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# 数学代写|抽象代数代写Abstract Algebra代考|MATH411 Ideals in ℤ and in 𝐹[𝑥]

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## 数学代写|抽象代数代写Abstract Algebra代考|Ideals in ℤ and in 𝐹[𝑥]

All of the ideals we’ve examined so far have been principal ideals of the form $\langle a\rangle=$ ${a \cdot r \mid r \in R}$, i.e., the set of all multiples of a fixed element $a \in R$. The next example shows that not every ideal is principal.

Example 31.31. Let $A$ be the set of all polynomials in $\mathbb{Z}[x]$ with an even constant term. For instance, consider $f(x)=3 x^{101}-171 x^{52}+x+12$ and $g(x)=5 x-21$, which are elements of $\mathbb{Z}[x]$. The constant terms of $f(x)$ and $g(x)$ are 12 (which is even) and $-21$ (which isn’t even), respectively. Therefore, $f(x) \in A$ and $g(x) \notin A$. In the exercises, you’ll show that (1) $A$ is an ideal of $\mathbb{Z}[x]$, but (2) $A$ is not a principal ideal; i.e., there does not exist an element $\alpha(x) \in \mathbb{Z}[x]$ such that $A=\langle\alpha(x)\rangle$.

Example 31.32. We’ve seen that $\langle 5\rangle=5 \mathbb{Z}$ is an ideal of $\mathbb{Z}$. Other ideals of $\mathbb{Z}$ include $\langle 2\rangle=2 \mathbb{Z},\langle 12\rangle=12 \mathbb{Z}$, and more generally, $\langle n\rangle=n \mathbb{Z}$, where $n$ is a fixed integer. These include the two extremes, i.e., the ring $\mathbb{Z}$ itself, $\langle 1\rangle=\mathbb{Z}$, and the trivial ideal $\langle 0\rangle={0}$.
It turns out that every ideal of $\mathbb{Z}$ is principal, as the following theorem states. We’ll start the proof but will leave it to you as an exercise to finish writing it.

## 数学代写|抽象代数代写Abstract Algebra代考|Big picture stuff

We continue to highlight the connections between the ring of integers $\mathbb{Z}$ and the polynomial ring $F[x]$, where $F$ is a field. As Theorems $31.33$ and $31.34$ indicate, both rings satisfy the condition that every ideal is principal. More generally, an integral domain whose ideals are all principal is called a principal ideal domain (or PID). And $\mathbb{Z}$ and $F[x]$ are classic examples of a PID.

For another connection, we restate an early observation in the language of principal ideals. In Chapter 3, Exercises #12 and #13, we proved the following:
Let $m, n \in \mathbb{Z}$. Then $n \mid m$ if and only if $\langle m\rangle \subseteq\langle n\rangle$.

We wrote $m \mathbb{Z}$ and $n \mathbb{Z}$ in Chapter 3 , but we saw in this chapter that those are equivalent to the principal ideals $\langle m\rangle$ and $\langle n\rangle$, respectively. Now, one of the exercises in this chapter states the following:
Let $f(x), g(x) \in F[x]$. Then $g(x) \mid f(x)$ if and only if $\langle f(x)\rangle \subseteq\langle g(x)\rangle$.
The actual proof is in $\mathbb{R}[x]$, but the argument remains the same in a more general setting of $F[x]$. Note how these statements are saying the same thing in two different rings $\mathbb{Z}$ and $F[x]$

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