Odd Set Notation (Square Brackets)

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$A$ is a ring such as $\Bbb Z$.

$\alpha$ is an algebraic integer such as $i$, whose minimal polynomial is $x^2+1$.

The extension $A[\alpha]$ has numbers of the form $c_0+c_1\alpha+c_2\alpha^2+...+c_{n-1}a^{n-1},$ with $c_j\in A$.

In the example with $\alpha=i,$ the elements of $\mathbb Z[i]$ have the form $c_0+c_1i$.

The $c_j$ are not related to the $a_i$.

Note that, if $\alpha$ is a root of an $n^{th}$ degree polynomial,

then $\alpha ^n$ can be expressed as a linear combination of $1, \alpha, \alpha^2, ..., \alpha^{n-1}$.

In the example with $\alpha=i$, $\alpha^2=-1(1)+0(\alpha)$.

That is why, in the sum for $p(x),$ the index goes up to $n$,

whereas in the sum for an element of $A[\alpha],$ the index goes up to $n-1$.

Operations inherited from $A$ means that when we add or multiply two elements of $A[\alpha]$,

say $(c_0+c_1\alpha+c_2\alpha^2+...+c_{n-1}\alpha^{n-1})+(d_0+d_1\alpha+d_2\alpha^2+...+d_{n-1}\alpha^{n-1}),$

the result is $(c_0+d_0)+(c_1+d_1)\alpha+(c_2+d_2)\alpha^2+...+(c_{n-1}+d_{n-1})\alpha^{n-1},$

where $c_j+d_j$ is computed in $A$.

And when we multiply $(c_0+c_1i)(d_0+d_1i),$

the result is $c_0d_0+(c_0d_1+c_1d_9)i+c_1d_1i^2=c_0d_0-c_1d_1+(c_0d_1+d_1c_0)i, $

where again products and sums of terms involving $c_j$ and $d_j$ are computed in $A$.

I just want to expand on "operations inherited from $A$." If the book only talks about infinite domains like $\mathbb Q$ and $\mathbb Z$ with the usual addition and multiplication, mentioning operation inheritance might be needlessly confusing.

Consider for example the finite ring $\mathbb Z_{10}$, consisting of only the integers 0 to 9. This would seem to have the usual addition and multiplication, since, for example, $1 + 1 = 1 \times 2 = 2$.

However, $7 + 7 = 7 \times 2$ but does not equal 14. Both of those operations "wrap around" to land back in $\mathbb Z_{10}$, giving 4 rather than 14 in this case.

Now consider $\mathbb Z_{10}[\sqrt{53}]$. Then $\sqrt{53} + \sqrt{53} = 2\sqrt{53}$ just like we expect. But $7 \times 2\sqrt{53}$ is not $14\sqrt{53}$ but $4\sqrt{53}$.

What would $(\sqrt{53})^2$ be? I'm not exactly sure, I myself am confused on this point. But hopefully I have given you a clearer understanding of how addition and multiplication might differ from what you're used to.

Rings of matrices might provide another good example.