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2. Jika \( R \) adalah sebuah Ring bilangan kompleks dan \( M \) adalah Ring matriks berordo \( 2 \times 2 \). Pemetaan \( \varphi: R \rightarrow M \) didefiniskan oleh \( \varphi(x+y i)=\left[\begin{array}{cc}a & b \\ -b & a\end{array}\right] \) dengan \( (x+y i) \in R \). Tunjukan bahwa \( \varphi \) adalah suatu isomorfisma.

Ask by Rojas Weber. in Indonesia
Jan 21,2025

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Answer

\(\varphi\) adalah isomorfisma antara ring bilangan kompleks \(R\) dan ring matriks \(2 \times 2\) \(M\).

Solution

Untuk menunjukkan bahwa pemetaan \(\varphi: R \rightarrow M\) didefinisikan oleh \(\varphi(x + y i) = \begin{bmatrix} x & y \\ -y & x \end{bmatrix}\) adalah isomorfisma, kita perlu membuktikan bahwa \(\varphi\) adalah homomorfisma ring yang bijektif. Berikut adalah langkah-langkah pembuktiannya: ### 1. **Homomorfisma Ring** Kita harus menunjukkan bahwa \(\varphi\) mempertahankan operasi penjumlahan dan perkalian, serta elemen identitas. #### a. **Penjumlahan** Untuk dua elemen \(z_1 = x_1 + y_1 i\) dan \(z_2 = x_2 + y_2 i\) di \(R\): \[ \varphi(z_1 + z_2) = \varphi((x_1 + x_2) + (y_1 + y_2)i) = \begin{bmatrix} x_1 + x_2 & y_1 + y_2 \\ - (y_1 + y_2) & x_1 + x_2 \end{bmatrix} \] Sementara itu: \[ \varphi(z_1) + \varphi(z_2) = \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} + \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 & y_1 + y_2 \\ -y_1 - y_2 & x_1 + x_2 \end{bmatrix} \] Maka, \(\varphi(z_1 + z_2) = \varphi(z_1) + \varphi(z_2)\). #### b. **Perkalian** Untuk dua elemen \(z_1 = x_1 + y_1 i\) dan \(z_2 = x_2 + y_2 i\) di \(R\): \[ z_1 \cdot z_2 = (x_1 x_2 - y_1 y_2) + (x_1 y_2 + x_2 y_1)i \] Sehingga: \[ \varphi(z_1 \cdot z_2) = \begin{bmatrix} x_1 x_2 - y_1 y_2 & x_1 y_2 + x_2 y_1 \\ - (x_1 y_2 + x_2 y_1) & x_1 x_2 - y_1 y_2 \end{bmatrix} \] Sementara itu: \[ \varphi(z_1) \cdot \varphi(z_2) = \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} = \begin{bmatrix} x_1 x_2 - y_1 y_2 & x_1 y_2 + y_1 x_2 \\ -y_1 x_2 - x_1 y_2 & x_1 x_2 - y_1 y_2 \end{bmatrix} \] Maka, \(\varphi(z_1 \cdot z_2) = \varphi(z_1) \cdot \varphi(z_2)\). #### c. **Elemen Identitas** Elemen identitas di \(R\) adalah \(1 + 0i\), dan: \[ \varphi(1 + 0i) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Yang merupakan elemen identitas di \(M\). ### 2. **Bijektivitas** Untuk menunjukkan bahwa \(\varphi\) adalah bijektif, kita harus menunjukkan bahwa \(\varphi\) adalah injektif dan surjektif. #### a. **Injektif** Misalkan \(\varphi(z_1) = \varphi(z_2)\), dimana \(z_1 = x_1 + y_1 i\) dan \(z_2 = x_2 + y_2 i\). Maka: \[ \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} = \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} \] Dari sini, kita peroleh \(x_1 = x_2\) dan \(y_1 = y_2\), sehingga \(z_1 = z_2\). Oleh karena itu, \(\varphi\) adalah injektif. #### b. **Surjektif** Ambil sembarang matriks \(A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in M\). Untuk \(A\) berada dalam citra \(\varphi\), harus ada \(z = x + y i \in R\) sehingga: \[ A = \begin{bmatrix} x & y \\ -y & x \end{bmatrix} \] Hal ini mensyaratkan bahwa \(c = -b\) dan \(a = d\). Karena setiap matriks di citra \(\varphi\) memenuhi kondisi ini, dan untuk setiap \(a, b \in \mathbb{C}\), terdapat \(z = a + b i \in R\) yang dipetakan ke \(A\), maka \(\varphi\) adalah surjektif. ### 3. **Kesimpulan** Karena \(\varphi\) adalah homomorfisma ring yang bijektif, maka \(\varphi\) adalah isomorfisma ring antara \(R\) (ring bilangan kompleks) dan \(M\) (ring matriks berordo \(2 \times 2\)).

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The Deep Dive

Untuk menunjukkan bahwa \( \varphi \) adalah suatu isomorfisma, kita perlu membuktikan dua hal: bahwa \( \varphi \) adalah homomorfisme (menghormati operasi penjumlahan dan perkalian) dan bahwa \( \varphi \) bijektif (yaitu, injektif dan surjektif). **Homomorfisme:** Untuk penjumlahan, misalkan \( z_1 = x_1 + y_1 i \) dan \( z_2 = x_2 + y_2 i \) di \( R \): \[ \varphi(z_1 + z_2) = \varphi((x_1 + x_2) + (y_1 + y_2)i) = \begin{bmatrix} x_1 + x_2 & y_1 + y_2 \\ -(y_1 + y_2) & x_1 + x_2 \end{bmatrix} \] Sedangkan, \[ \varphi(z_1) + \varphi(z_2) = \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} + \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 & y_1 + y_2 \\ -(y_1 + y_2) & x_1 + x_2 \end{bmatrix} \] Jadi, \( \varphi(z_1 + z_2) = \varphi(z_1) + \varphi(z_2) \). Untuk perkalian, kita punya: \[ \varphi(z_1 z_2) = \varphi((x_1 + y_1 i)(x_2 + y_2 i)) = \varphi((x_1 x_2 - y_1 y_2) + (x_1 y_2 + x_2 y_1)i) = \begin{bmatrix} x_1 x_2 - y_1 y_2 & x_1 y_2 + x_2 y_1 \\ -(x_1 y_2 + x_2 y_1) & x_1 x_2 - y_1 y_2 \end{bmatrix} \] Sedangkan: \[ \varphi(z_1) \varphi(z_2) = \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} = \begin{bmatrix} x_1 x_2 - y_1 y_2 & x_1 y_2 + x_2 y_1 \\ -(x_1 y_2 + y_2 x_1) & x_1 x_2 - y_1 y_2 \end{bmatrix} \] Maka, \( \varphi(z_1 z_2) = \varphi(z_1) \varphi(z_2) \). **Bijektif:** Pemetaan \( \varphi \) adalah injektif, jika \( \varphi(z_1) = \varphi(z_2) \) maka \( \begin{bmatrix} x_1 & y_1 \\ -y_1 & x_1 \end{bmatrix} = \begin{bmatrix} x_2 & y_2 \\ -y_2 & x_2 \end{bmatrix} \), yang mengarah pada \( x_1 = x_2 \) dan \( y_1 = y_2 \), sehingga \( z_1 = z_2 \). Surjektif, setiap matriks bentuk \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \) dapat dihasilkan dari elemen \( a + b i \) di \( R \). Dengan demikian, \( \varphi \) adalah isomorfisma antara \( R \) dan \( M \).

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