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# 数学代写|数论代写Number Theory代考|Polynomial congruences

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## 数学代写|数论代写Number Theory代考|Polynomial congruences

Throughout this section, $F$ denotes a field.
Specializing the congruence notation introduced in $\S 9.3$ for arbitrary rings to the ring $F[\mathrm{X}]$, for polynomials $a, b, n \in F[\mathrm{X}]$, we write $a \equiv b(\bmod n)$ when $n \mid(a-b)$. Because of the division with remainder property for polynomials, we have the analog of Theorem 2.1:

Theorem 17.12. Let $n \in F[\mathrm{x}]$ be a non-zero polynomial. For every $a \in$ $F[\mathrm{X}]$, there exists a unique $b \in F[\mathrm{X}]$ such that $a \equiv b(\bmod n)$ and $\operatorname{deg}(b)<n$, namely, $b:=a \bmod n$.

For a non-zero $n \in F[\mathrm{X}]$, and $a \in F[\mathrm{X}]$, we say that $a^{\prime} \in F[\mathrm{X}]$ is a multiplicative inverse of $a$ modulo $n$ if $a a^{\prime} \equiv 1(\bmod n)$.

All of the results we proved in $\S 2.2$ for solving linear congruences over the integers carry over almost identically to polynomials. As such, we do not give proofs of any of the results here. The reader may simply check that the proofs of the corresponding results translate almost directly.

Theorem 17.13. Let $a, n \in F[\mathrm{X}]$ with $n \neq 0$. Then a has a multiplicative inverse modulo $n$ if and only if $a$ and $n$ are relatively prime.

Theorem 17.14. Let a, $n, z, z^{\prime} \in F[\mathrm{X}]$ with $n \neq 0$. If a is relatively prime to $n$, then $a z \equiv a z^{\prime}(\bmod n)$ if and only if $z \equiv z^{\prime}(\bmod n)$. More generally, if $d:=\operatorname{gcd}(a, n)$, then $a z \equiv a z^{\prime}(\bmod n)$ if and only if $z \equiv z^{\prime}(\bmod n / d)$.
Theorem 17.15. Let $a, b, n \in F[\mathrm{X}]$ with $n \neq 0$. If a is relatively prime to $n$, then the congruence $a z \equiv b(\bmod n)$ has a solution $z$; moreover, any polynomial $z^{\prime}$ is a solution if and only if $z \equiv z^{\prime}(\bmod n)$.

As for integers, this theorem allows us to generalize the ” $\bmod$ ” operation as follows: if $n \in F[\mathrm{X}]$ is a non-zero polynomial, and $s \in F(\mathrm{X})$ is a rational function of the form $b / a$, where $a, b \in F[\mathrm{X}], a \neq 0$, and $\operatorname{gcd}(a, n)=1$, then $s \bmod n$ denotes the unique polynomial $z$ satisfying
$$a z \equiv b(\bmod n) \text { and } \operatorname{deg}(z)<\operatorname{deg}(n) .$$
With this notation, we can simply write $a^{-1} \bmod n$ to denote the unique multiplicative inverse of $a$ modulo $n$ with $\operatorname{deg}(a)<\operatorname{deg}(n)$.

## 数学代写|数论代写Number Theory代考|Polynomial quotient algebras

Throughout this section, $F$ denotes a field.
Let $f \in F[\mathrm{X}]$ be a monic polynomial, and consider the quotient ring $E:=F[\mathrm{x}] /(f)$. As discussed in Example 17.4, we may naturally view $E$ as an $F$-algebra via the map $\tau$ that sends $c \in R$ to $[c]_f \in E$. Moreover, if $\operatorname{deg}(f)>0$, then $\tau$ is an embedding of $F$ in $F[\mathbf{x}] /(f)$, and otherwise, if $f=1$, then $E$ is the trivial ring, and $\tau$ maps everything to zero.

Suppose that $\ell:=\operatorname{deg}(f)>0$. Let $\eta:=[\mathrm{X}]_f \in E$. Then $E=F[\eta]$, and as an $F$-vector space, $E$ has dimension $\ell$, with $1, \eta, \ldots, \eta^{\ell-1}$ being a basis (see Examples $9.34,9.43,14.3$, and 14.22). That is, every element of $E$ can be expressed uniquely as $g(\eta)$ for $g \in F[\mathrm{X}]$ of degree less than $\ell$.

Now, if $f$ is irreducible, then every polynomial $a \not \equiv 0(\bmod f)$ is relatively prime to $f$, and hence invertible modulo $f$; therefore, it follows that $E$ is a field. Conversely, if $f$ is not irreducible, then $E$ cannot be a field – indeed, if $g$ is a non-trivial factor of $f$, then $g(\eta)$ is a zero divisor.

If $F=\mathbb{Z}_p$ for a prime number $p$, and $f$ is irreducible, then we see that $E$ is a finite field of cardinality $p^{\ell}$. In the next chapter, we shall see how one can perform arithmetic in such fields efficiently, and later, we shall also see how to efficiently construct irreducible polynomials of any given degree over a finite field.

Minimal polynomials. Now suppose that $E$ is any $F$-algebra, and let $\alpha$ be an element of $E$. Consider the polynomial evaluation map $\rho: F[\mathrm{X}] \rightarrow E$ that sends $g \in F[\mathrm{X}]$ to $g(\alpha)$. The kernel of $\rho$ is an ideal of $F[\mathrm{X}]$, and since every ideal of $F[\mathrm{X}]$ is principal, it follows that there exists a polynomial $\phi \in F[\mathrm{X}]$ such that $\operatorname{ker}(\rho)$ is the ideal of $F[\mathrm{X}]$ generated by $\phi$; moreover, we can make the choice of $\phi$ unique by insisting that it is monic or zero. The polynomial $\phi$ is called the minimal polynomial of $\alpha$ (over $F$ ). If $\phi=0$, then $\rho$ is injective, and hence the image $F[\alpha]$ of $\rho$ is isomorphic (as an $F$-algebra) to $F[\mathrm{X}]$. Otherwise, $F[\alpha]$ is isomorphic (as an $F$-algebra) to $F[\mathrm{X}] /(\phi)$; moreover, since any polynomial that is zero at $\alpha$ is a polynomial multiple of $\phi$, we see that $\phi$ is the unique monic polynomial of smallest degree that is zero at $\alpha$.

If $E$ has finite dimension, say $n$, as an $F$-vector space, then any element $\alpha$ of $E$ has a non-zero minimal polynomial. Indeed, the elements $1_E, \alpha, \ldots, \alpha^n$ must be linearly dependent (as must be any $n+1$ vectors in a vector space of dimension $n$ ), and hence there exist $c_0, \ldots, c_n \in F$, not all zero, such that
$$c_0 1_E+c_1 \alpha+\cdots+c_n \alpha^n=0_E$$
and therefore, the non-zero polynomial $g:=\sum_i c_i \mathrm{X}^i$ is zero at $\alpha$.

## 数学代写|数论代写Number Theory代考|Polynomial congruences

$z \equiv z^{\prime}(\bmod n / d)$

$$a z \equiv b(\bmod n) \text { and } \operatorname{deg}(z)<\operatorname{deg}(n)$$

## 数学代写|数论代写Number Theory代考|Polynomial quotient algebras

$$c_0 1_E+c_1 \alpha+\cdots+c_n \alpha^n=0_E$$

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