# Why is the block size chosen to be q-1 for Reed-Solomon codes?

Consider a Reed-Solomon code over a finite field of $$\mathbb{F}_q$$. Why is the typical block size chosen to be $$q-1$$ [1][2][3]? The reasoning I saw around this is that in order to maximize the rate for a fixed minimum distance. But wouldn't choosing $$q$$ as the code size would be even slightly better? Does it have to do something with how BCH decoders work? Wikipedia lists both $$q$$ and $$q-1$$ as a possible good choice for the block size, but doesn't cite any reference.

[1]: Manz: Fehlerkorrigierende Codes
[2]: Pretzel: Error-correcting Codes and Finite Fields
[3]: Hoffman: Coding Theory: The Essentials

• did you see my updated answer? Dec 19, 2023 at 13:26

Reed Solomon codes are cyclic. A cyclic code [over whatever Galois field] of length $$n$$ is defined with respect to reduction modulo $$x^n-1$$ to get the cyclic property.
Now, the generator polynomial $$g$$ and the parity check polynomials $$h$$ are defined by $$g(x)h(x)=x^n-1,$$ more exactly one is chosen and then the other one is determined using this equation.
The factorization of $$x^n-1$$ is crucial in determining the properties of the code. The nonzero elements of a finite field $$GF(q)$$ obey the equation $$x^{q-1}-1=0,$$ and form the multiplicative group $$GF(q)^{\ast}.$$
Thus the only choice is to make the length of the code either $$q-1$$ or a divisor of $$q-1.$$ This is because, it is known that $$x^n-1$$ is divisible by $$x^m-1$$ if and only if $$n$$ is divisible by $$m.$$ Usually we take $$n=q-1,$$ for efficiency since it gives the longest code for a fixed alphabet size.
Remark: For Reed-Solomon which have alphabet $$GF(q)$$ the factorization is trivial (no cyclotomic cosets need to be computed which are used to determine minimum distance for binary codes). So we take $$g$$ to have a sequence of consecutive powers of $$\alpha$$ the primitive element as its roots.