Answer to Question #106438 in Abstract Algebra for Garima Ahlawat

Question #106438
Use the ring Z(underroot -2) to show that (1) the quotient ring of a ufd need not be a UFD. (2)an irreducible element of a UFD need not be a prime element
1
Expert's answer
2020-03-27T13:35:50-0400

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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