Proof: $S = \{[2m]_{1000} : m = 0, 1, ... , 500\}$ is an Ideal of $R = \mathbb{Z}_{1000}$

Question

It is given in the question that $S \subset R$, so it is not necessary to prove this, in my understanding I only need to show that $S$ is closed under addition and multiplication with elements from $R$.

So, I have the following proof, which makes sense to me but seems a bit long winded. My question is, is this proof strong enough, and also any suggestions on how to simplify the proof.

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$$ \begin{align} S &= \{[2m]_{1000} : m = 0, 1, ... , 500\}\\ &= \{[n]_{1000} : n = 0, 2, ... , 1000\}\\ &= \{k \in \mathbb{Z} : k = 0, 2, ... , 996, 998\} \end{align} $$ Let $a , b \in S$. Then, I claim $[a+b]_{1000} \in S$.

As $a,b$ can be written as $a = 2m$, $b=2n$ for arbitrary $\{m,n \in \mathbb{Z} : m = 0, 1,...,499 \;, n = 0, 1,...,499\}$, we have $a+b = 2(m+n) = 2k$ for $k = 0, 2, ..., 998$. Thus, $$ \begin{align} [a+b]_{1000} &= [2k]_{1000}, \; k=0,2,...,998\\ &= [2k]_{1000}, \; k=0,2,...,998, 1000 \;\; ([2\cdot0]_{1000} = [2\cdot1000]_{1000} = 0)\\ &= [h]_{1000}, \; h=0,2,...,998,1000 \;\; (h=2k)\\ &= [2j]_{1000}, \; j=0,1,...,499, 500 \;\; (j = h/2) \; \in S \end{align} $$

Thus, $[a+b]_{1000} \in S$, and so $S$ is closed under addition.

Now, consider $a \in S$ and $\alpha \in R$. I claim $[a\cdot\alpha]_{1000} \in S$.

$a$ can be written as $a = 2m$ for $m=0,1,...,499$. So, $$ \begin{align} [a\cdot\alpha]_{1000} &= [2m\alpha]_{1000} \;\; \alpha=0,1,...,999\;,m=0,1,...,499\\ &= [2k]_{1000} \;\; k = 0,1,...,499, 500,...,(499\cdot999)\\ &= [h]_{1000} \;\; h = 0, 2,...,998, 1000,..., 2(499\cdot999)\;\; (h=2k)\\ &= [h]_{1000} \;\; h = 0, 2,...,998, 1000\\ &= [2j]_{1000} \;\; j = 0,1,...499, 500 \;\; (j = h/2) \; \in S \end{align} $$ Thus $[a\cdot\alpha]_{1000} \in S$, and so S is closed under multiplication from $R$.

Therefore, S is an ideal of R.

Answer 1

Consider $\phi:\Bbb{Z}_{1000}\to \Bbb{Z_2}$ defined by $$\phi([a]_{1000})=[a]_{ 2}$$

1. $\phi$ is a ring homomorphism. $\begin{align}\phi([a]_{1000}+[b]_{1000})&=\phi([a+b]_{1000})\\&=[a+b]_2\\&=[a]_2+[b_2]\\&=\phi([a]_{1000})+\phi([b]_{1000})\end{align}$

$\begin{align}\phi([a]_{1000}[b]_{1000})&=\phi([ab]_{1000})\\&=[ab]_2\\&=[a]_2[b_2]\\&=\phi([a]_{1000})\phi([b]_{1000})\end{align}$

2. $\ker\phi=S$

$\begin{align}ker \phi&=\{[a]_{1000}\in \Bbb{Z}_{1000}: \phi([a]_{1000}=[0]_2\}\\&=\{[a]_{1000}\in\Bbb{Z}_{1000} :[a]_2=[0]_2\}\\&=\{[a]_{1000}\in\Bbb{Z}_2:2\mid a\}\\&=\langle [2]_{1000}\rangle\end{align}$