# Abstract Nonsense

## The Chinese Remainder Theorem (Pt. II)

Point of Post: This is a continuation of this post.

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Let’s see if we can give an example of the CRT which doesn’t involve $\mathbb{Z}$. In particular, we shall show that the CRT can help identify $F[x]$ mod certain ideals. In particular, we have the following

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Theorem: Let $F$ be a field and $z_1,\cdots,z_n\in F$ be $n$-distinct non-zero numbers. Then, $F[x]/[(x-z_1)\cdots (x-z_n)]\cong F^n$ (where the denominator is the product of the ideals $(x-z_1),\cdots,(x-z_n)$ not the ideal generated by $(x-z_1)\cdots(x-z_n)$ although, this is a completely artificial distinction, since, they are equal).

Proof: What we claim first is that $(x-z_i),(x-z_j)$ are coxmaximal for each distinct $i,j\in[n]$. Indeed, we merely note that $\displaystyle x-z_i+-(x-z_j)=z_i-z_j\in (x-z_i)+(x-z_j)$ and since $z_i-z_j\ne 0$ we have that $z_i-z_j$ is a unit and so $(x-z_i)+(x-z_j)=F[x]$. It thus frollows from the CRT that $F[x]/[(x-z_1)\cdots (x-z_n)]\cong F[x]/(x-z_1)\times\cdots\times F[x]/(x-z_n)$. That said, recall that $F[x]/(x-z_j)\cong F$ by the first isomorphism theorem and the epimorphism $p\mapsto p(z_n)$ (we technically don’t know that $p(x)$ has a zero at $z_j$ implies that $p$ is divisible by $x-z_j$, but it’ll soon be proven, and not hard to believe). The conclusion follows. $\blacksquare$

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Corollary: Let $a_1x-z_1,\cdots,a_nx-z_n\in F[x]$ be such that $\displaystyle a_1,\cdots,a_n$ are non-zero and $\displaystyle \frac{z_1}{a_1},\cdots,\frac{z_n}{a_n}$ are distinct numbers. Then, $F[x]/[(a_1x-z_1)\cdots(a_nx-z_n)]\cong F^n$.

Proof: Merely note that $\displaystyle (a_jx-z_j)=\left(a_j(x-\frac{z_j}{a_j})\right)=\left(x-\frac{z_j}{a_j}\right)$. $\blacksquare$

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Number Theoretic Implications

Not only does the CRT have important pure ring theoretic implications but it also implies a few classic number theoretic facts. Most obviously is the refinement of the one given in the motivation of this post:

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Theorem: Let $n_1,\cdots,n_m$ be pairwise coprime numbers in $\mathbb{Z}$. Then, the system of simultaneous linear equations

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$\left\{\begin{array}{c}x\equiv a_1\text{ mod }n_1\\ \vdots\\ x\equiv a_m\text{ mod }n_m\end{array}\right\}$

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has a solution $x\in\mathbb{Z}$ which is unique modulo $n_1\cdots n_m$.

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That said, it has much more satisfying, applicable number theoretic applications. For example:

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Theorem: Let $n$ be a natural number with prime factorization $p_1^{a_1}\cdots p_m^{a_m}$ and $P(x)\in\mathbb{Z}[x]$. Then, the equation $P(x)\equiv 0\text{ mod }n$ has a solution if and only if $P(x)\equiv 0\text{ mod }p_j^{a_j}$ has a solution for each $j=1,\cdots,m$. Moreover, if $N_k(P)$ (for some generic integer $k$) denotes the number of solutions $P(x)\equiv 0\text{ mod }k$ then $N_P(n)=N_P(p_1^{a_1})\cdots N_P(p_m^{a_m})$.

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Indeed, this follows immediately from the CRT used in conjunction with following lemma:

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Theorem: Let $R$ and $S$ be rings and $a_0+\cdots+a_nx^n=P(x)\in R[x]$ then if $f:R\to S$ is a monomorphism and $Q(x)\in S[x]$ is given by $Q(x)=f(a_0)+\cdots+f(a_n)x^n$ then the map $F:\mathbb{V}_R(P)\to\mathbb{V}_S(Q)$ (where $\mathbb{V}_R(P)=\left\{x\in R:P(x)=0\right\}$ and $\mathbb{V}_S(Q)$ is defined similarly) given by $x\mapsto f(x)$ is an injection.

Proof: To see that $F$ is really well-defined (in the sense that $F$ maps $\mathbb{V}_R(P)$ into $\mathbb{V}_S(P)$) we merely note that if $P(x)=0$ then $f(P(x))=0$, but since $f$ is a morphism, we know that $f(P(x))=P(f(x))$ and so $P(f(x))=0$. Since $f$ is injective evidently we have that $F$ is injective. $\blacksquare$

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Corollary: Since evidently if $f$ is an isomorphism the map $G:\mathbb{V}_S(P)\to\mathbb{V}_R(P)$ given by $x\mapsto f^{-1}(x)$ is a two-sided inverse to $F$ we have that if $f$ is an isomorphism then $\#(\mathbb{V}_R(P))=\#(\mathbb{V}_S(P))$.

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To see why this proves the stated theorem we first note that since $(P(x_1),\cdots,P(x_n))$ equals (by definition) $P((x_1,\cdots,x_n))$ and so

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$\#(\mathbb{V}_{\mathbb{Z}/(p_1^{a_1})\times\cdots\times\mathbb{Z}/(p_m^{a_m})}(P))=\#(\mathbb{V}_{\mathbb{Z}/(p_1^{a_1})}(P))\times\cdots\times\#\mathbb{V}_{\mathbb{Z}/(p_m^{a_m})}(P))=N_P(p_1^{a_1})\cdots N_P(p_m^{a_m})$

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But, by the above corollary since $\mathbb{Z}/(n)\cong\mathbb{Z}/(p_1^{a_1})\times\cdots\times\mathbb{Z}/(p_m^{a_m})$ we have that

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$N_P(n)=\#(\mathbb{V}_{\mathbb{Z}/(n)}(P))=\#(\mathbb{V}_{\mathbb{Z}/(p_1^{a_1})\times\cdots\times\mathbb{Z}/(p_m^{a_m})})$

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From where the second assertion of the theorem follows. The first assertion is immediate from the second  since it clearly is equivalent to $N_P(n)=0$ if and only if $N_P(p_1^{a_1})=\cdots=N_P(p_m^{a_m})=0$.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.