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Replace p with m notation
p_i should only be used for primes, while CRT works for any coprime m_i (esp. case of prime powers p^e)
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src/algebra/chinese-remainder-theorem.md

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@@ -10,36 +10,36 @@ The Chinese Remainder Theorem (which will be referred to as CRT in the rest of t
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## Formulation
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Let $p = p_1 \cdot p_2 \cdots p_k$, where $p_i$ are pairwise relatively prime. In addition to $p_i$, we are also given a set of congruence equations
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Let $m = m_1 \cdot m_2 \cdots m_k$, where $m_i$ are pairwise relatively prime. In addition to $m_i$, we are also given a set of congruence equations
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$$\begin{align}
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a &\equiv a_1 \pmod{p_1} \\\\
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a &\equiv a_2 \pmod{p_2} \\\\
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a &\equiv a_1 \pmod{m_1} \\\\
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a &\equiv a_2 \pmod{m_2} \\\\
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&\ldots \\\\
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a &\equiv a_k \pmod{p_k}
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a &\equiv a_k \pmod{m_k}
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\end{align}$$
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where $a_i$ are some given constants. The original form of CRT then states that the given set of congruence equations always has *one and exactly one* solution modulo $p$.
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where $a_i$ are some given constants. The original form of CRT then states that the given set of congruence equations always has *one and exactly one* solution modulo $m$.
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### Corollary
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A consequence of the CRT is that the equation
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$$x \equiv a \pmod{p}$$
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$$x \equiv a \pmod{m}$$
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is equivalent to the system of equations
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$$\begin{align}
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x &\equiv a_1 \pmod{p_1} \\\\
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x &\equiv a_1 \pmod{m_1} \\\\
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&\ldots \\\\
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x &\equiv a_k \pmod{p_k}
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x &\equiv a_k \pmod{m_k}
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\end{align}$$
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(As above, assume that $p = p_1 p_2 \cdots p_k$ and $p_i$ are pairwise relatively prime).
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(As above, assume that $m = m_1 m_2 \cdots m_k$ and $m_i$ are pairwise relatively prime).
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## Garner's Algorithm
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Another consequence of the CRT is that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. From calculations we can see that $p$ has around $3000$ digits.
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Another consequence of the CRT is that we can represent big numbers using an array of small integers. For example, let $p$ be the product of the first $1000$ primes. From calculations we can see that $P$ has around $3000$ digits.
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Any number $a$ less than $p$ can be represented as an array $a_1, \ldots, a_k$, where $a_i \equiv a \pmod{p_i}$. But to do this we obviously need to know how to get back the number $a$ from its representation. In this section, we discuss Garner's Algorithm, which can be used for this purpose. We seek a representation on the form
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