I have Hamming code $Ham(3,4)$ which is $[21,18,3]$ code. What are the parameters for $Ham(3,4)\mid_{F_2}=Ham(3,4)\cap F_2^{21}$?
Parameters of subfield subcode of Hamming code
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This is not an answer but rather an extended comment on the last part of @JyrkiLahtonen's answer.
The columns of the parity check matrix of an arbitrary (not necessarily cyclic) Hamming code with redundancy $3$ over $\mathbb F_4$ are $21$ nonzero $3$-tuples over $\mathbb F_4$ with the property that no column is a scalar multiple of another column. In fact, one method of choosing the columns is to start with the set $S_{63}$ of $63$ nonzero $3$-tuples, pick one for the first column and delete its two scalar multiples from $S_{63}$ leaving a set $S_{60}$ of $60$ nonzero $3$-tuples; pick a $3$-tuple from $S_{60}$ as the second column and delete its two scalar multiples from $S_{60}$ leaving $S_{57}$; $\ldots$ and so on, till we have chosen $21$ columns. This gives an arbitrary linear $[21,18,3]_4$ Hamming code.
Jyrki wonders whether the subfield subcode of every linear $[21,18,3]_4$ Hamming code is necessarily a $[21,15,3]_2$ code (as happens with the cyclic Hamming code). Might the subfield subcode be a $[21,15,4]_2$ code for some noncyclic $[21,18,3]_4$ code? An alternative speculation is: Might the subfield subcode be a $[21,16,3]_2$ code in for some noncyclic $[21,18,3]_4$ code? Note that shortening a $[31,26,3]_2$ Hamming code results in a $[21,16,3]_2$ single-error-correcting code, that is, a $[21,16,3]_2$ code is not a figment of the imagination whose existence is prohibited by some bound or the other. Similarly, a $[21,15,4]_2$ code can be produced by shortening the extended $[32,26,4]_2$ Hamming code. The question is: are binary codes with these parameters obtainable as subfield subcodes of some suitably chosen $[21,18,3]_4$ code? Like Jyrki, I have no ready answer.
I assume that $Ham(3,4)$ means the linear code with alphabet $\Bbb{F}_4$ with a check matrix having as columns some maximal set of $21$ pairwise linearly independent vectors of $\Bbb{F}_4^3$ (points of a projective plane over $\Bbb{F}_4$ if you like). Otherwise I cannot make sense out of the question.
Even so, the answer may to some extent depend on the choice of particular set of $21$ columns. I only consider the following. Let $\beta\in\Bbb{F}_{64}$ be a fixed primitive root of unity of order $21$. One exists, because $21\mid (64-1)$. Define $Ham(3,4)$ as the $\Bbb{F}_4$-linear code determined by the check matrix $$ H=(1,\beta,\beta^2,\ldots,\beta^{20}). $$ This has the benefit of turning the resulting code into a cyclic one, and we do the familiar business of viewing codewords as (cosets of) polynomials in the ring $\Bbb{F}_4[x]/\langle x^{21}-1\rangle$. Then we easily see the following facts (leaving the details to you):
I am not sufficiently familiar with this business of subfield subcodes to be able to tell right away, whether you always get a $[21,15,3]$ code no matter how you select the $21$ columns of $H$. It would be my guess that this is the case though. After all, a single check equation over $\Bbb{F}_4$ amounts to two check equations over ${}{}{}{}\Bbb{F}_2$. However, it may be possible to select $H$ in such a way that there will be no words of weight three in the subfield subcode. Not sure whether the resulting code should then be called an alternant code or something else.