Proving that set of all polynomials over $F$ form a vector space I'm trying to prove that the set of polynomials, $\mathrm{Poly}(F)$, with coefficients in the field $F$ form a vector space. The only definition of addition I have is "in the usual way," which I assume means that I would add corresponding coefficients and then leave the coefficients for higher-degree monomials. For example,
$$
(x^3 + x^2 + x + 1) + (2x^2 + 1) = x^3 + (2+1)x^2 + (1 + 0)x + (1+1)x^0. 
$$
I'm trying to find a way to rigorously write this so that I can prove, for example, commutativity or associativity. The problem is that I may have to add polynomials of different degrees, which complicates things. I can take polynomials of degrees $n,m$ where $n \leq m$ and "embed" the smaller degree polynomial into the set of higher-degree polynomials by adding additional coefficients of $0$, and then continue the algebra in the higher-degree set. Is that the trick? Otherwise, I can't think of a way to do this that involves quite a bit of casework over the relative sizes of the degrees of the polynomials.
 A: One way to handle the issue of "how to write these polynomials and define their operations rigorously" would be as so: let
$$\begin{align*}&\text{Poly}(F) \\
&:= \left\{ p(x) := \sum_{n=0}^\infty a_n x^n \, \middle| \, a_n \in F, \exists N_p \in \mathbb{N} \text{ s.t. } a_{N_p} \ne 0 \;\&\; n > N_p \Rightarrow a_n = 0 \right\} \cup \{\mathcal{Z}\}
\end{align*}$$
and we say $N_p = \deg(p)$, i.e. $N_p$ is the degree of $p$ (for $p \ne \mathcal{Z}$). Above, $\mathcal{Z}$ is understood to be the zero polynomial, i.e. $\mathcal{Z}(x) \equiv 0$ for all $x$. It is often taken that $\deg(\mathcal{Z}) = -\infty$ to mesh well with later observations. You can take
$$\mathcal{Z}(x) := \sum_{n=0}^\infty 0x^n$$
to make it work better in terms of operations.
Notice in particular how this retains our intuitive ideas of polynomials: sure, we have an upper bound of $\infty$, so it's more like power series, but for a particular $p \ne \mathcal{Z}$ and corresponding $N_p$ (always finite!) we have the equivalence
$$p(x) := \sum_{n=0}^\infty a_n x^n  = \sum_{n=0}^{N_p} a_n x^n$$
Then you can define our operations quite simply. Take $p,q \in \text{Poly}(F)$ and $\alpha \in F$ with
$$\begin{align*}
p(x) &:= \sum_{n=0}^\infty p_n x^n \\
q(x) &:= \sum_{n=0}^\infty q_n x^n \\
N_p &= \deg(p) \\
N_q &= \deg(q)
\end{align*}$$
Then we define
$$\begin{align*}
(p+q)(x) &:= \sum_{n=0}^\infty (p_n + q_n)x^n \\
(p\cdot q)(x) &:= \sum_{n=0}^\infty \left( \sum_{i+j=n} p_i q_j \right) x^n \\
(\alpha p)(x) &:= \sum_{n=0}^\infty \alpha p_n x^n
\end{align*}$$
We can ensure that
$$\begin{align*}
\deg(p+q) &= \max \{N_p,N_q\} \\
\deg(p \cdot q) &= N_p + N_q \\
\deg(\alpha p) &= \begin{cases}
\deg(p) & \alpha \ne 0 \\
-\infty & \alpha = 0 \text{ (and hence } \alpha p = \mathcal{Z}) \end{cases} \end{align*}$$
where we assume $-\infty + k = -\infty$ for any finite $k$ (and, of course, $-\infty < k$ for any finite $k$).
Using this alongside the field axioms can be used to show, indeed, $p+q,p\cdot q,\alpha p \in \text{Poly}(F)$.
