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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?

Solution:

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.

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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.

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  • $\begingroup$ 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
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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 $d\mid n$.

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  • $\begingroup$ 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
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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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