field

**Sampras** · October 27th 2009, 02:29 PM

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?

**tonio** · October 27th 2009, 02:39 PM

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

**proscientia** · October 28th 2009, 09:39 AM

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?

**tonio** · October 28th 2009, 11:00 AM

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?

.

**NonCommAlg** · October 28th 2009, 12:19 PM

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

the ideal $\mathcal{N}=<x^2+1>$ 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.

**aliceinwonderland** · October 28th 2009, 06:25 PM

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 $<p>= \{a+bi: p|a \ \text{and} \ p|b \}$ are maximal ideals in $R=\mathbb{Z}[i]$ (For example, <3>, <7>, <11>, <19>, ..., are maximal ideals in $R=\mathbb{Z}[i]$ ).

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