I am reading Emil Artin's "Galois Theory" now.
There is a small gap in this book:
We shall now try to imitate the set $E_0$ without having an extension field $E$ and an element $a$ at our disposal. We shall assume only an irreducible polynomial $$f(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_0$$ as given. We select a symbol $\xi$ and let $E_1$ be the set of all formal polynomials $$g(\xi) = c_0 + c_1 \xi + \cdots + c_{n-1} \xi^{n-1}$$ of a degree lower than $n$. This set forms a group under addition. We now introduce besides the ordinary multiplication a new kind of multiplication of two elements $g(\xi)$ and $h(\xi)$ of $E_1$ denoted by $g(\xi) \times h(\xi)$. It is defined as the remainder $r(\xi)$ of the ordinary product $g(\xi) h(\xi)$ under division by $f(\xi)$. We first remark that any product of $m$ terms $g_1(\xi), g_2(\xi), \cdots, g_m(\xi)$ is again the remainder of the ordinary product $g_1(\xi) g_2(\xi) \cdots g_m(\xi)$. This is true by definition for $m = 2$ and follows for every $m$ by induction if we just prove the easy lemma: The remainder of the product of two remainders (of two polynomials) is the remainder of the product of these two polynomials.
And I tried to fill the gap.
Is the following proof ok or not?
$a, b, f$ is a polynomial.
$\textrm{Mod}[a, f]$ is the remainder on division of $a$ by $f$.
$\textrm{Quotient}[a, f]$ is the polynomial quotient of $a$ and $f$.
$a \times b = a \times \textrm{Mod}[b, f]$
Proof:
$a b = \textrm{Quotient}[a b,f] f + \textrm{Mod}[a b,f]$
and
$b = \textrm{Quotient}[b,f] f + \textrm{Mod}[b,f]$.
$a b = a \textrm{Quotient}[b,f] f + a \textrm{Mod}[b,f]$.
$a \textrm{Mod}[b,f] = \textrm{Quotient}[a \textrm{Mod}[b,f],f] f + \textrm{Mod}[a \textrm{Mod}[b,f],f]$.
$a b = a \textrm{Quotient}[b,f] f + \textrm{Quotient}[a \textrm{Mod}[b,f],f] f + \textrm{Mod}[a \textrm{Mod}[b,f],f] = (a \textrm{Quotient}[b,f] + \textrm{Quotient}[a \textrm{Mod}[b,f],f]) f + \textrm{Mod}[a \textrm{Mod}[b,f],f].$
$\therefore \textrm{Mod}[a b,f] = \textrm{Mod}[a \textrm{Mod}[b,f],f].$
$\therefore a \times b = a \times \textrm{Mod}[b, f]$.
$a \times b = \textrm{Mod}[a, f] \times \textrm{Mod}[b, f]$
Proof:
$a \times b = a \times \textrm{Mod}[b, f] = \textrm{Mod}[b, f] \times a = \textrm{Mod}[b, f] \times \textrm{Mod}[a, f] = \textrm{Mod}[a, f] \times \textrm{Mod}[b, f]$.
$(g_1(\xi) \times g_2(\xi)) \times g_3(\xi) = g_1(\xi) \times (g_2(\xi) \times g_3(\xi)) = \textrm{Mod}[g_1(\xi) g_2(\xi) g_3(\xi), f]$
Proof:
$\textrm{Mod}[g_1(\xi) g_2(\xi) g_3(\xi), f] = (g_1(\xi) g_2(\xi)) \times g_3(\xi) = \textrm{Mod}[g_1(\xi) g_2(\xi), f] \times g_3(\xi) = (g_1(\xi) \times g_2(\xi)) \times g_3(\xi).$
$\textrm{Mod}[g_1(\xi) g_2(\xi) g_3(\xi), f] = g_1(\xi) \times (g_2(\xi) g_3(\xi)) = g_1(\xi) \times \textrm{Mod}[g_2(\xi) g_3(\xi), f] = g_1(\xi) \times (g_2(\xi) \times g_3(\xi)).$
$g_1(\xi) \times g_2(\xi) \times \cdots \times g_m(\xi) \times g_{m+1}(\xi) = \textrm{Mod}[g_1(\xi) g_2(\xi) \cdots g_m(\xi) g_{m+1}(\xi), f]$
Proof:
I prove by induction.
$g_1(\xi) \times g_2(\xi) \times \cdots \times g_m(\xi) \times g_{m+1}(\xi) = \textrm{Mod}[g_1(\xi) g_2(\xi) \cdots g_m(\xi), f] \times g_{m+1}(\xi) = (g_1(\xi) g_2(\xi) \cdots g_m(\xi)) \times g_{m+1}(\xi) = \textrm{Mod}[(g_1(\xi) g_2(\xi) \cdots g_m(\xi)) g_{m+1}(\xi), f] = \textrm{Mod}[g_1(\xi) g_2(\xi) \cdots g_m(\xi) g_{m+1}(\xi), f].$