The given ring is "\\Z[\\sqrt{-2}]" "=\\{ a+b\\sqrt{-2} :a,b\\in \\Z \\}" .
Claim: The Quotient ring of a UFD need not be a UFD.
Since "R" is an Euclidean domain ,so "R" is PID.
Again , we known that every PID is UFD.
Hence, "R" is a UFD.
Now ,consider the ideal "<6>=\\{" "6(a+(\\sqrt{-2})b \\ ):a,b\\in \\Z \\}"
Then ,"\\frac{R}{I}=\\{ (a+b(\\sqrt{-2}))+<6> :a,b\\in \\Z \\}"
Clearly,"2\\notin <6>," and "3\\notin <6>" but "6\\in<6>"
Hence ,"\\frac{R}{I}" is not a domain.
But we known that ,to became a UFD ,it must be first integral domain.
Hence , it is not a unique factorization domain.
(Proved).
Since "R" is a PID , Every irreducible elements is prime .
Hence, 2nd part of the question is wrong. That is ,we can't find a irreducible element in the given ring R which is not a prime.
But in place of "R" if you take "R'=" "\\Z [\\sqrt{-5}]" then there are beautiful results,
For example:
The ring "\\Z[x]" is a UFD but when we quotient by the ideal"I=<x^2+5>" then,
"\\frac{R'}{I}=\\frac{ \\Z[x]}{I}\\approx\\Z[\\sqrt{-5}]"
Which is not a UFD.
Since ,"46=2.23"
"=(1+3\\sqrt{-5})(1-3\\sqrt{-5})"
We can easily verify that "2,23,(1+3\\sqrt{-5})(1-3\\sqrt{-5})" are irreducible.
Define a norm on "\\Z[\\sqrt{-5}]" by "N(a+b\\sqrt{-5})=a^2+5b^2"
Then "x\\in R' \\ is \\ a \\ unit \\ iff \\ N(x)=1"
Suppose "x=a+b\\sqrt{-5}"is a unit "\\iff" there exit "y=a'+b'\\sqrt{-5}" such that "xy=1 \\"
Suppose "x \\ is \\ a \\ unit \\implies\\exists y \\in R' s.t . \\ xy=1"
So,"N(xy)=N(1)\\implies \\ N(x)N(y)=1"
Hence ,"N(x)=1. \\ since" norm gives non negative integer.
Now ,assume "N(x)=1\\implies\\ a^2+5b^2=1"
Since,"a^2 \\ and \\ 5b^2 are \\ positive \\ number\\implies a=+1 \\ or \\ -1"
Hence,"x=a"
Which is a unit .
Also it follows that the only unit in "R'" are +1 and -1.
If possible let "2=xy" "for \\ some \\ x,y\\in R'" ,where neither x nor y is a unit.
Then "N(2)=N(x)N(y)\\implies N(x)N(y)=4"
"\\therefore N(x)=N(y)=2" ,Since "N(x) \\ and \\ N(y) are" non negative integer.
Hence,"2=4"
Which is not possible .Hence 2 is irreducible.
Similarly 23 is not irreducible.
If possible let "1+3\\sqrt{-5}=xy" ,for some "x,y\\in R'" where neither x nor y is a units.
Then "N(xy)=N(1+3\\sqrt{-5})=46=N(x)N(y)"
"N(x)N(y)=2\u00d723\\implies \\ either N(x)=2 \\ or \\ N(x)=23"
Therefore, "N(x)=a^2+5b^2=2 \\ or \\ 23"
Since,"a^2,5b^2 \\ are \\ positive \\ number \\ so N(x)\\neq2"
If N(x)=23 then "5b^2=23-a^2\\geq0"
Which is not possible because a is an integer .
Hence, "1+3\\sqrt{-5}" is irreducible.
Similarly,"1-3\\sqrt{-5}" is irreducible.
Also ,we can prove "(1+3\\sqrt{-5})" is irreducible but not a prime
just using the definition of prime in an integral domain.
If possible ,let this number is a prime.
Then,
"(1+3\\sqrt{-5})(1-3\\sqrt{-5})=46=2.23"
"\\implies \\ (1+3\\sqrt{-5}) |2.23"
But neither "(1+3\\sqrt{-5})|2 \\ nor \\ (1+3\\sqrt{-5})|23"
Suppose,"(1+3\\sqrt{-5})\\mid 2" then there exit a number "(a+b\\sqrt{-5} ) \\ such \\ that \\ (a+b\\sqrt{-5})(1+3\\sqrt{-5})=2"
"\\implies a+15b=2 \\ and \\ a+b=0"
Which is impossible.
Hence"(1+3\\sqrt{-5})\\nmid 2."
Similarly,"(1+3\\sqrt{-5})\\nmid23"
Therefore,"(1+3\\sqrt{-5})" is not a prime.
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