Polynomial Forms over Field form Integral Domain/Formulation 2

Theorem
Let $\left({F, +, \circ}\right)$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $\Bbb F$ be the set of all polynomials over $\left({F, +, \circ}\right)$ defined as sequences.

Let polynomial addition and polynomial multiplication be defined as:


 * $\forall f = \left \langle {a_k}\right \rangle = \left({a_0, a_1, a_2, \ldots}\right), g = \left \langle {b_k}\right \rangle = \left({b_0, b_1, b_2, \ldots}\right) \in \Bbb F$:
 * $f \oplus g := \left({a_0 + b_0, a_1 + b_1, a_2 + b_2, \ldots}\right)$
 * $f \otimes g := \left({c_0, c_1, c_2, \ldots}\right)$ where $\displaystyle c_i = \sum_{j \mathop + k \mathop = i} a_j \circ b_k$

Then $\left({\Bbb F, \oplus, \otimes}\right)$ is an integral domain.

Proof
As $\left({F, +, \circ}\right)$ is a field, it is also by definition a ring.

Thus from Polynomial Ring of Sequences is Ring we have that $\left({\Bbb F, \oplus, \otimes}\right)$ is a ring.

From Field is Integral Domain, a field is also by definition an integral domain.

Let $f, g \in \Bbb F$ such that neither $f$ nor $g$ are the null polynomial.

Let:
 * $\deg f = m, \deg g = n$

where $\deg$ denotes the degree of $f$ and $g$ respectively.

By Degree of Product of Polynomials over Integral Domain, the degree of $f \times g$ is $m + n$.

Then by definition of polynomial multiplication, its leading coefficient is $a_m \circ b_n$.

As by definition an integral domain has no proper zero divisors:
 * $a_m \circ b_n \ne 0_F$.

So, by definition, $f \otimes g$ has a leading coefficient which is not $0_F$.

That is, $f \otimes g$ is not the null polynomial

The result follows by definition of integral domain.