Is $\mathbb{Z_5}[x]/ \langle x^2+3,3 \rangle$ a field?

Question

Is $R=\mathbb{Z_5}[x]/ \langle x^2+3,3 \rangle$ a field?

Answer: If $I=\langle x^2+3,3 \rangle=\{f_1(x)(x^2+3)+f_2(x)3\in\mathbb{Z_5}[x]: f_1(x),f_2(x)\in\mathbb{Z_5}[x]\}$

and if $J=\langle x^2+3 \rangle + \langle 3\rangle=\{g_1(x)+g_2(x): g_1(x)\in\langle x^2+3 \rangle,g_2(x)\in\langle 3\rangle\}=\{a(x)(x^2+3)+b(x)3: a(x),b(x)\in\mathbb{Z_5}[x]\} $.

So can we claim that $I=J$? If this is true then we have that $$J=\langle gcd(x^2+3,3)\rangle=\langle1\rangle=\mathbb{Z_5}[x]$$

And that means $R\cong \{0_5\}$.

But I believe that this is wrong. Could anybody help me please?

General Question: In which cases $\langle f(x),g(x) \rangle = \langle f(x) \rangle + \langle g(x) \rangle $?

Answer 1

I'm not sure what you denote by $\;0_5\;$ (the zero element in the field $\;\Bbb F_5\cong\Bbb Z_5\;?$), but observe that

$$1=2\cdot3\in I:=\langle\,x^2+3,\,3\,\rangle\implies I=R=\Bbb F_5[x]\;$$

and thus we indeed get

$$R/I\cong\{0\}\;$$

which is certainly not a field (if we adscribe to widely agreed on condition that a field must have at least two elements)