What does $p(\theta)$ mean in this theorem?

Question

Theorem Let $p(x) \in F[x]$ be an irreducible polynomial of degree $n$. Then $K=F[x]/(p(x))$ is a field in which $\theta=\bar x=x+(p(x))$ that satisfies $p(\theta)=0$. Furthermore, the elements $1, \theta, \theta^2, ..., \theta^{n-1}$ form a basis of $K$ as a vector space over $F$. So $[K:F]=n$ and $K=F[\theta]$.

This is a theorem from Lovett. "Abstract Algebra." p. 325. (Chapter 7. Field Extensions.) I know that $(p(x))$ means the smallest ideal in $F[x]$ containing $p(x)$. But if $\theta=x+(p(x))$, then $p(\theta)$ is $p(x+(p(x)))$, and I have no idea what this means. Please help me with this.

Answer 1

There is an identification being made between the field $F$ and the field $\tilde{F} := \{ a + (p(x)) : a \in F \}$, which is a subfield of $K$. We have that $F \cong \tilde{F}$ under the natural isomorphism $a \mapsto a + (p(x))$. (Henceforth, let $I = (p(x))$ for shorthand.)

This isomorphism induces an isomorphism between $F[x]$ and $\tilde{F}[x]$, where the polynomial $$f(x) = a_0 + a_1 x + \dots + a_n x^n \in F[x]$$ is mapped to $$\tilde{f}(x) = (a_0 + I) + (a_1 + I)x + \dots + (a_n + I)x^n \in \tilde{F}[x].$$

Rather than introducing many cumbersome notations, the author chooses to call the field $\tilde{F}$ as $F$ itself, and the polynomial $\tilde{f}(x)$ as $f(x)$ itself.

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For the sake of clarity, at least initially, let us retain the earlier notation with the $\sim$'s. Now, it can be easily checked that for every $\tilde{\alpha} = \alpha + I \in K$, $$\tilde{f}(\tilde{\alpha}) = f(\alpha) + I.$$

So, what does $p(\theta)$ mean for $\theta = x + I$? Clearly, the author is actually talking about $\tilde{p}(x + I)$. And, we have that $$\tilde{p}(x + I) = p(x) + I = 0 + I.$$

Thus, "$p(\theta) = 0$" when making the appropriate identifications.

Answer 2

Suppose $p(x)=\sum_{j=0}^{n}a_j x^j$ where the $a_j\in K$.

You need to use the following facts about operating with cosets:

$$ \overline{p_1(x)+p_2(x)}=\overline{p_1(x)}+\overline{p_2(x)}$$ and $$ \overline{p_1(x)\cdot p_2(x)}=\overline{p_1(x)}\cdot\overline{p_2(x)}$$ and $$ \overline{c\cdot p_1(x)}=c\cdot \overline{p_1(x)}.$$

Then calculate:

$$p(\theta)=p(\bar{x})=\sum_{j=0}^{n}a_j \bar{x}^j=\overline{\sum_{j=0}^{n}a_j x^j}=\overline{p(x)}=\bar{0}.$$

Answer 3

Elements of the quotient set $K$ are equivalence classes by the relation $a\sim b\Leftrightarrow a-b\in(p(x))$. Polynomials conserve this equivalence, so if $a\sim b$ so $q(a) \sim q(b)$ and we can then talk about $q(\bar a)$, with $\bar a$ is the equivalence class of $a$ and $b$.

As $p(x)-0\in (p(x))$ so $p(\bar x)=\overline{p(x)}=0$ that what means $p(\theta)=0$.