the following. If $L$ and $L'$ are any two full lattices inside $\Bbb{R}^n$, i.e. both discrete subsets and rank $n$ free abelian groups, then there exists a linear isomorphism $\sigma:\Bbb{R}^n\to \Bbb{R}^n$ such that $\sigma(L)=L'$. The proof is easy. Tha bases for both $L$ and $L'$ are also bases of the ambient real vector space. Meaning that we can just send basis elements of $L$ to basis elements of $L'$, and extend linearly. The embedding you probably have in mind has special metric properties coming from the embeddings of $K$ into $\Bbb{C}$ $\endgroup$ Commented Mar 14, 2018 at 17:33

2 Answers 2


Disclaimer: I can't be sure that all of this is completely correct, since I actually haven't been able to find it explicitly spelled out anywhere, and I myself am also not super familiar with tensor products. So this answer is mainly to share how I see things, and hopefully you can find something you are looking for within. Let me know if you find anything that seems like a mistake.

Firstly, on the number theory side of things, I'm aware of two slightly different perspectives you can take when viewing $\mathcal{O}_K$ as a lattice in some vector space, so even though you say you are familiar with this, let me just briefly lay it out how I see it so that we are on the same page.

Let $K$ be a number field of degree $n$. There are $n$ embeddings of $K$ into $\mathbb{C}$ which we can denote by $\tau_1, \ldots, \tau_n$. It is also often convenient to come up with a labeling of these embeddings that discriminates the real embeddings from the non-real ones. So we will also use the labeling scheme where the totally real embeddings are denoted by $\rho_1, \ldots , \rho_r$. The non-real embeddings (which I'll just call complex embeddings) come in conjugate pairs, so we can denote these by $\sigma_1, \ldots ,\sigma_s, \overline{\sigma}_1, \ldots, \overline{\sigma}_s$. I'll freely use both of these labeling schemes, depending on which makes the most sense. Define the following map. The notation $(\alpha_\tau)_\tau$ just means that I am indexing the components of the column vector by the $n$ embeddings.

$$ \begin{align} j:&K \to \mathbb{R}^n \\ &\alpha \mapsto (\alpha_\tau)_\tau \end{align} $$

where $\alpha_\rho = \rho(\alpha)$ and $\alpha_\sigma = \Re(\sigma(\alpha))$ and $\alpha_{\overline{\sigma}} = \Im(\sigma(\alpha))$. Then, the set $j(\mathcal{O}_K)$ is a rank $n$ lattice in $\mathbb{R}^n$. This is an explicit way to write the map that yields a lattice. You are probably familiar with this or something very close to it.

Now for the tensor product. It seems (to me) like you are looking for an explicit explanation of how the embedding into the tensor product agrees with the map I've written above, so this is what I'll show.

For aesthetic reasons I actually like to write the tensor product in question as $\mathbb{R}\otimes_\mathbb{Q}K$ rather than the other way around (it doesn't matter because for $R$ modules $M$ and $N$ the products $M\otimes N$ and $N\otimes M$ are unique up to unique isomorphism). I am thinking of this object as being produced by extension of scalars. That is, we start out with the $\mathbb{Q}$ vector space $K$, but we would like an $\mathbb{R}$ vector space in which we can view the ring of integers as a lattice. So we need to extend the multiplication from $\mathbb{Q}$ to $\mathbb{R}$. Because of the universal property of the tensor product, the vector space $\mathbb{R}\otimes_\mathbb{Q}K$ is the "best" way that we can do this. Thus, to achieve our goal of embedding $K$ into an $\mathbb{R}$ vector space, it is natural to look for an embedding into this object $\mathbb{R}\otimes_\mathbb{Q}K$. By the way, Dummit and Foote has a very clear and excellent exposition about extension of scalars. If you are shaky with intuition for tensor products, I recommend reading it.

Now, say that $\{\alpha_i, \ldots , \alpha_n \}$ is a $\mathbb{Q}$ basis for $K$. As Matt Samuel mentions in his comment, which you said you agree with, we know that $\{1\otimes\alpha_1, \ldots, 1\otimes\alpha_n \}$ is a basis for $\mathbb{R}\otimes_\mathbb{Q}K$. (This is now an $\mathbb{R}$ basis. If you want, Lang's Algebra has a proof of this for free modules in general somewhere in the chapter about tensor products). Now take an arbitrary element $\alpha = q_1\alpha_1 + \cdots + q_n\alpha_n$ in $K$ and define the map

$$ \begin{align} f:K &\to \mathbb{R}\otimes_\mathbb{Q}K \\ \alpha &\mapsto \sum_{i = 1}^n (q_i\otimes \alpha_i) \end{align} $$

We know that the object on the right is indeed in $\mathbb{R}\otimes_\mathbb{Q}K$ because it is just a linear combination of basis elements.

Now, take an arbitrary element in the tensor product $\sum_{i = 1}^n (x_i\otimes \alpha_i)$. Here $x_i \in \mathbb{R}$. We can define a map from the tensor product into the more familiar space $\mathbb{R}^n$.

$$ \begin{align} g: &\mathbb{R}\otimes_\mathbb{Q}K \to \mathbb{R}^n \\ &\sum_{i = 1}^n (x_i\otimes \alpha_i) \mapsto \sum_{i = 1}^n \left[ x_i\cdot j(\alpha_i)\right] \end{align} $$

The map $g$ is actually an isomorphism (we obviously know such an isomorphism exists without even writing this down if we agree that the tensor product here is an $n$ dimensional $\mathbb{R}$ vector space).

With these maps defined, note that $g\circ f = j$. So, this is an explicit way of writing down what you already know. Namely, you know before going through any of this that $\mathcal{O}_K$ embeds as a lattice in $\mathbb{R}^n$, and since we know $\mathbb{R}\otimes_\mathbb{Q}K\cong \mathbb{R}^n$ then clearly $\mathcal{O}_K$ must embed as a lattice in $\mathbb{R}\otimes_\mathbb{Q}K$ as well. It is perhaps more natural to think of the embedding as going initially into $\mathbb{R}\otimes_\mathbb{Q}K$ rather than into $\mathbb{R}^n$ because intuitively $\mathbb{R}\otimes_\mathbb{Q}K$ should be the "best" $\mathbb{R}$ vector space that we can embed $K$ into. It just so happens that in this case that vector space is indeed $\mathbb{R}^n$.


Here's the main thing of what's going on. Let $K$ and $L$ be fields containing a common field $k$. Consider the polynomial ring $k[t]$ over $k$. A basic fact about tensor products is that $k[t] \otimes_k L = L[t]$.

Assume $K$ is a finite separable extension of $k$. We can find $a \in K$ such that $K = k(a)$. If $f \in k[t]$ is the minimal polynomial of $a$ over $k$, then $K = k[t]/(f)$.

Taking tensor products is compatible with quotients. Combining this compatibility with the fact that $k[t] \otimes_k L = L[t]$, we get

$$K \otimes_k L = k[t]/(f) \otimes_k L = L[t]/(f)$$

where on the right, $(f)$ is now the ideal generated by $f$ in $L[t]$. In general, $f$ is not going to remain irreducible in $L[t]$. It will factor as $p_1 \cdots p_r$ for $p_i \in L[t]$ distinct and irreducible. Then by the Chinese remainder theorem,

$$L[t]/(f) = L[t]/(p_1) \times \cdots \times L[t]/(p_r)$$

where each $L[t]/(p_i)$ is a field extension of $L$.

Now take $k = \mathbb{Q}$ and $L = \mathbb{R}$, and go through the above construction. The only irreducible polynomials $p$ in $\mathbb{R}[t]$ are linear polynomials (in which case $\mathbb{R}[t]/(p) = \mathbb{R}$) or quadratic polynomials (in which case $\mathbb{R}[t]/(p) = \mathbb{C}$).

This shows that

$$K \otimes_{\mathbb{Q}} \mathbb R = \mathbb{R} \times \cdots \times \mathbb{R} \times \mathbb{C} \times \cdots \times \mathbb{C}$$

and if we are just consider this as a real vector space instead of a ring, each $\mathbb{C}$ can be identified with $\mathbb{R} \times \mathbb{R}$, so the tensor product becomes just a product of copies of $\mathbb{R}$.

  • $\begingroup$ In your first paragraph, you note that it's "a basic fact about tensor products" that $k[t]\otimes_k L=L[t]$. Should that make sense to me, or do you recommend I just remember it and try to apply it? $\endgroup$ Commented Mar 15, 2018 at 1:59
  • $\begingroup$ It's easy to convince yourself that it's true. Every element in there looks like $$1 \otimes a_0 + t \otimes a_1 + \cdots + t^n \otimes a_n$$ for $a_0, ... , a_n \in L$. $\endgroup$
    – D_S
    Commented Mar 16, 2018 at 15:22

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