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An additive subgroup \(I\) of a ring \(R\) satisfiying \(aI\subset I, Ib\subset I\) for all \(a,b\in R\) is an ideal of \(R\). Especially, \(I\) is a subring.
\(a\) 和 \(b\) 被 \(I\) 吸收了！
For \(r\in R\) and \(x\in I\), \(rx, xr\in I\).
The kernel of a ring homomorphism is an ideal.
Proof
By this lemma.
Let \(H\) be a subring of a ring \(R\). The map
\[\times:(a+H, b+H)\mapsto (ab)+H\]is a well-defined multiplication of additive cosets of \(H\) iff \(H\) is an ideal.
Proof
(待補)
Let \(I\) be an ideal of a ring \(R\). The additive cosets of \(I\) form a ring \(R/I\), with addition and multiplication given by
\[\begin{align*} +:(a+I, b+I)\mapsto(a+b)+I, \\ \times:(a+I,b+I)\mapsto(ab)+I. \end{align*}\]
\(R/I\) 讀做 the quotient ring (or factor ring) of \(R\) by \(I\).
Let \(R\) be a commutative ring, and \(a\) be an element of \(R\). Then the set \(\{ra\mid r\in R \}\) is an ideal, called the principal ideal generated by \(a\), and is denoted by \(\langle a\rangle\) or \(Ra\).
小心！\(Ra\) 不一定包含 \(a\)，因為 \(1_R\) 不一定存在（沒有 unity）。
\(\langle a\rangle \subset \langle b\rangle \iff a=bc\) for some \(c\in R\). In this case, we also say \(b\) divides \(a\), denoted by \(b\mid a\).
Let \(I\) be an ideal of a ring \(R\). The map \(\gamma: R\to R/I\) defined by \(\gamma: a\mapsto a+I\) is a ring homomorphism with kernel \(I\).
Proof
Use this theorem.
Let \(\phi: R\to R'\) be a ring homomorphism. Then we have \(R/\text{ker}\phi \cong \phi(R)\).
Proof
Let \(I=\text{ker}\phi\). Define \(\phi^*: R/I \to \phi(R)\) by \(\phi^*(a+I) = \phi(a)\). By the first isomorphism theorem for groups, we have proved that \(\phi^*\) is a well-defined additive group isomorphism. It remains to show that \(\phi^*((a+I)(b+I)) = \phi^*(a+I)\phi^*(b+I)\). This follows from the definition of \(\phi^*\). ◼
處理 ring 時，當 additive group 搞定後，再把乘法的結合、分配解決就好了！
Let \(\phi: R\to R'\) be a ring homomorphism with kernel \(\text{ker}\phi\). Then there is a bijection between ideals of \(R\) containing \(\text{ker}\phi\) and ideals of \(\phi(R)\).
Proof
Let \(K=\text{ker}\phi\). Suppose \(I\) is an ideal of \(R\) containing \(K\). Then
\[I/K = \{K, r_1+K, \cdots, r_{m-1}+K \},\]for some \(r_i\in R\). That is to say, in the set of all ideals containing \(K\), each of them can be distinguished by the cosets \(r_i + K\) of their own. Moreover, \(I/K\) is also an additive subgroup of \(R/K\), and even an ideal of \(R/K\), for the addition and multiplication defined above!
Then, by the ring isomorphism theorem, \(R/K \cong \phi(R)\). Since the two rings are isomorphic, their ideals must have a bijective correspondence. Therefore the theorem is proved. ◼
發現 \(I/K\) 是 \(R/K\) 的 ideal 非常關鍵！
在定理敘述中，如果 \(I\) 不包含 kernel 的話，\(I/K\) 就不存在了，因為 quotient 根本無法成立。
In a commutative ring \(R\), there are three basic operations on ideals: (Let \(I\) and \(J\) be two ideals of \(R\).)
- Sum: \(I+J=\{a+b\mid a\in I, b\in J\},\)
- Intersection: \(I\cap J\), and
- Product: \(IJ = \big\{\sum_{i=1}^na_ib_i\mid n\in \Bbb N, a_i\in I, b_i\in J\big\}\).
The results are also ideals of \(R\).
In particular, we have
\[IJ \subset (I\cap J) \subset I,J \subset (I+J).\]
How can we see that \(IJ\) is a subset of \(I\) and \(J\)? Since \(IJ\) must maintain its closedness, all \(i\in I\) and \(j\in J\) in \(IJ\) must also in \(I\cap J\). Thus \(IJ\subset (I\cap J)\subset I, J\).
對於 group 來說，任意 \(g\in G\) 都能在 normal subgroup 左右移動；對於 ring 來說，任意 \(r\in R\) 皆被 ideal 吸收。而且，group homomorphism 的 kernel 是 normal subgroup；ring homomorphism 的 kernel 正是 ideal！