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  1. Let $\Bbb R$ be the set of real numbers and $r$ a fixed element of $\Bbb R.$ Let $T=\{ar\mid a\in\Bbb Q\}$ where $\Bbb Q$ is the set of rational numbers. Is $T$ a subring of $\Bbb R$?
  2. Let $R$ be a ring and $a$ be a fixed element in $R$ satisfying $a^2=a.$ Let $T=\{na\mid n\in\Bbb Z\}$ where $\Bbb Z$ is the set of integers. Is $T$ a subring of $R$?

I know that both of the proofs are similar to the one for:

Let $R$ be a ring and $b$ a fixed element of $R.$ Let $T=\{rb\mid r\in R\}.$ Prove $T$ is a subring of $R.$

Which is:

Since $R$ is a ring, $0_R\in R.$ Thus, $0_Rb = 0_R\in T.$ Hence, $T$ is nonempty.

For any $\overline{a},\overline{b}\in T,$ there exist $r_1, r_2\in R$ such that $\overline{a} =r_1b$ and $\overline{b}=r_2b.$

$$\overline{a}-\overline{b} = r_1b-r_2b = (r_1-r_2)b\in T,$$ since $r_1-r_2\in R,$ a ring. Hence, $T$ is closed under subtraction.

$$\overline{a}\overline{b}=(r_1b)(r_2b) = (r_1br_2)b\in T,$$ since $r_1br_2\in R$ ($r_1, b, r_2\in R,$ a ring). Hence, $T$ is closed under multiplication.

I know the proofs for #1 and #2 are similar to this proof but I'm not sure what needs to be changed.

For #2 I can't figure out what the differences in the proof are because of $a^2=a.$

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1) Let $t_1,t_2\in T$. Then $t_1=a_1r$ and $t_2=a_2r$ for $a_1,a_2\in \mathbb{Q}$. Clearly $t_1-t_2=(a_1-a_2)r$ is again in $T$, but $t_1t_2=a_1a_2r^2$ may not be of the form $ar$, for example for $r=2^{1/3}$ and $a_1=a_2=1$. So it is not a subring.

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The first question that comes to my mind is this: How do you know that the two statements are provable (that is, true)--much less that the proofs are similar to that of the other statement--if you aren't sure how to prove them? That should be a red flag for you: it lets you know that perhaps you're making too many assumptions, and that you aren't even aware of them all.

For example, the proof of #1 would be (effectively) the same if #1 had instead said the following: "Let $T=\{ar∣a∈\Bbb R\}.$" Unfortunately, that's not what it said, so the first claim isn't even true!

As for what relevance the "$a^2=a$" assumption holds for #2, it turns out that the conclusion fails if this assumption is removed from the hypothesis. That is, the following is false:

Let $R$ be a ring and $a$ be a fixed element in $R.$ Let $T=\{na∣n∈Z\}$ where $\Bbb Z$ is the set of integers. Is $T$ a subring of $R$?

Otherwise, the proof does turn out to be mostly similar to the one you've seen before. Try to prove it as I've stated above, and you'll hopefully see where the assumption $a^2=a$ comes into play. If not--or if you simply want to run your proof attempt by someone--feel free to leave a comment.

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