I perfectly understand the tensor product of vector spaces over finite fields. But when I regard these vector spaces as finite fields I get confused.

Let the vector spaces $\mathbb{F}_p^m$ and $\mathbb{F}_p^n$ over the finite field $\mathbb{F}_p$ be given. Then their tensor product $\mathbb{F}_p^m\otimes \mathbb{F}_p^n$ is the vector space $\mathbb{F}_p^{mn}$.

The vector spaces $\mathbb{F}_p^m$ and $\mathbb{F}_p^n$ can also be considered as finite fields $\mathbb{F}_{p^m}$ and $\mathbb{F}_{p^n}$. For that, let $r,s\in\mathbb{F}_p[X]$ be irreducible with $\deg(r)=m$ and $\deg(s)=n$. Hence, $\mathbb{F}_{p^m}\cong\mathbb{F}_p[X]/r$ and $\mathbb{F}_{p^n}\cong\mathbb{F}_p[X]/s$.

Is there a canonical way to express the tensor product of these fields in terms of $r$ and $s$? Something like $\mathbb{F}_p[X]/r\otimes \mathbb{F}_p[X]/s\cong\mathbb{F}_p[X]/(r\otimes s)$. How can $r\otimes s$ be defined? Is it the insertion of $r$ into $s$, because $deg(r(s(x)))=mn$? But is the insertion of irreducible polynomials into each other always irreducible? Does anyone know a piece of literature dealing with this problem?




2 Answers 2


A bit more detailed look at what Qiaochu said. Unfortunately my answer won't really be expressed in terms of $r$ and $s$. I hope it still helps you in some way.

We know that $\Bbb{F}_p[X]\otimes \Bbb{F}_{p^n}\cong\Bbb{F}_{p^n}[X]$ and that $\Bbb{F}_{p^n}$ is a flat $\Bbb{F}_p$-module. Let us consider the short exact sequence $$0\to\Bbb{F}_p[X]\to\Bbb{F}_p[X]\to\Bbb{F}_p[X]/\langle r\rangle\to0,$$ where the first map is multiplication by $r$, and the last module is isomorphic to $\Bbb{F}_{p^m}$. Upon tensoring with $\Bbb{F}_{p^n}$ this gives rise to the short exact sequence $$0\to\Bbb{F}_{p^n}[X]\to\Bbb{F}_{p^n}[X]\to\Bbb{F}_{p^n}[X]/\langle r\rangle\to0.$$ Therefore a comparison of the last modules shows that $$ \Bbb{F}_{p^m}\otimes \Bbb{F}_{p^n}\cong \Bbb{F}_{p^n}[X]/\langle r\rangle. $$

The polynomial $r$ has no multiple zeros in $\overline{\Bbb{F}_p}$, so over $\Bbb{F}_{p^n}$ it factors into a product of distinct factors $$ r=\prod_{i=1}^t r_i $$ for some irreducible polynomials $r_i\in\Bbb{F}_{p^n}[X]$. Because these factors are distinct, the Chinese remainder theorem tells us that $$ \Bbb{F}_{p^n}[X]/\langle r\rangle\cong\bigoplus_i \Bbb{F}_{p^n}[X]/\langle r_i\rangle. $$

Note that everything above applies equally well to any finite extension of fields $L/K$. There is no need for the fields $L,K$ to be finite. We only needed the polynomial $r$ to be separable, so that we avoided the possibility of repeated factors.

The next step is, as Qiaochu pointed out, specific to Galois extensions. Namely, we can also deduce that the factors $r_i$ are Galois conjugates of each other. Most notably they all have the same degree. In the case of finite fields we can see this more concretely, because we know that the Galois group consists of powers of the Frobenius automorphism $F:x\mapsto x^p$. The zeros of $r$ are $$ \alpha,\alpha^p,\alpha^{p^2},\ldots,\alpha^{p^{m-1}} $$ where $\alpha$ is some (fixed) zero of $r$. For example $\alpha=X+\langle r\rangle$. The roots of one of the factors $r_i$ are then lists like $$ \alpha^{p^i},\alpha^{p^{i+n}},\alpha^{p^{i+2n}},\ldots $$ because we get such lists of conjugates by applying powers of $F^n$ to one of them.

The original list of $m$ roots consisted of a single orbit of the Galois group $G=\langle F\rangle$. This list is now partitioned into orbits of the subgroup $H=\langle F^n\rangle$. Basic facts about actions of cyclic groups tells us that the $H$-orbits all have size $m/\gcd(m,n)$, and that there are $\gcd(m,n)$ of them. Therefore we get $$ \Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}\cong\bigoplus_{i\in D}\Bbb{F}_{p^n}(\alpha^{p^i}), $$ where the set $D=\{0,1,\ldots,\gcd(m,n)-1\}$ consists of representatives of those orbits. It is easy to see that all those fields $$\Bbb{F}_{p^n}(\alpha^{p^i})\cong \Bbb{F}_{p^\ell}$$ with $\ell=\operatorname{lcm}(m,n)$.

Summary: $\Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}$ is isomorphic to a direct sum of $\gcd(m,n)$ copies of $\Bbb{F}_{p^\ell}$ where $\ell=\operatorname{lcm}(m,n)$. In particular, $\Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}$ is a field if and only if $\gcd(m,n)=1$.

  • $\begingroup$ Thank you very much for this detailed explanation. It will take me some more time and reading to fully understand it. But I really appreciate it. The summary was nice too. $\endgroup$
    – Chris
    Nov 16, 2015 at 12:01
  • $\begingroup$ This leads me to a further question. How can I represent or calculate elementary tensors. Let $m=2,n=3,p=3$. If I consider a polynomial representation of $\mathbb{F}_{3^2}$ and $\mathbb{F}_{3^3}$. How can I express or calculate the elementary tensor, lets say $(x+1)\otimes(x^2+1)$? $\endgroup$
    – Chris
    Nov 16, 2015 at 12:06

The tensor product is $\mathbb{F}_p[X, Y]/(r(X), s(Y))$. More generally, tensor products of commutative rings can be computed by "concatenating" their presentations.

This tensor product will usually fail to be a field. For example, $\mathbb{F}_{p^n} \otimes \mathbb{F}_{p^n}$ turns out to be the direct product $\prod_{i=1}^n \mathbb{F}_{p^n}$. This is a special case of a more general fact about Galois extensions.


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