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# 数学代写|交换代数代写Commutative Algebra代考|Math523 Ideals

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## 数学代写|交换代数代写Commutative Algebra代考|Ideals

Definition (1.4.1). – Let A be a ring and let I be a subgroup of A (with respect to the addition). One says that $\mathrm{I}$ is a left ideal if for any $a \in \mathrm{I}$ and any $b \in \mathrm{A}$, one has $b a \in \mathrm{I}$. One says that $\mathrm{I}$ is a right ideal if for any $a \in \mathrm{I}$ and any $b \in \mathrm{A}$, one has $a b \in \mathrm{I}$. One says that $\mathrm{I}$ is a two-sided ideal if it is both a left and right ideal.

In a commutative ring, it is equivalent for a subgroup to be a left ideal, a right ideal or a two-sided ideal; one then just says that it is an ideal. Let us observe that in any ring $\mathrm{A}$, the subsets 0 and $\mathrm{A}$ are two-sided ideals. Moreover, an ideal I is equal to A if and only if it contains some unit, if and only if it contains 1

For any $a \in \mathrm{A}$, the set $\mathrm{A} a$ consisting of all elements of $\mathrm{A}$ of the form $x a$, for $x \in \mathrm{A}$, is a left ideal; the set $a \mathrm{~A}$ consisting of all elements of the form $a x$, for $x \in \mathrm{A}$, is a right ideal. When $\mathrm{A}$ is commutative, this ideal is denoted by $(a)$ To show that a subset I of $A$ is a left ideal, it suffices to prove the following properties:
(i) $0 \in \mathrm{I}$;
(ii) for any $a \in \mathrm{I}$ and any $b \in \mathrm{I}, a+b \in \mathrm{I}$;
(iii) for any $a \in \mathrm{I}$ and any $b \in \mathrm{A}, b a \in \mathrm{I}$.
Indeed, since $-1 \in \mathrm{A}$ and $(-1) a=-a$ for any $a \in \mathrm{A}$, these properties imply that $\mathrm{I}$ is a subgroup of $\mathrm{A}$; the third one then shows that it is a left ideal.
The similar characterization of right ideals is left to the reader.

## 数学代写|交换代数代写Commutative Algebra代考|Intersection

Let I and $\mathrm{J}$ be left ideals of $A$; their intersection $\mathrm{I} \cap \mathrm{J}$ is again a left ideal. More generally, the intersection of any family of left ideals of $\mathrm{A}$ is a left ideal of $\mathrm{A}$.

Proof. – Let $\left(\mathrm{I}s\right){s \in \mathrm{S}}$ be a family of ideals of $\mathrm{A}$ and let $\mathrm{I}=\bigcap_{\mathrm{S}} \mathrm{I}s$. (If $\mathrm{S}=\emptyset$, then $\mathrm{I}=\mathrm{A}$.) The intersection of any family of subgroups being a subgroup, $\mathrm{I}$ is a subgroup of $\mathrm{A}$. Let now $x \in \mathrm{I}$ and $a \in \mathrm{A}$ and let us show that $a x \in \mathrm{I}$. For any $s \in S, x \in \mathrm{I}{\mathrm{s}}$, hence $a x \in \mathrm{I}{\mathrm{s}}$ since $\mathrm{I}{\mathrm{s}}$ is a left ideal. It follows that $a x$ belongs to every ideal $\mathrm{I}_s$, hence $a x \in \mathrm{I}$.

We leave it to the reader to state and prove the analogous statements for right and two-sided ideals.

## 数学代写|交换代数代写Commutative Algebra代考|ldeals

(ii) 对于任何 $a \in \mathrm{I}$ 和任何 $b \in \mathrm{I}, a+b \in \mathrm{I}$;
(iii) 对于任何 $a \in \mathrm{I}$ 和任何 $b \in \mathrm{A}, b a \in \mathrm{I}$.

## MATLAB代写

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