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  1. Is $\mathbb{F}_5$ a subfield of $\mathbb{F}_7$? I can think of the answer 'yes' because they have the same set op operations $+ \cdot$ and the answer 'no' because in $\mathbb{F}_5: 2\cdot3=1$ and in $\mathbb{F}_7: 2\cdot3=6$.

  2. When I consider the finite field with four elements $\mathbb{F}_4$: $\{0,1,\omega,\omega^2=\omega+1\}$ as being $\mathbb{F}_2 \times \mathbb{F}_2$ how do I prove or know that in this field $1+1=0$ like in $\mathbb{F}_2$?
    EDIT: by $\mathbb{F}_2 \times \mathbb{F}_2$ I mean that the product may be defined in a complicated way, e.g. $(a,b)\cdot(c,d)=(ac+bd,ad+bc+bd)$. Unfortunately I don't know the correct notation.

  3. Can it be proved also for the field with 8 elements $\mathbb{F}_8 = \mathbb{F}_2 \times \mathbb{F}_2\times \mathbb{F}_2$?

  4. Is it possible to enumerate the elements of $\mathbb{F}_8$ like an extension of the elements of $\mathbb{F}_4$: $\{0,1,\omega,\omega^2=\omega+1, \gamma, \gamma^2, \ldots, \delta, \ldots\} $

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  • $\begingroup$ In what way do $\mathbb{F}_5$ and $\mathbb{F}_7$ have the same operations? $\endgroup$ – Tobias Kildetoft Jun 27 '15 at 20:29
  • $\begingroup$ Note that in $\mathbb F_2\times \mathbb F_2$ you have $(0,1)\cdot (1,0)=(0,0)$ so it isn't a field. $\endgroup$ – Mark Bennet Jun 27 '15 at 20:39
  • $\begingroup$ Notice my edits: you had the curly braces outside of MathJax, thus: {$0,1$}. I put them inside, thus: $\{0,1\}$. That avoids mismatches in fonts and assures proper spacing and horizontal alignment. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jun 27 '15 at 20:41
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    $\begingroup$ @MarkBennet see my edit $\endgroup$ – Gerard Jun 27 '15 at 20:56
  • $\begingroup$ @Gerard: All these are valid questions, great that you are asking. $\endgroup$ – Orest Bucicovschi Jun 27 '15 at 21:43
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  1. No. A finite field $\mathbf F_{p^m}$ is a subfield of the finite field $\mathbf F_{q^n}$ if and only if $p=q$ and $m\mid n$.
  2. and 3. $\;\mathbf F_2$ is (isomorphic to) a subfield of each of $\mathbf F_{2^m}$, and if $1\cdots 2=0$ in $\mathbf F_2$, it remains true in all $\mathbf F_{2^m}$.

    1. $\;\mathbf F_4 \,$ is not a subfield of $\;\mathbf F_8$, so your question is meaningless.
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    $\begingroup$ $\mathbf F_8=\mathbf F_{2^3}$. $\mathbf F_4$ is a subfield of $\mathbf F_{16}=\mathbf F_{2^4}$. $\endgroup$ – Bernard Jun 27 '15 at 21:15
  • $\begingroup$ Oops! :) Sorry! Bad arithmetic here! :) I'll delete my comment shortly... $\endgroup$ – paul garrett Jun 27 '15 at 21:30
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  1. If $K\le L$ fields, then $L$ is a vector space over $K$, in particular, so $|L|=|K|^{d}$ where $d=\dim_KL$.
  2. Start adding $1$ with itself. In a finite ring there's a smallest $n$ such that $n\cdot 1=0$. If the ring has no zero divisors then the smallest $n$ must be prime. Note that in case of a field, this set $\{0,1,1+1,\dots\}$ will be a subfield.
  3. Yes.
  4. Not exactly like that. For $\Bbb F_8$ you need to find an irreducible polynomial of degree $3$, and adjoin its root formally.

Note also that, as rings (or fields) we don't have $\Bbb F_4\cong\Bbb F_2\times\Bbb F_2$ or $\Bbb F_8\cong \Bbb F_2\times\Bbb F_2\times\Bbb F_2$, this is only valid for their underlying additive group.

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$\mathbb F_5$ is definitely not a subfield of $\mathbb F_7$, for the reason you mention. The operations are not the same operations if they don't have the same values when given the same arguments.

In a field with four elements, you have $1\ne0$, but if $1+1\ne0$, then the set $\{0,1,1+1\}$ contains three of the four elements. It cannot be a subgroup because $3$ does not divide $4$, or, to put it another way, it would have to have at least one coset consisting of three other elements, and there aren't that many other elements. So you'd have to have $1+1+1$ as another non-zero element. Then you would have the problem of what the multiplicative inverse of $1+1$ is. Notice that $(1+1)^2 = 1+1+1+1$ (by the distributive law), and that $=0$. If the square of some element is $0$, can that element have a multiplicative inverse?

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