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A ring \(\langle R,+,\cdot \rangle\) is a set \(R\) together with two binary operations \(+\) and \(\cdot\), called addition and multiplication, such that the following axioms are satisfied:
- \(\langle R,+\rangle\) is an abelian group.
- Multiplication is associative.
- For all \(a,b,c\in R\), the left distributive law \(a\cdot(b+c) = (a\cdot b)+(a\cdot c)\) and the right distributive law \((a+b)\cdot c = (a\cdot c) + (b\cdot c)\) hold.
Ring 可以沒有乘法單位元素（unity）！如果有，叫 a ring with unity。
Then \(\langle \text{End}(V), +, \circ \rangle\) is a ring called the endomorphism ring of \(V\).
By convention we set \(0 \cdot a = 0_R\), where the \(0\) on the left-hand side is the integer \(0\), while \(0_R\) on the right is the additive identity element of \(R\).
A ring with a multiplicative identity \(1_R\) is a ring with unity.
Let \(R\) be a ring with unity \(1\not = 0\).
A commutative division ring is a field.
Field 在這裡終於出現了！
What if \(1_R = 0_R\)? If that is the case, let \(1_R=0_R=e\). For any \(a\in R\), we have
\[a + e = a\cdot e = a, \\ a = a\cdot e = a(e-e) = e.\]Thus \(R = \{e\}\), which is trivial.
Let \(R\) be a ring. A subset \(S\) of \(R\) is a subring of \(R\) if
which is denoted by \(S<R\).
可以先說明 the additive identity of \(R\) is …（例如零矩陣 \(O\)），再說明 \(O\) 在 \(S\) 中。
Subfield 多說明 \(a^{-1}\) 也在其中。
For rings \(R\) and \(R'\), \(\phi: R\to R'\) is a (ring) homomorphism if
- \(\phi(a+b) = \phi(a) + \phi(b)\),
- \(\phi(ab) = \phi(a)\phi(b)\).
for all \(a,b \in R\).
The kernel of \(\phi\) is
\[\text{ker}(\phi) = \{a\in R\vert \phi(a) = 0\}.\]映到加法單位元素才是 kernel！
A ring homomorphism is one-to-one if and only if its kernel is \(\{0\}\).
Let \(\phi: R\to R'\) be a ring homomorphism. Then \(\phi(R)\) is a subring of \(R'\).
Let \(G\) be a group. A homomorphism from \(G\) to itself is an endomorphism.
Let \((G,+)\) be an abelian group. The set \(\text{End}(G)\) of all endomorphisms of \(G\) forms a ring under function addition and function composition.
The function \(\phi: \mathbb{Z}\to \mathbb{Z}_n\) defined by \(\phi(a) = \bar a\) is a ring homomorphism.
加法和乘法都要檢查！
Let \(R = \mathbb{Q}[x]\) be the set of all polynimails over \(\mathbb{Q}\). Given \(a\in \mathbb{R}\), define \(\phi_a:\mathbb{Q}[x]\to \mathbb{R}\) by
\[\phi_a(f(x)) = f(a).\]Then \(\phi_a\) is an evalutation homomorphism.
For \(a=\sqrt{2}\), we have
\[\mathbb{Q}[\sqrt{2}] := \phi_{\sqrt{2}}(\mathbb{Q}[x]) = \{f(\sqrt{2})\vert f[x]\in\mathbb{Q}[x] \},\]which is a subring of \(R\).
\(\mathbb{Q}[\sqrt{2}]\) 不再是多項式的集合，而是數值的集合。
一如往常：
given by \(\rho(\bar a) = (\bar a, \bar a)\) is a group isomorphism and is also a ring isomorphism.
Let \(\alpha\) be a basis of \(n\)-dimensional vector space \(V\) over \(F\). Then the matrix representation \(\text{Rep}_\alpha:\text{End}(V) \to M_n(F)\) is a ring isomorphism.
\(\text{End}(V)\): see this.
任意 linear trans. 都可以用矩陣表示！
Regard \(\mathbb{C}\) as a vector space over \(\mathbb{R}\) with the basis \(\alpha = \{1, i\}\). Consider the map
\[\rho: \mathbb{C} \to \text{End}(\mathbb{C})\]given by \(\rho(x) = \rho_x\), and \(\rho_x(y) = \rho_xy\).
\(\rho_x\) is a linear tranformation.
It can be shown that \(\rho\) is an injective ring homomorphism. Moreover, let
\[\phi: \mathbb{C} \to \Bigg\{\begin{pmatrix} a & -b \\ b & a \end{pmatrix} \Bigg\vert a, b \in \mathbb{R}\Bigg\}\]given by
\[\phi(a+bi) = \begin{pmatrix} a & -b \\ b & a \end{pmatrix},\]which is a ring isomorphism. We will show that \(\phi\) is equal to \(\text{Rep}_{\alpha} \circ \rho\). Note that for \(a,b \in \mathbb{R}\),
\[\rho_{a+bi}(1) = a\cdot 1 + b\cdot i, \\ \rho_{a+bi}(i) = -b\cdot 1 + a\cdot i.\]記得 \(\rho_{a+bi}\) 是線性變換！而 \(1\) 和 \(i\) 是 basis 中的 vector。
Therefore,
\[\text{Rep}_\alpha(\rho_{a+bi}) = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \phi(a + bi).\]\(\rho_{a+bi} = \rho(a+bi)\).
將 basis 從 \(\{1, i\}\) 變換到 \(\{\rho_{a+bi}(1), \rho_{a+bi}(i) \}\)。
For two positive coprime integers \(m\) and \(n\), we have a ring isomorphism
\[\begin{align*} \rho: \Bbb Z_{mn} &\to \Bbb Z_m \times \Bbb Z_n \\ x \bmod{mn} &\mapsto (x\bmod m, x\bmod n). \end{align*}\]This means that the system of congruent equation
\[\begin{cases} x \equiv a \bmod m \\ x \equiv b \bmod n \end{cases}\]has a unique solution upto a multiple of \(mn\). To solve this, it is sufficient to find
\[x_1 = \rho^{-1}(1, 0), x_2 = \rho^{-1}(0, 1),\]which are called the fundamental solutions. Then we have
\[\rho^{-1}(a, b) = ax_1 + bx_2.\]Since \(x_1\equiv 0 \bmod n\), we can write \(x_1=kn\) for some integer \(k\). Thus,
\[x_1 = kn \equiv 1 \bmod m.\]This implies that \(k\) is the multiplicative inverse of \(n\) modulo \(m\), which can be solved easily. Generalize it, we have the following theorem:
Consider the system of equations
\[\begin{cases} x &\equiv a_1 \bmod b_1 \\ x &\equiv a_2 \bmod b_2 \\ &\vdots\\ x &\equiv a_k \bmod b_k \end{cases}\]If any pair of \(b_1, \cdots b_k\) are coprime, then the above equation has a unique solution upto a multiple of \(b_1\cdots b_k\). Moreover, if \(x_1,\cdots,x_k\) are fundamental solutions, the general solution is
\[a_1x_1 + \cdots a_kx_k \pmod {b_1\cdots b_k}.\]Moreover, if we let \(B = \prod_i b_i\) and let \(B_i = B/b_i\), we have
\[x_i = B_i(B_i^{-1} \mod{b_i}).\]Thus, the general solution is
\[\sum_{i} a_iB_i(B_i^{-1} \mod{b_i}) \pmod{B}.\]