Question

Show that the map f : Z+iZ → Z2, defined by f(a+ib) = (a−b) (mod 2), is an onto ring homomorphism. Describe kef f. Is it a maximal ideal? Justify your answer.

Answer 1

Answer on Question #44732 – Math – Abstract Algebra

Show that the map $f: \mathbb{Z} + i\mathbb{Z} \to \mathbb{Z}_2$, defined by $f(a + ib) = (a - b) \pmod{2}$, is an onto-ring homomorphism. Describe $\text{Ker}(f)$. Is it a maximal ideal? Justify your answer.

Solution.

We need to prove that $\forall x, y \in \mathbb{Z} + i\mathbb{Z}: f(x + y) = f(x) + f(y), f(x \cdot y) = f(x)f(y)$.

$$\begin{aligned} x, y \in \mathbb{Z} + i\mathbb{Z} &\Rightarrow x = a_1 + i b_1, y = a_2 + i b_2, a_1, a_2, b_1, b_2 \in \mathbb{Z}; \\ f(x + y) &= f(a_1 + i b_1 + a_2 + i b_2) = f\big(a_1 + a_2 + i(b_1 + b_2)\big) = \\ &= (a_1 + a_2 - b_1 - b_2) \pmod{2} = (a_1 - b_1 + a_2 - b_2) \pmod{2} = f(x) + f(y); \\ f(x \cdot y) &= f\big((a_1 + i b_1)(a_2 + i b_2)\big) = f\big(a_1 a_2 - b_1 b_2 + i(a_1 b_2 + a_2 b_1)\big) = \\ &= (a_1 a_2 - b_1 b_2 - a_1 b_2 - a_2 b_1) \pmod{2} = (a_1 a_2 - b_1 b_2 - a_1 b_2 - a_2 b_1 + 2 b_1 b_2) \pmod{2} = \\ &= (a_1 a_2 + b_1 b_2 - a_1 b_2 - a_2 b_1) \pmod{2} = \big(a_1(a_2 - b_2) - b_1(a_2 - b_2)\big) \pmod{2} = \\ &= \big((a_1 - b_1)(a_2 - b_2)\big) \pmod{2} = f(x) f(y); \end{aligned}$$

So, $f$ is a homomorphism of rings.

$$f(0) = (0 - 0) \pmod{2} = 0;$$ $$f(1) = (1 - 0) \pmod{2} = 1;$$

All elements of $\mathbb{Z}_2$ are images of $f$, so $f$ is a homomorphism onto.

Find $\text{Ker}(f)$:

$$f(a + ib) = 0 \Rightarrow a - b \equiv 0 \pmod{2} \Rightarrow a \equiv b \pmod{2};$$

Hence:

$$\text{Ker}(f) = \{a + ib \mid a \equiv b \pmod{2}\}.$$

Prove that $\text{Ker}(f)$ is a maximal ideal.

Assume the contrary, i.e. there exists ideal $J \neq \text{Ker}(f)$ such that $1 \notin J$ and $\text{Ker}(f) \subset J$.

$$J \neq \text{Ker}(f) \Rightarrow \exists a + ib \in J: a + ib \notin \text{Ker}(f);$$ $$a + ib \notin \text{Ker}(f) \Rightarrow a - b \equiv 1 \pmod{2} \Rightarrow \begin{bmatrix} a = 2n, b = 2m + 1 \\ a = 2n + 1, b = 2m \end{bmatrix};$$

Hence:

$$\begin{bmatrix} 2n + i(2m + 1) \in J \\ 2n + 1 + i2m \in J \end{bmatrix} \Rightarrow \begin{bmatrix} 2n + i(2m + 1) - (2n - 1 + i(2m + 1)) \in J \\ 2n + 1 + i2m - (2n + i2m) \in J \end{bmatrix} \Rightarrow$$ $$\Rightarrow \left[ \begin{array}{l} 1 \in J \\ 1 \in J \end{array} \right] \Rightarrow 1 \in J - \text{contradiction}.$$

So, our assumption doesn't hold and $\text{Ker}(f)$ is a maximal ideal.

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