Shouldn't the yellow marked $a_0$ be $a_0+\langle p(x)\rangle?$

Question

I'm having problem in getting the proof from Gallian text in the higlighted region:

Shouldn't the yellow marked $a_0$ be $a_0+\langle p(x)\rangle?$

Edited: Shouldn't the $a_i$'s in the equation be $a_i+\langle p(x)\rangle?$

Answer 1

The polynomial $p(x)$ has coefficients in $F$. In order to evaluate it on elements of $E$, as Gallian is doing in the last set of equations, you need to identify $F$ as a subfield of $E$, which he did in the previous paragraph. But once you've done that, there's no need to write an element $a \in F$ as $a + \left<p(x)\right>$ when considering it as an element of $E$; you can just write $a$.

Two further notes:

• If you didn't adopt this convention, you'd have to add equivalence classes to all the coefficients of $p$; that is, you'd write the leading summand as $$ (a_n + \left<p(x)\right>)(x + \left<p(x)\right>)^n $$ and so on. This would be more verbose than illuminating.

• By writing the coefficients of $p(x)$ without equivalence classes, Gallian is emphasizing that he really is evaluating the original polynomial $p(x) \in F[x]$; doing it the other way would make it look like he was evaluating some polynomial in $E[x]$ (which of course $p$ is, but that's not the point of this proof).

Answer 2

No, it shouldn't. What you're doing is switching every single instance of $x$ with $x + \langle p(x)\rangle$. Which means that $a_nx^n$ becomes $a_n(x + \langle p(x)\rangle)^n$ and so on. But there is no $x$ near the $a_0$ term. Or you could say that there is, we have $a_0 = a_0x^0$ which becomes $a_0(x + \langle p(x)\rangle )^0 = a_0 \cdot 1 = a_0$.

As for where $\ldots + \langle p(x)\rangle$ in the last lines comes from, that's just collecting the terms from all the exponentiations, like this: $$ a_n(x + \langle p(x)\rangle)^n = a_n \sum_{i = 0}^n\binom{n}{i}x^i\langle p(x)\rangle^{n - i} = a_n \left(x^n + \sum_{i = 0}^{n -1}\binom{n}{i} x^i\langle p(x)\rangle^{n-i}\right)\\ = a_n \left(x^n + \langle p(x)\rangle\sum_{i = 0}^{n -1}\binom{n}{i} x^i\langle p(x)\rangle^{n-i - 1}\right) \\ = a_n (n^n + \langle p(x)\rangle) = a_nx^n + a_n\langle p(x)\rangle = a_nx^n + \langle p(x)\rangle $$ then as you gather all the different degree terms together, you get $n$ copies of $\langle p(x)\rangle$ added together, which is just $\langle p(x)\rangle$, and that is the $\langle p(x)\rangle$ term in the last two lines.

Answer 3

Nope. This is saying nothing more in essence than that if you have a polynomial $$ p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots a_1x + a_0,$$ and you translate by some amount $h$, you get $$ p(x + h) = a_n(x+h)^n + a_{n-1}(x + h)^{n-1} + \ldots a_1(x + h) + a_0.$$

Notice that the last term doesn't have an $h$.

All of the remaining cancellation comes from the fact that $\langle p(x) \rangle$ is an ideal, and thus is closed under multiplication under $F$ and addition within its own ideal.