1. ## field

1. Prove that $R = \{a+bi: a,b \in \mathbb{Z} \}$ is a subring of $\mathbb{C}$ and that $M = \{a+bi: 3|a \ \text{and} \ 3|b \}$ is a maximal ideal in $R$.

Showing it's a subring is obvious. Just show closure under addition and multiplication and that $0_{\mathbb{C}} \in R$. Also show that the solution of the equation $a+x = 0_{\mathbb{C}}$ is in $R$.

To show that $M$ is maximal consider the following: $M \subset J \subset R$ means that there is some $j \in J$ such that $j \notin M$ where $j = r+si$. So $3 \not| \ |j| = r^2+s^2 = (r+si)(r-si)$. And any ideal that contains $r+si$ must contain $1$ so that $J = R$.

Is this correct?

2. To show that $R/M$ is a field we know this from the fact that $M$ is maximal. How do you show it has 9 elements?

2. Originally Posted by Sampras
1. Prove that $R = \{a+bi: a,b \in \mathbb{Z} \}$ is a subring of $\mathbb{C}$ and that $M = \{a+bi: 3|a \ \text{and} \ 3|b \}$ is a maximal ideal in $R$.

Showing it's a subring is obvious. Just show closure under addition and multiplication and that $0_{\mathbb{C}} \in R$. Also show that the solution of the equation $a+x = 0_{\mathbb{C}}$ is in $R$.

To show that $M$ is maximal consider the following: $M \subset J \subset R$ means that there is some $j \in J$ such that $j \notin M$ where $j = r+si$. So $3 \not| \ |j| = r^2+s^2 = (r+si)(r-si)$. And any ideal that contains $r+si$ must contain $1$ so that $J = R$.

Is this correct?

No! Why would J = R if J contains r + si = j??? J = R iff 1 is in J, and this is what you must show.

2. To show that $R/M$ is a field we know this from the fact that $M$ is maximal. How do you show it has 9 elements?

Try to find 9 different elements in the quotient ring. It isn't that hard as you may think at first, and it'll give some hours of leisure...and learning.

Tonio

3. Originally Posted by Sampras
To show that $M$ is maximal consider the following: $M \subset J \subset R$ means that there is some $j \in J$ such that $j \notin M$ where $j = r+si$. So $3 \not| \ |j| = r^2+s^2 = (r+si)(r-si)$. And any ideal that contains $\color{red}r+si$ must contain $\color{red}1$ so that $J = R$.
Why is this true?

4. Originally Posted by Sampras
1. Prove that $R = \{a+bi: a,b \in \mathbb{Z} \}$ is a subring of $\mathbb{C}$ and that $M = \{a+bi: 3|a \ \text{and} \ 3|b \}$ is a maximal ideal in $R$.

Showing it's a subring is obvious. Just show closure under addition and multiplication and that $0_{\mathbb{C}} \in R$. Also show that the solution of the equation $a+x = 0_{\mathbb{C}}$ is in $R$.

To show that $M$ is maximal consider the following: $M \subset J \subset R$ means that there is some $j \in J$ such that $j \notin M$ where $j = r+si$. So $3 \not| \ |j| = r^2+s^2 = (r+si)(r-si)$. And any ideal that contains $r+si$ must contain $1$ so that $J = R$.

Is this correct?

Not as far as I can see. The element $j= r+si$ is an arbitrary element in $J$; how does containing this element makes an ideal the whole ring?!
I think you must show that the norm of such an element j is coprime with the norm of any element in $M$ and thus...

Read here, in particular 10.2: it is exactly what you need and want for this and the next question:
http://www.fen.bilkent.edu.tr/~franz/nt/ch10.pdf

Tonio

2. To show that $R/M$ is a field we know this from the fact that $M$ is maximal. How do you show it has 9 elements?
.

5. in the following i'll use $\mathbb{F}_3$ for the field of order 3 instead of $\mathbb{Z}/3\mathbb{Z}$:

the ideal $\mathcal{N}=$ is a maximal ideal of $\mathbb{F}_3[x]$ because $x^2+1$ is irreducible. thus $\frac{\mathbb{F}_3[x]}{\mathcal{N}}$ is a field of order 9. define $f: R \longrightarrow \frac{\mathbb{F}_3[x]}{\mathcal{N}}.$ by $f(a+bi)=a+bx + \mathcal{N}.$

see that $f$ is a surjective ring homomorphism and clearly: $\ker f = \{a+bi : \ 3 \mid a, \ 3 \mid b \}=\mathcal{M}.$ therefore $\frac{R}{\mathcal{M}} \cong \frac{\mathbb{F}_3[x]}{\mathcal{N}}$ and you're done.

6. Originally Posted by Sampras
1. Prove that $R = \{a+bi: a,b \in \mathbb{Z} \}$ is a subring of $\mathbb{C}$ and that $M = \{a+bi: 3|a \ \text{and} \ 3|b \}$ is a maximal ideal in $R$.

Showing it's a subring is obvious. Just show closure under addition and multiplication and that $0_{\mathbb{C}} \in R$. Also show that the solution of the equation $a+x = 0_{\mathbb{C}}$ is in $R$.

To show that $M$ is maximal consider the following: $M \subset J \subset R$ means that there is some $j \in J$ such that $j \notin M$ where $j = r+si$. So $3 \not| \ |j| = r^2+s^2 = (r+si)(r-si)$. And any ideal that contains $r+si$ must contain $1$ so that $J = R$.

Is this correct?

2. To show that $R/M$ is a field we know this from the fact that $M$ is maximal. How do you show it has 9 elements?
This is another approach.

It is clear that every odd prime in $\mathbb{Z}$ has a form either 1(mod 4) or 3(mod 4) and an even prime 2=(1+i)(1-i) is reducible in $R=\mathbb{Z}[i]$. If $p \equiv \text{1(mod 4)}$, then $p=x^2 + y^2$ with x and y integers by Fermat's two squares theorem. Further, $p=x^2 + y^2=(x + yi)(x - yi)$ and neither $x+yi$ nor $x-yi$ is a unit because their norms are bigger than 1. Thus $p \equiv \text{1(mod 4)}$ is reducible.

Since $R=\mathbb{Z}[i]$ is an integral domain (not a field) and a PID, we have $p \equiv \text{3(mod 4)}$ as irreducible elements and $

= \{a+bi: p|a \ \text{and} \ p|b \}$

are maximal ideals in $R=\mathbb{Z}[i]$ (For example, <3>, <7>, <11>, <19>, ..., are maximal ideals in $R=\mathbb{Z}[i]$).