Deduce that if $A$ is a unique factorization domain (UFD) of char- acteristic zero, then $A[X, Y ]/(X^2+Y^2−1)$ is an integral domain.
Let $B = A[X, Y]/(X^2 + Y^2 - 1)$, and suppose that $f(X, Y)$ and $g(X, Y)$ are elements of $B$ such that $f(X, Y)g(X, Y) = 0$ in $B$. We need to show that either $f(X, Y) = 0$ or $g(X, Y) = 0$ in $B$.
Since $A$ is a $UFD$, it is an integral domain. Let $p(X)$ and $q(X)$ be the numerators of $f(X, Y)$ and $g(X, Y)$ respectively, so that $f(X, Y) = p(X)/h(X)$ and $g(X, Y) = q(X)/k(X)$ for some polynomials $h(X)$ and $k(X)$ in $A[X]$ that are coprime with $X^2$ + $Y^2$ - $1$.
Multiplying the equation $f(X, Y)g(X, Y) = 0$ in $B$ by $h(X)k(X)$ gives $p(X)q(X) = 0$ in $A[X]$, which implies that either $p(X) = 0$ or $q(X) = 0$ in $A[X]$, since $A$ is an integral domain. Without loss of generality, assume that $p(X) = 0$ in $A[X]$.
Now let $r(Y)$ be the content of $h(X)$ in $A$, so that $h(X) = r(Y)s(X)$, where $s(X)$ is a primitive polynomial in $A[X]$. Since $h(X)$ is coprime with $X^2$ + $Y^2$- $1$, $s(X)$ is also coprime with $X^2$ + $Y^2$ - $1$. Therefore, we can write $s(X) = u(X) + v(X)Y$ for some polynomials $u(X)$ and $v(X)$ in $A[X]$. Substituting this expression into $h(X) = r(Y)s(X)$ and setting $X$ = $\sqrt(-1)$ gives $h(\sqrt(-1)) = 0$ in $A$, since $X^2$+ $Y^2$ - $1$ = $0$ in $B$ when $X$ = $\sqrt(-1)$ and $Y = 0$.
Since $A$ is a UFD of characteristic zero, it is also a Euclidean domain, and hence a principal ideal domain. Thus, we can write $r(Y) = a*d(Y)$, where $a$ is a unit in $A$ and $d(Y)$ is a monic irreducible polynomial in $A[Y]$. Since $h(\sqrt(-1)) = 0$ in $A$, we have $d(\sqrt(-1)) = 0$ in $A$, which implies that $d(Y)$ divides $Y^2$ + $1$ in $A[Y]$. However, since $A$ has characteristic zero, $Y^2$ + $1$ is irreducible in $A[Y]$, and hence $d(Y)$ is either $1$ or $Y^2$ + $1$.
If $d(Y) = 1$, then $h(X) = a$ for all $X$ in $A$, which implies that $f(X, Y) = p(X)/h(X)$ is a well-defined element of $A[X, Y]/(X^2 + Y^2 - 1)$, and hence $f(X, Y) = 0$ in $B$ implies that $p(X) = 0$ in $A[X]$. This in turn implies that $f(X, Y) = 0$ in $A[X, Y]/(X^2 + Y^2 - 1)$, since $p(X)$ and $h(X)$ are coprime in $A[X]$.
If $d(Y)$ = $Y^2$ + $1$, then $h(X)$ is irreducible in $A[X]$ and has degree at most two, since it divides $X^2$ + $Y^2$ - $1$, which is irreducible in $A[X, Y]$. Therefore, $h(X)$ is either $X$ - $\sqrt(-1)$, $X$ + $\sqrt(-1)$, or $X^2$ + $Y^2$ - $1$, up to associates in $A[X]$. In the first two cases, we can write $h(X)$ = $X$ - $\sqrt(-1)$ or $h(X)$ = $X$ + $\sqrt(-1)$, and hence $f(X, Y)$ = $p(X)$/($X$ - $\sqrt(-1)$) or $f(X, Y) = p(X)/(X + \sqrt(-1))$. In either case, we can write $f(X, Y) = a(X, Y) + b(X, Y)Y$, where $a(X, Y)$ and $b(X, Y)$ are polynomials in $A[X, Y]$.
Since $f(X, Y)$ is well-defined in $B$, we have $f(\sqrt(-1), 0) = a(\sqrt(-1), 0) = 0$ in $A$, which implies that $a(X, Y)$ is divisible by $X^2$ + $Y^2$ - $1$ in $A[X, Y]$. Therefore, we can write $a(X, Y) = (X^2 + Y^2 - 1)c(X, Y)$ for some polynomial $c(X, Y)$ in $A[X, Y]$. Substituting thisexpression into $f(X, Y) = a(X, Y) + b(X, Y)Y$ gives $f(X, Y) = (X^2 + Y^2 - 1)c(X, Y) + b(X, Y)Y$. Multiplying this equation by $h(X)$ gives $p(X) = (X^2 + Y^2 - 1)c(X, Y)s(X) + b(X, Y)Yr(Y)s(X)$, which implies that $p(X)$ is divisible by $X^2 + Y^2 - 1$ in $A[X]$, since $s(X)$ and $r(Y)$ are coprime with $X^2 + Y^2 - 1$. This contradicts the assumption that $p(X) = 0$ in $A[X]$, and hence we must have $d(Y) = 1$ in this case as well.
Therefore, we have shown that if $A$ is a UFD of characteristic zero, then $A[X, Y]/(X^2 + Y^2 - 1)$ is an integral domain.
Please cheak this...or give another solution...
Also 2nd prt is
If the characteristics of $A$ is not zero .let the characteristics of $A$ is positive $p$. Then Give a necessary and sufficient condition on $p$ such that $A[X,Y]/(X²+Y²-1)$ is integral domain.
And now if $A$ is replaced by set of integers $ℤ$ then give examples of prime ideals in $ℤ[X,Y]/(X²+Y²-1)$ which are not maximal and give examples of maximal ideals in $ℤ[X,Y]/(X²+Y²-1)$.
Thank you
I think you only need the assumption that $\operatorname{char}(A)\neq 2$.
If $A$ is a UFD, so is $A[X]$ and thus also $A[X,Y]=(A[X])[Y]$. An element $p\in A$ is prime iff $A/(p)$ is an integral domain. Moreover, in a UFD it holds that irreducible $\iff$ prime (in general only $\Leftarrow$ holds). Finally, for a UFD $R$ the Eisenstein-criterion holds: Given a monic polynomial $p=X^{n+1}+r_nX^n+...+r_0 \in R[X]$ and a prime ideal $\mathfrak p\subseteq R$ such that $r_i \in \mathfrak p$ for all $i$ and $r_0 \notin \mathfrak p^2$, then $p$ is irreducible.
Now consider the specific case $$R=A[X],\;\;p=Y^2+(X^2-1)\in R[Y],\;\;\mathfrak{p}=(X-1)\subseteq A[X].$$ We have $(X-1)^2 = (X^2 - 2X+1)$ and since $\operatorname{char}(A)\neq 2$ we find $$X^2-1 = (X-1)(X+1) \in (X-1) \setminus (X-1)^2 = \frak{p\setminus p^2}.$$ By the Eisenstein criterion $p$ is irreducible / prime in $R[Y] = A[X,Y]$ making $A[X,Y]/(X^2+Y^2-1)$ an integral domain.
If $\operatorname{char}(A) = 2$ then $$(X+Y+1)^2 = X^2+2XY+Y^2+2(X+Y)+1 = X^2+Y^2-1$$ (using that $2=0$ and $1=-1$), so $X^2+Y^2-1$ is not prime, in particular $A[X,Y]/(X^2+Y^2-1)$ is not an integral domain.