If A={x/x^2= -1, x ∈ R}, then n(p(A)) ...plz give the answer i will mark the correct answer as brainliest. plz don't copy the answers from google or any other browser.
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Answer:
Let d∈Z be a square free integer. R=Z[d−−√] ={a+bd−−√|a,b∈Z}. Overall, I'm trying to show that every prime ideal P⊂R is a maximal ideal.
So far I showed that I⊂R is finitely generated. I={(x,s+yd−−√)} And now I'm trying to show that R/P, for some prime ideal is a finite ring with no zero divisors. From there it would follow that R/P is a field and any prime ideal is maximal.
I know R could also be written as R=Z[x]/(x2−d). How could I show that R/P is a quotient of Z/nZ[x]/(x2−d)?
Answer:
A Cartesian product of two sets A and B, written as A×B, is the set containing ordered pairs from A and B. That is, if C=A×B, then each element of C is of the form (x,y), where x∈A and y∈B:
A×B={(x,y)|x∈A and y∈B}.
For example, if A={1,2,3} and B={H,T}, then
A×B={(1,H),(1,T),(2,H),(2,T),(3,H),(3,T)}.
Note that here the pairs are ordered, so for example, (1,H)≠(H,1). Thus A×B is not the same as B×A.
If you have two finite sets A and B, where A has M elements and B has N elements, then A×B has M×N elements. This rule is called the multiplication principle and is very useful in counting the numbers of elements in sets. The number of elements in a set is denoted by |A|, so here we write |A|=M,|B|=N, and |A×B|=MN. In the above example, |A|=3,|B|=2, thus |A×B|=3×2=6. We can similarly define the Cartesian product of n sets A1,A2,⋯,An as
A1×A2×A3×⋯×An={(x1,x2,⋯,xn)|x1∈A1 and x2∈A2 and ⋯xn∈An}.
The multiplication principle states that for finite sets A1,A2,⋯,An, if
|A1|=M1,|A2|=M2,⋯,|An|=Mn,
then
∣A1×A2×A3×⋯×An∣=M1×M2×M3×⋯×Mn.
An important example of sets obtained using a Cartesian product is Rn, where n is a natural number. For n=2, we have
R2 =R×R
={(x,y)|x∈R,y∈R}.
Thus, R2 is the set consisting of all points in the two-dimensional plane. Similarly, R3=R×R×R and so on.