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Find a generator polynomial of a 16-ary Reed-Solomon code of dimension 10.

My work so far:

First, $$k=q-1-\left(\delta -1\right)$$

$$10=15-1-\left(\delta -1\right)$$

$$\implies \delta =6 $$

and then noting that the generator polynomial $g(x)$ of a Reed-Solomon code will have degree $\delta-1$, which means $g(x)$ will have degree 5.

I then calculated $g(x)$ by noting that any $q$-ary Reed-Solomon code of length $q-1$ will be generated by a $g(x)$ such that

$$g(x) = \prod^{a+\delta -1}_{i=a+1}\left(x-\alpha^{i}\right), a \geq0, 2\leq\delta\leq q-1 $$

Setting $a=0$ and $\delta=6$, and letting $\alpha$ be a root of the the minimal polynomial $x^4+x+1$,

$$\prod^{5}_{i=1}\left(x-\alpha^i\right)=\left(x-\alpha\right)\left(x-\alpha^2\right)\left(x-\alpha^3\right)\left(x-\alpha^4\right)\left(x-\alpha^5\right)$$

expanding this I have

$$x^5-x^4\left(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha\right)-x^3\left(\alpha^9+\alpha^8+2\alpha^7+2\alpha^6+2\alpha^5+\alpha^4+\alpha^3\right)-x^2\left(\alpha^{12}+\alpha^{11}+2\alpha^{10}+2\alpha^9+2\alpha^8+\alpha^7+\alpha^6\right)=x\left(\alpha^{14}+\alpha^{13}+\alpha^{12}+\alpha^{11}+\alpha^{10}\right)-\alpha^{15}$$

and finally after using a log table for the primitive elements of GF$\left(2^4\right)$, and reducing everything modulo 2, I have $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$

So my question is, does this look like the correct generator for this code?

Update: I used the comments below to find $g(x)$ to be $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$
Is the final step to find a value $\alpha$ satisfying $\alpha^{3}= \alpha +1$?

Chairman Meow
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    The principle is correct. Cannot check all the details because you didn't tell us what is the minimal polynomial of $\alpha$. If we have $\alpha^4+\alpha+1=0$, then $$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha=\alpha^3+\alpha+1.$$ The other alternative is $\alpha^4+\alpha^3+1=0$ when we would have $$\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha=\alpha^3+\alpha^2.$$ Implying a problems with the coefficient of $x^4$. Mind you, plus or minus makes no difference when doing addition in characteristic two :-) – Jyrki Lahtonen Apr 24 '21 at 10:40
  • Thanks for the feedback. I think I have corrected my mistakes, and updated the question accordingly. – Chairman Meow Apr 24 '21 at 20:22
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    Using that $\alpha$ I got $$g(x)=x^5+(\alpha^3+\alpha+1)x^4+\alpha^2x^3+(\alpha^2+\alpha)x^2+\alpha x+1.$$ Does your table of powers of $\alpha$ match with mine? My $\gamma$ is your $\alpha$. I computed the linear and degree four terms by hand, and all of it with Mathematica (getting the same result). – Jyrki Lahtonen Apr 24 '21 at 21:10
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    Anyway, to answer the question in your last sentence. No! This is it. The element $\alpha$ is a constant. We fixed it when we described the field $GF(16)$. Asking about that is not unlike asking: Do we want to find a value of $2$ such that $2^3=2+1$? Why do you think we should try to do something like that? – Jyrki Lahtonen Apr 24 '21 at 21:11
  • I will check . . . it is very likely that I made a calculation error. This is my first attempt at this and I am trying to do it all by hand. Ahhh I see. That does not make much sense. I think that my hunch to do that was based off of an example in my text where they found a value of $\alpha$ using primitive roots mod 7 for a different (but similar) Reed Solomon code. – Chairman Meow Apr 24 '21 at 21:52
  • Yes, that table does match yours and I did make a mistake. Thanks for pointing that out, and for all of the help! You have no idea how much I appreciate it. – Chairman Meow Apr 24 '21 at 21:59

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After working through all of the comments a generator polynomial for this [15, 10, 6] Reed-Solomon code, where $\alpha$ is a primitive element of $x^4+x+1$, is $$g(x)=x^5+x^4\left(\alpha^3+\alpha+1\right)+x^3\left(\alpha^2\right)+x^2\left(\alpha^2+\alpha\right)+x\left(\alpha\right)+1$$

Chairman Meow
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