Prove that $\mathbb{F}_{2^4}$ is not isomorphic to a subring of $\mathbb{F}_{2^5}$.

So my attempt at proving this is as follows:

Proof: Since $\mathbb{F}_{2^5}$ is finite, every subgroup has finite order, and by Lagrange's Theorem we have that any subgroup can have order $2,4,8,16$ or $32$.

Proceeding from here is my trouble. I feel as if showing that no subgroup of order 16 is exists is the only plausible way to go, but showing this is what I am unsure about. Any hints/suggestions?


Proof: Consider the unit group of $\mathbb{F}_{2^5}$. The order of this unit group is $2^5 -1= 31$. Clearly, $31$ is prime, thus by Lagrange's Theorem we have that the only two subgroups are the trivial subgroups. Thus, adding the additive identity to the group of order one, we have $\mathbb{F}_2$, and similarly, adding the additive identity to the group of order 31, we have $\mathbb{F}_{2^5}$. Clearly, $\mathbb{F}_{2^4} \ncong \mathbb{F}_2 , \mathbb{F}_{2^5}$ since an isomorphism between finite fields can exists if they have the same number of elements.

Thank you to everyone's comments, they really helped.


Rather than looking at the additive group structure, you could perhaps look at the multiplicative group structure. Note that the multiplicative group of units in $\mathbb F_{2^n}$ only contains $2^n - 1$ elements, since it excludes the zero element.

  • $\begingroup$ Well, this implies that there are only 2 subgroups. Trivial subgroups. $\endgroup$ – El Spiffy Mar 15 '17 at 0:19
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    $\begingroup$ Yes! Indeed, $\mathbb F_{2^5}$ only has two subfields: $\mathbb F_2$ and $\mathbb F_{2^5}$ itself. $\endgroup$ – Kenny Wong Mar 15 '17 at 0:20
  • $\begingroup$ Hold on, I'm quite confused. If there is two trivial subgroups, how does this lead me to these two subfields? $\endgroup$ – El Spiffy Mar 15 '17 at 1:03
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    $\begingroup$ $\mathbb F_{2^4}$ is clearly not isomorphic to $\mathbb F_2$ or $\mathbb F_{2^5}$ because they contain different numbers of elements! $\endgroup$ – Kenny Wong Mar 15 '17 at 1:31
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    $\begingroup$ Slick. It avoids the delicate argument via field extension degrees, and I think one may see how it generalizes to other cases. Again, slick. $\endgroup$ – Lubin Mar 15 '17 at 3:10

You cannot show that since there are, in fact, several sub-groups of $\mathbb{F}_{2^5}$ of order $16$; note that, as a group, $$\mathbb{F}_{2^5}\cong C_2\times C_2\times C_2\times C_2\times C_2$$ where $C_2$ is the finite cyclic group of order $2$. However, there are no sub-rings of $\mathbb{F}_{2^5}$ of order $16$.

Here is a general guide:

  • Prove that there is a unique subring, let's call it $A$, of $\mathbb{F}_{2^5}$ that is isomorphic to $\mathbb{F}_2$ (a.k.a. $\mathbb{Z}/2\mathbb{Z}$), and that any subring of $\mathbb{F}_{2^5}$ must contain $A$. Hint: there is a unique ring homomorphism $\mathbb{Z}\to R$, for any ring $R$. Let $A$ be the image of the unique ring homomorphism $\mathbb{Z}\to\mathbb{F}_{2^5}$.

  • Prove that any subring of $\mathbb{F}_{2^5}$ is actually a field (more generally, any finite integral domain is a field).

  • Prove that if $K$ is a subfield of $L$ and $L$ is a subfield of $M$, then $[M:K]=[M:L][L:K]$.

  • Prove that $[\mathbb{F}_{2^5}:A]=5$. If $R$ is any subring of $\mathbb{F}_{2^5}$, what can the factorization $$5=[\mathbb{F}_{2^5}:A]=[\mathbb{F}_{2^5}:R][R:A]$$ look like, considering that $5$ is prime?

More generally, it's true that $\mathbb{F}_{p^d}$ is isomorphic to a subfield of $\mathbb{F}_{p^n}$ if and only if $d\mid n$.

  • $\begingroup$ I have a small notation barrier with this answer. Notation in the bracket means? $\endgroup$ – El Spiffy Mar 15 '17 at 0:50
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    $\begingroup$ Really nice answer. @ElSpiffy: if $L$ is a subfield of $M$, then $M$ is naturally a vector space over $L$ (why?). The notation $[M:L]$ is the dimension of $M$ as an $L$-vector space. We say that $M$ is a field extension of $L$, and $[M:L]$ is called the degree of the extension. $\endgroup$ – Alex Wertheim Mar 15 '17 at 1:05
  • $\begingroup$ @AlexWertheim Thank you. I have know what these terms are but have never encountered these notations. $\endgroup$ – El Spiffy Mar 15 '17 at 1:08
  • $\begingroup$ Right you are. That’s the proof I would have given. $\endgroup$ – Lubin Mar 15 '17 at 3:12

If $\mathbf F_{16}$ were a subfield of $\mathbf F_{32}$, the answer is easy : the order of $\mathbf F^*_{16}$ would divide that of $\mathbf F^*_{32}$ (Lagrange), which is impossible. One could also note directly that $\mathbf F_{p^m} \subset \mathbf F_{p^n}$ if and only if $m$ divides $n$ (multiplication of degrees in a tower of extensions).

To reduce to this situation, just show that a finite domain $D$ is necessarily a field : because $D$ is finite, for any non zero $x \in D$, there necessarily exist exponents $n > m$ such that $x^m = x^n$, or equivalently $x^m (x^{n-m} -1)=0$; because $D$ has no zero divisor, this implies that $x$ is invertible.

NB. After posting my answer, I realize that it is just a rephrasing of that of @Zev Chonoles. Note however that the original problem does not make much sense. Because if $f$ is a ring isomorphism $K \to R \subset L $, where $K$ is a field and $R$ is a subring of a field $L$, necessarily $f(0)=0$ (exploiting the additive group structure), which implies $f(1)=1$ (by the above argument since, as a subring of a field, $R$ is automatically a domain), from which it follows that $R$ is a field (just because if $xy=1$ in $K$, then $f(x)f(y)=f(1)=1$ in $R$) .


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