Question

Let S = { matrix [ a 0] | a,b ∈ Z } .
[0 b ]

i) Check that S is a subring of M(subscript2)(R) and it is a commutative ring with identity.
ii) Is S an ideal of M(subscript2)(R)? Justify your answer.
iii) Is S an integral domain? Justify your answer.
iv) Find all the units of the ring S.
v) Check whether
I = { matrix [ a 0] | a,b ∈ Z, 2 | a } .
[0 b]
is an ideal of S.
vi) Show that S is congruent to Z×Z where the addition and multiplication operations are componentwise addition and multiplication.

Answer 1

Answer on Question #44738 – Math – Abstract Algebra:

Let $S = \left\{\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \mid a, b \in \mathbb{Z}\right\}$.

(a) Check that $S$ is a subring of $M_2(\mathbb{R})$ and it is a commutative ring with identity.

(b) Is $S$ an ideal of $M_2(\mathbb{R})$? Justify your answer.

(c) Is $S$ an integral domain? Justify your answer.

(d) Find all the units of the ring $S$.

(e) Check whether $I = \left\{\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \in S \mid 2|a\right\}$ is an ideal of $S$.

(f) Show that $S$ is congruent to $\mathbb{Z} \times \mathbb{Z}$ where the addition and multiplication operations are componentwise addition and multiplication.

Solution.

(a)

We need to check the following properties:

$$\forall A, B \in S: A + B, A \cdot B \in S;$$ $$\forall A, B \in S: A \cdot B = B \cdot A;$$ $$\exists e \in S: \forall A \in S: A \cdot e = A;$$

So:

$$A, B \in S \Longrightarrow A = \left( \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right), B = \left( \begin{array}{cc} c & 0 \\ 0 & d \end{array} \right) \Longrightarrow \left\{ \begin{array}{c} A + B = \left( \begin{array}{cc} a + c & 0 \\ 0 & b + d \end{array} \right) \in S \\ A \cdot B = \left( \begin{array}{cc} a c & 0 \\ 0 & b d \end{array} \right) \in S \end{array} ; \right.$$ $$B \cdot A = \left( \begin{array}{cc} c a & 0 \\ 0 & d b \end{array} \right) = \left( \begin{array}{cc} a c & 0 \\ 0 & b d \end{array} \right) = A \cdot B;$$ $$e = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \in S, A = \left( \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right) \Longrightarrow A \cdot e = \left( \begin{array}{cc} a \cdot 1 & 0 \\ 0 & b \cdot 1 \end{array} \right) = A.$$

So, $S$ is a commutative ring with identity $e$.

(b)

Prove that $S$ is not ideal of $M_2(\mathbb{R})$.

Assume the contrary.

$$S \text{ is an ideal } \Longrightarrow \forall A \in M_2(\mathbb{R}), \forall X \in S: A \cdot X \in S;$$ $$e \in S \Longrightarrow \forall A \in M_2(\mathbb{R}): A \cdot e = A \in S \Longrightarrow S = M_2(\mathbb{R});$$

But, for example, $A = \left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array} \right) \notin S$.

So, our assumption doesn't hold and $S$ is not ideal.

(c)

$S$ is not an integral domain, because:

$$A = \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right) \neq 0, B = \left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right) \neq 0, A \cdot B = \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right) = 0.$$

(d)

Assume that $x = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}$ is a unit of the ring $S$. So:

$$\exists y = \left( \begin{array}{cc} c & 0 \\ 0 & d \end{array} \right) \in S: x \cdot y = e;$$ $$x \cdot y = \left( \begin{array}{cc} a c & 0 \\ 0 & b d \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right) \Longrightarrow a c = b d = 1;$$ $$a, c \in \mathbb{Z}, a c = 1 \Longrightarrow \left[ \begin{array}{c} a = c = 1 \\ a = c = -1 \end{array} \right];$$

Hence:

$$a, b = \pm 1 \Rightarrow x \in \left\{\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right), \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right), \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right), \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array} \right) \right\}.$$

So, there are 4 units in $S: \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$.

(e)

$$A \in S, B \in I \Rightarrow A = \left( \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right), B = \left( \begin{array}{cc} 2n & 0 \\ 0 & m \end{array} \right) \Rightarrow A \cdot B = \left( \begin{array}{cc} 2an & 0 \\ 0 & bm \end{array} \right) \in I.$$

So, $I$ is an ideal of $S$.

(f)

We need to show that the mapping $f \colon S \to \mathbb{Z} \times \mathbb{Z}, f\left( \begin{array}{cc}a & 0\\ 0 & b \end{array} \right) = (a,b)$ is an isomorphism of rings.

Note that $f$ is bijective. Show that $f$ saves both operations:

$$A, B \in S \Rightarrow A = \left( \begin{array}{cc} a & 0 \\ 0 & b \end{array} \right), B = \left( \begin{array}{cc} c & 0 \\ 0 & d \end{array} \right);$$ $$f(A \cdot B) = f \left( \begin{array}{cc} ac & 0 \\ 0 & bd \end{array} \right) = (ac, bd) = (a, b)(c, d) = f(A)f(B);$$ $$f(A + B) = f \left( \begin{array}{cc} a + c & 0 \\ 0 & b + d \end{array} \right) = (a + c, b + d) = (a, b) + (c, d) = f(A) + f(B);$$ $$f(0_S) = (0,0) = 0_{\mathbb{Z} \times \mathbb{Z}};$$ $$f(e_S) = (1,1) = e_{\mathbb{Z} \times \mathbb{Z}}.$$

So, $f$ is an isomorphism of rings.

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