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# 数学代写|现代代数代考Modern Algebra代写|MATH611 The cyclic prime fields Zp

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## 数学代写|现代代数代考Modern Algebra代写|The cyclic prime fields Zp

Since division doesn’t work well with congruence, we can’t expect $\mathbf{Z}_n$ to always have reciprocals, so we don’t expect it to be a field. Let’s first see when an element in $\mathbf{Z}_n$ is a unit. The term unit in a ring refers to an element $x$ of the ring that does have a reciprocal. 1 is always a unit in a ring, and every nonzero element in a field is a unit.
Theorem 2.10. An element $k$ in $\mathbf{Z}_n$ is a unit if and only if $k$ is relatively prime to $n$.
Proof. First, suppose that $k$ is a unit in $\mathbf{Z}_n$. That means there exists $l$ such that $k l \equiv$ $1(\bmod n)$. Then $n \mid(k l-1)$, and hence $n$ is relatively prime to $k$.

Second, suppose that $k$ is relatively prime to $n$. Then, by the extended Euclidean algorithm, their greatest common divisor, 1, is a linear combination of $k$ and $n$. Thus, there are integers $x$ and $y$ so that $1=x k+y n$. Then $1 \equiv x k(\bmod n)$, and $k$ does have a reciprocal, namely $x$, in $\mathbf{Z}_n$. Thus $k$ is a unit in $\mathbf{Z}_n$. $\quad$ Q.E.D.
Recall from definition $1.39$ that the totient function $\varphi(n)$ denotes the number of positive integers less than $n$ that are relatively prime to $n$.
Corollary $2.11$ (Units in $\mathbf{Z}_n$ ). The number of units in $Z_n$ is $\phi(n)$.
Theorem 2.12. The cyclic ring $\mathbf{Z}_n$ is a field if and only if $n$ is prime.
Proof. Part of this theorem is a direct corollary of the previous one. Suppose $n$ is prime. Then every nonzero element of $\mathbf{Z}_n$ is relatively prime to $n$. Therefore, $\mathbf{Z}_n$ is a field.

Next we’ll show that if $n$ is composite, the ring is not a field. Let $n$ be the product of two integers $m$ and $k$, both greater than 1. Then neither $m$ nor $k$ can have a reciprocal in $\mathbf{Z}_n$. Why not? Suppose that $m^{-1}$ did exist in $\mathbf{Z}_n$. Then
\begin{aligned} \left(m^{-1} m\right) k & \equiv 1 k \equiv k(\bmod n) \ m^{-1}(m k) & \equiv m^{-1} n \equiv 0(\bmod n) \end{aligned}
But $k \not \equiv 0(\bmod n)$, a contradiction. So $m^{-1}$ doesn’t exist. Therefore, $\mathbf{Z}_n$ is not a field.
Q.E.D.

## 数学代写|现代代数代考Modern Algebra代写|Characteristics of fields, and prime fields

The characteristic of a ring was defined above, so we already have the definition for the characteristic of a field.

Those fields that have characteristic 0 all have $\mathbf{Q}$ as a subfield. The flawed proof we saw earlier included the mistaken assumption that all the elements $0,1,2, \ldots$ were distinct, which, as we’ve seen with these finite fields, isn’t always the case. But we can correct the flawed proof to validate the following theorem. First, a definition.

Definition 2.18. A prime field is a field that contains no proper subfield. Equivalently, every element in it is a multiple of 1.

Theorem 2.19. Each field $F$ has exactly one of the prime fields as a subfield. It will have $\mathbf{Z}_p$ when it has characteristic $p$, but it will have $\mathbf{Q}$ if it has characteristic 0 .

The Frobenius endomorphism. Exponentiation to the power $p$ has an interesting property when a commutative ring $R$ has prime characteristic $p$ :
$$(x+y)^p=x^p+y^p$$
There are various ways to prove this. For instance, you can show that the binomial coefficient $\left(\begin{array}{l}p \ k\end{array}\right)$ is divisible by $p$ when $1<k<p$. Since $\left(\begin{array}{l}p \ k\end{array}\right)=\frac{p !}{k !(n-k) !}$ and $p$ divides the numerator but not the denominator, therefore it divides $\left(\begin{array}{l}p \ k\end{array}\right)$.

This function $\varphi: R \rightarrow R$ defined by $\varphi(x)=x^p$ also preserves 1 and multiplication: $1^p=1$ and $(x y)^p=x^p y^p$. Therefore, it is a ring endomorphism, called the Frobenius endomorphism.
We’re most interested in the endomorphism when the ring is a field $F$ of characteristic p. It’s not particularly interesting when $F$ is the prime field $\mathbf{Z}_p$ because it’s just the identity function then. For other finite fields of characteristic $p$ it will be an automorphism-it’s a bijection since it’s an injection on a finite set-and it’s not the identity function for those fields.

# 现代代数代写

## 数学代写|现代代数代考Modern Algebra代写|The cyclic prime fields Zp

$$\left(m^{-1} m\right) k \equiv 1 k \equiv k(\bmod n) m^{-1}(m k) \equiv m^{-1} n \equiv 0(\bmod n)$$

QED prime fields

## 数学代写|现代代数代考Modern Algebra代写|Characteristics of fields, and prime fields

Frobenius 自同态。指数的权力 $p$ 当交换环时有一个有趣的性质 $R$ 具有主要特征 $p$ :
$$(x+y)^p=x^p+y^p$$

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## MATLAB代写

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