# 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?

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1. 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.
2. $\mathbb{Z}$ is a ring, $\mathbb{Z}/2\mathbb{Z}$ is a quotient ring (not an ideal).
Okay, so suppose I have a function $\phi : R[x] \to R$ and it maps each polynomial in the domain to its constant term. Clearly it is a surjective homomorphism and not one-to-one since we can have $\mathbb{R}\_{2\mathbb{R}}[x]$ and $\mathbb{R}\_{3\mathbb{R}}[x]$ and pick two elements and show it is operation preserving. Correct? –  Don Larynx Oct 28 '13 at 6:16
It isn't injective, because $1+x$ and $1+2x$ map to the same element, but what do you mean by $\mathbb{R}_2\mathbb{R}[x]$? –  Prahlad Vaidyanathan Oct 28 '13 at 7:07