Let $I＝(a＋bx)$ be an nonzero ideal of quotient ring $R＝\mathbb Z[x]/(x^2＋5)$, here $(a＋bx)$'s $x$ means image of $x$ in quotient ring $\mathbb Z[x]/(x^2＋7)$. Then, why $I＝(2,x＋1)$ is not principal?
My proof: Suppose $I$ is principle, then $I＝(f)$ for some $f∈R$. Then, $2∈I$, so, $2$ need to be written like a form $fg$, $f,g∈R$. But I cannot proceed from here.. This is a quotient ring, so I cannot say $f,g$ are constants.
Thank you in advance.
I will write formally $w=\sqrt{-5}$ instead of $x$ modulo the ideal generated in $\Bbb Z[x]$ by $(x^2+5)$. So $w^2=-5$. Let $R$ be the quotient ring $\Bbb Z[x]/(x^2+5)$. (We may formally write now $R=\Bbb Z[w]=\Bbb Z[\sqrt{-5}]$...)
Observe that there is a ring isomorphism $R\to R$, $a+bw\to \overline{a+bw}:=a-bw$, induced by $x\to -x$. (Since the generator $(x^2+5)$ of the quotient ideal is mapped into an element, in fact the same element, of the quotient idea.) Here and below, $a,b\in \Bbb Z$. Then there is a multiplicative "norm" map $R\to \Bbb Z_{\ge 0}$ defined by $$ N(a+bw) = (a+bw)(a-bw)=a^2+5b^2\ . $$ Assume now that the given ideal is principal, and let us write $2=fg$, $1+w=fh$. We pass to the norms and get: $$ \begin{aligned} 4 &= 2^2+5\cdot 0^2 = N(2)=N(fg)=N(f)\; N(g)\ ,\\ 6 &= 1^2+5\cdot 1^2 = N(1+w)=N(fh)=N(f)\; N(h)\ , \end{aligned} $$ so $N(f)$ divides in $\Bbb Z$ both $4$ and $6$, so it non-zero, positive, and further either $1$ or $2$. If $N(f)=1$, then $f\bar f=1$, so $f$ is a unit, contradiction, since $1$ is not in the given ideal $(2,x+1)$, as seen by computing $\Bbb Z[x]/(x^2+5)/(2)$. So $N(f)=2$. But there is no solution to $a^2+5b^2=2$, since this immediately implies $b=0$ (else $5b^2>2$ already), and $a^2=2$ has no solution in $\Bbb Z$.
The other two answers are fine, but I wanted to provide a proof which uses no machinery at all.
Because $x^2 + 5 = 0$ in $R$, all elements of $R$ have the form $ax+b$.
So, suppose for a contradiction that $f$ exists and write $f = ax+b$. Note that $2\in (f)$ and $(f)\neq R$ by hypothesis, so the next proposition gives us the only possibility for $f$.
Proposition: If $2\in (f)$ and $(f)\neq R$, then $f = \pm 2$.
Since $2\in(f)$, we must be able to find $g = cx + d$ with $fg = 2$.
This gives $$2 = (ax+b)(cx+d) = acx^2 + (ad+bc)x +bd = bd-5ac + (ad+bc) x.$$ which leads to the system of equations $$\begin{cases} bd-5ac &= 2\\ ad+bc &= 0\end{cases}.$$
If $b = 0$, the first equation gives $-5ac = 2$, giving an obvious contradiction. Thus, we must have $b\neq 0$. Since $(f) = (-f)$, we may therefore assume that $b > 0$. The second equation can now be solved for $c$, giving $c = - \frac{ad}{b}$. Substituting this into the first equation, we find $bd + 5a\frac{ad}{b} = 2$, which simplifies to $(b^2 + 5a^2) d = 2b$.
This tells you that $b^2 + 5a^2$ divides $2b$. However, $b^2 > 2b$ unless $b\leq 2$. Thus, we have two cases to consider: $b=1$ or $b=2$.
If $b=1$, we find $(1+5a^2)d = 2$ which obviously implies that $a = 0$. But then $f = 1$, so is a unit, so $(f) = R$, giving a contradiction.
If $b=2$, we find $(4+5a^2)d = 4$, which again implies $a=0$. Thus, $f=2$. $\square$
So, we now know that $f = \pm 2$. Can $x+1\in (f)$? Well if so, this means that there is an $h = Cx + D$ with $fh = \pm2(Cx+D) = x+1$. This gives $2C = \pm 1$ and $2D = \pm 1$, which obviously has no solution. Thus, the case $f=\pm 2$ is ruled out. Having ruled out all possible choices for $f$, the ideal $(2,x+1)\subseteq R$ must be not principal.
With $I=(2,x+1)$, we have $I^2=(4,2x+2,x^2+2x+1)$. This can be simplified as follows: $$\begin{align}I^2&=(4,2x+2,x^2+2x+1)\\&=(4,2x+2,x^2+2x+5)\\&=(4,2x+2,2x)\\&=(4,2,2x)\\&=(2,2x)\\&=(2)\end{align}$$
Thus $I^2=(2)$. Suppose that $I=(\alpha)$ is principal, then $(\alpha^2)=(2)$. This means that $2\mid \alpha^2$, and $\alpha^2\mid 2$, so $\alpha^2=2u$ for some unit $u$. Taking the norm, we get $N(\alpha)^2=4$, so $N(\alpha)=2$. But this is impossible, as $$N(\alpha)=a^2+5b^2\neq 2$$
Question: "Then, why $I＝(2,x＋1)$ is not principal?"
Answer: There is a general result (1) classifying maximal ideals in $A:=\mathbb{Z}[x]$. An ideal $I\subseteq A$ is maximal iff there is a (non-zero) prime $p\in \mathbb{Z}$ and an irreducible polynomial $f(x)\in B:=\mathbb{Z}/p\mathbb{Z}[x]$ with $I=(p,F(x))$ where $f(x)$ is the reduction of $F(x)$ in $B$. Hence no maximal ideal in $A$ is principal. Your ideal $I$ has the property that $A/I \cong \mathbb{F}_2$ is a field, hence $I$ is maximal. The proof of (1) may be found in a(ny) book on field theory I believe. You may also find online notes with an elementary proof.
