Difference between $F_2^n$ and $\Bbb F_2^n$ for a field

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!

• 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? – kodlu Mar 14 '18 at 21:56
• How is this about cryptography? – fkraiem Mar 15 '18 at 2:01

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.

• 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. :) – Silvia Rossi Mar 14 '18 at 20:32

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 $\mathbb F_{2^n}$ .