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A polynomial $f(x)\in \mathbb{Z}_q[x]$ is said to be square-free if it has no repeated roots.

The square-free factorisation of $f$ is

$$f(x)=\prod_{i=1}^k f_i(x)^i,$$ where each $f_i(x)$ is square free polynomial and $\gcd(f_i(x), f_j(x))$ is $1$ for $i\neq j$.

My basic doubt is why exponent $i$ is appearing in the factorisation? Why can't we then say that $f_2(x)^2$ divides $f(x)$ (suppose it is not $1$), which means $f$ is not square free.

PAMG
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  • That looks like a typo. – Sammy Black Sep 24 '24 at 16:39
  • If it were some other symbol, let's say $\epsilon_i$, then it could make sense each exponent were restricted to $\epsilon_i = 0$ or $\epsilon_i = 1$ for each $i$. This can be useful when comparing two or more factorizations, since we might like to name any factor that appears in any of the polynomials. – Sammy Black Sep 24 '24 at 16:41
  • But same symbol is used at other references also (e.g. on wikipedia). – PAMG Sep 24 '24 at 16:43
  • Oh! You're right. The factors $f_i$ are square-free, not the product $f$, in the definition given there of a square-free factorization. – Sammy Black Sep 24 '24 at 16:50
  • Oh, I'm wrong (and have deleted my incorrect comment). You aren't requiring that the $f_i$ be irreducible, so the expression makes sense. You can just gather together all the factors that have the same exponents. here is a reference. – lulu Sep 24 '24 at 16:56

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If you read the definition closely, it's a statement about factoring an arbitrary polynomial in a particular way involving increasing powers of square-free polynomials. The factors are square-free.

This is just a statement about sorting. Basically, $f_1$ is a product of irreducible factors that only appear once, $f_2^2$ is a product of irreducible factors that appear exactly twice, etc.

There's even an efficient algorithm due to Yun to construct these factors.

Sammy Black
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