where (sqrt(10),2) is the ideal generated by those two elements.
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Let R is a ring and J and ideal; then the ring R/J is an integral domain if and only if J is a prime ideal.
All you have to do is show the ideal J=(sqrt(10),2) is prime in R=Z[sqrt(10)]. I'm going to write i=sqrt(10).
Let x,y be elements in R and suppose xy is in the ideal J. Description of elements in R is clear, since it's just the polynomial expressions in i with integer coefficients. Then x=a+bi and y=c+di, for integers a,b,c,d. Now xy=ac+10bd+i(bc+ad). But xy is in J by assumption so xy=si+2t for integers s,t. So we have
ac+10bd+i(bc+ad)=si+2t
In particular 2t=ac+10bd ((since i is irrational)).
This means ac+10bd is even; 10bd is clearly even, so we must have ac must be even. a,c are integers; if they were both odd then ac would be odd, so one of a,c must be even.
Then one of x,y is of the form 2n+im, for integers n,m, which is an element in J. Then the ideal J is prime and R/J is an integral domain
All you have to do is show the ideal J=(sqrt(10),2) is prime in R=Z[sqrt(10)]. I'm going to write i=sqrt(10).
Let x,y be elements in R and suppose xy is in the ideal J. Description of elements in R is clear, since it's just the polynomial expressions in i with integer coefficients. Then x=a+bi and y=c+di, for integers a,b,c,d. Now xy=ac+10bd+i(bc+ad). But xy is in J by assumption so xy=si+2t for integers s,t. So we have
ac+10bd+i(bc+ad)=si+2t
In particular 2t=ac+10bd ((since i is irrational)).
This means ac+10bd is even; 10bd is clearly even, so we must have ac must be even. a,c are integers; if they were both odd then ac would be odd, so one of a,c must be even.
Then one of x,y is of the form 2n+im, for integers n,m, which is an element in J. Then the ideal J is prime and R/J is an integral domain