0
$\begingroup$

I am confused between the notation $F_2^n$ and $\Bbb F_2^n$ for a field in regards to codes.

I thought that $F_2^n$ and $\Bbb F_2^n$ were both fields composed by codes of length n and entries in mod 2, but I now know that this cannot be true. What is the difference between them?

For example, if code C is consists of all the codewords in $F_n^2$ that have even weight, to show that C is a linear code, do I use the fact that it is a subspace of $F_2^n$ or $\Bbb F_2^n$?

Thank you!

$\endgroup$
  • $\begingroup$ A field has to have multiplicative inverses. These are vector spaces, only scalar multiplication is defined, and no inverses exist, what would be the multiplicative inverse for a vector? $\endgroup$ – kodlu Mar 14 '18 at 21:56
  • $\begingroup$ How is this about cryptography? $\endgroup$ – fkraiem Mar 15 '18 at 2:01
2
$\begingroup$

Some authors write $F_q$ for the finite field of order $q$. Some authors write $\mathbb F_q$. Some authors write $\mathbf F_q$. Unless there's a typo, or unless your pet author is particularly confusing and as a personal peculiarity makes a nonstandard semantic distinction between them, which one you use is a matter of taste. And as we all know—de gustibus non disputandum est: only with gusto can taste be argued.

What parts of the text you are reading are leading you to a state of confusion? Maybe you can ask a question about those in particular.

$\endgroup$
  • $\begingroup$ Thank you for your answer. What lead to my confusion was that my professor wrote down in the notes that $F= \Bbb F_q$ and that $F^n= \Bbb F_q^n$, so I thought that $\Bbb F$ simply corresponded to the finite version of the field $F$. But then in the homework used $F$ and $\Bbb F$ interchangeably for finite fields. Now I think that my professor simply picked homework problems from different sources leading to inconsistencies in the notation. :) $\endgroup$ – Silvia Rossi Mar 14 '18 at 20:32
2
$\begingroup$

Your $F_2^n$ and $\mathbb F_2^n$ are exactly the same object, just typeset differently!

They denote the vector space of dimension $n$ (the exponent) over the binary field $\mathbb F_2$ with two elements. Sometimes $GF(2)$ is also used.

Compare with vector space of dimension 2 over the reals, the $xy$plane, which is denoted $\mathbb R^2.$

Now, since the elements of $\mathbb F_{2^n}$ which is the field of $2^n$ elements can also be written in the polynomial representation $$ a_0+a_1 x+ \cdots+ a_{n-1} x^{n-1},\quad a_i \in \mathbb F_2, $$ modulo an irreducible polynomial of degree $n$, one can set up a one to one correspondence between $(a_0,\ldots, a_{n-1}) \in \mathbb F_2^n$ and the field $\mathbb F_{2^n}$ .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.