Show a subring contains certain elements. Show that the set of all real numbers of the form
  $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$
  with $n≥0$ and $a_i ∈ \mathbb{Z}$ is a subring of $R$ that contains $\mathbb{Z}$ and $\pi$.


Proof. We show that the set of all real numbers of the form
$a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ (call it the set $A$) contains $\mathbb{Z}$, $\pi$, and the set is closed under subtraction and multiplication of its elements.

Take $a_1 = 1$ and $(a_n)_{n \in \{0\} \cup \mathbb{N} \setminus \{1\}}$. Then $\pi$ is in $A$.

Consider the set $D = \{\pi^n : n \in \mathbb{N}\}$. It becomes evident this can be re-written as $aD = Da : a \in A$. $A$ clearly has a bijection with $\mathbb{Z}$.

Another way to show this is to construct a function $g$ such that it sends $\pi^n$ to $1$. Thus we will make $g: D \twoheadrightarrow 1$. In this fashion I want to show I can use the function $g$ on $C$ to get it to correspond to $c_0 + c_1 + c_2 + \dots + c_n$. In this sense we can correspond to each $c_i$ a natural number. Since the naturals share the same cardinality with the integers, we are done. 

Comment: I am annoyed at how I had to use set theory to arrive at my answer. Perhaps there is a better way, algebraically?

To show that $A$ is closed under subtraction, take any real number $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ in $A$ and subtract $b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n$ (also in $A$) from it to get $$(a_0 - b_0) + (a_1 - b_1)\pi + (a_2 - b_2)\pi^2 +\cdots+ (a_n - b_n)\pi^n$$

Clearly subtraction of integers yields an integer, and since $\mathbb{Z}$ is contained in $A$, $A$ is closed under subtraction. 

For multiplication, I have no idea. It gets quite messy in my head.



Questions. First off, what is it called when a set contains a ring? Is it called an ideal? Or a coset? For example, when I said "We show that the set of all real numbers of the form
$a_0 + a_1\pi + a_2\pi2 +\cdots+ a_n\pi^n$ (call it the set $A$) contains $Z$". I know $a_i \in Z$ and I have seen examples like $bZ = Zb : b \in B$ but I don't know what that's called.

Secondly, what is $\mathbb{Z}$ called? A ring? What is $\mathbb{Z} \mod 2$ called? An ideal in the ring? Just making sure.

Lastly, how was my proof written - subjectively, what did you like and dislike about the proof? Is it correct in its entirety?
You seem to be confused about various definitions :


The set
$$
A := \{a_0 + a_1\pi + a_2\pi^2 + \ldots + a_n\pi^n : a_i \in \mathbb{Z}\}
$$
forms a ring because, as you say, it forms a group under addition, and for any
$$
a = \sum_{i=0}^n a_i\pi^i, \text{ and } b:= \sum_{j=0}^m b_j\pi^j
$$
you have
$$
ab = \sum_{k=0}^{n+m} c_k\pi^k, \text{ where } c_k = \sum_{i=0}^k a_ib_{k-i}
$$
This is called a "Cauchy product" of two polynomials.
$\mathbb{Z}$ is a ring, $\mathbb{Z}/2\mathbb{Z}$ is a quotient ring (not an ideal).
For subtraction, you did it perfectly. For multiplication you need to do some algebra: 

we know that $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ is in our set and, if we take $b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n$ to be another member of our set,
we get that :
\begin{align}(a_0 + a_1\pi + a_2\pi^2 +&amp; \cdots+ a_n\pi^n)(b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n)\\
&amp;=
(a_0b_0) + \pi(a_0 b1 + b_0 a_1) + \pi^2(a_2 b_0 + b_2a_0 + a_1b_1) + \cdots + (a_nb_n)\pi^{2n}
\end{align}

Verify that the coefficients of this product are integers, and that the product is in the set, and you are finished