Question: Consider R = {a + bi | a, b Z2} = Z2[i], the ring of Gaussian integers

Question

Question: Consider R = {a + bi | a, b Z2} = Z2[i], the ring of Gaussian integers modulo 2. Find all the units in R. Find all the zero divisors in R. Find all the maximal ideals in R. Find all the prime ideals in R. What is the characteristic of R? Is R an integral domain? Is R a field?

Answer: R = {0,1,i,1+i}. 1 and i are units in R, [How do I find these units] 1+i is a zero divisor in R. [How do I find this zero divisor] In particular, R is not an integral domain or field. [How do I know and check?] {0, 1 + i} is the only nontrivial proper ideal in R. [How do I find and realize that this is a proper ideal?] It is maximal and prime. [How do I check and know and can realize that it is?] ({0, 1+i} is also principal, {0, 1+i} = (1 + i)R) [What is a principal?]

Please explain the answer to me and walk me through my trouble. PLEASE. Most of the time there is something that isn't legible please explain, it is because I really don't understand.

Explanation / Answer

Answer: R = {0, 1, i, 1+i}. 1 and i are units in R, [How do I find these units: Ans: Since 0.a=0 for all a in R, 0 can not be a unit; and (1+i).(1+i)=1-1+2i=2i=0 in R and hence (1+i) can not be a unit in R. Since 1.1=1, i.i=-1=1 in R, 1 and i are units in R.]

1+i is a zero divisor in R. [How do I find this zero divisor: Ans: Since (1+i).(1+i)=1-1+2i=2i=0 in R and hence 1+i is a zero divisor in R.]

In particular, R is not an integral domain or field. [How do I know and check?: Ans: Since (1+i) is nonzero but (1+i).(1+i)=0 in R, R is not an intergral domain and hence R is not a field also (because all the fields are integral domain.]

I={0, 1 + i} is the only nontrivial proper ideal in R. [How do I find and realize that this is a proper ideal?: Ans: Since a.0=0 in I for all a in R, and 0.(1+i)=0 in I, 1.(1+i)=(1+i) in I, i.(1+i)=i-1=1+i in I, and (1+i)(1+i)=0 in I, I is an ideal in R. Also, it is the only proper ideal because 1 and i are unit in R, and hence any ideal J in R containing 1 or i contains all the elements in R, that is, J=R; and if (1+i) does not belong to J, and if J is a proper ideal, then J={0}, that is J is the trivial ideal in R.]

It is maximal and prime. [How do I check and know and can realize that it is? Ans: From the last paragraph, we obtain that I={0, (1+i)} is the maximal ideal in R because any other ideal J containing I propely also contains one of the units is R, and hence J=R; and since all the maximal ideals are also prime ideal, I={0, (1+i)} is also a prime ideal in R]

I={0, 1+i} is also principal, I={0, 1+i} = (1 + i)R [What is a principal?: Ans: By definition, an Ideal I is said to be a principal ideal in a ring R, if there exists an element a in R such that I=aR. Since (1+i)R={0, 1+i}=I, I is a principal ideal in R.]

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