$I=(3,x^2+1)$ be an ideal of $\mathbb{Z}[x]$. Then $I$ is a proper ideal of $\mathbb{Z}[x]$ This is from an old exam.
Here is the problem.
Let $I=(3,x^2+1)$ be an ideal of $\mathbb{Z}[x]$. Then $I$ is a proper ideal of $Z[x]$
How does one go about showing that this is a proper ideal. Should I consider $Z[x]/I$ and then show that it is neither $Z[x]$ nor $0$ thus showing that $I$ must be a proper ideal or is there some other easier way to proceed?
My attempt: Since $I=\{ 3p(x)+(x^2+1)q(x): p(x), q(x) \in Z[x] \} $ we know that the constant term of $f(0)$ of any element $f(x) \in I$ must be of the form $3n+m$ for $m,n \in \mathbb{Z}$. Does this help me in anyway?. Am I making this too complicated than it should be?
Can you kindly help?
Thank you in advance for your answers.
 A: Boring way: show that a single element, say 1, is not in your ideal.  Do this by contradiction --- what can you say about p(x) and q(x) if 1 is in $I$?
Fun way:  set $R = \mathbb{Z}[x]/(x^2+1)$, and show that
1)  R is isomorphic to the ring of Gaussian integers $\mathbb{Z}[i]$, and
2) the quotient $\mathbb{Z}[x]/I$ is isomorphic to $\mathbb{Z}[i]/(3)$
Then use your favorite method to show that $3$ is not a unit in $\mathbb{Z}[i]$, so generates a proper ideal.
A: How far are you with the ring isomorphism theorems to show that $\mathbb{Z}[X]/(3, X^2+1) \cong \mathbb{F}_{3^2} $? Hint: $X^2+1$ is irreducible over $\mathbb{F}_{3}$.
A: If $1$ were of the form $3p(x) + (x^2+1)q(x)$, then plugging in $x=i$, you'd get $p(i) = 1/3$, but $p(i)$ must be of the form $a+bi$ with $a$ and $b$ integers.
A: The field $\mathbb F_9$ has a square root of $-1$. Let us call it $i$.
There is a unique map $f:\mathbb Z[X]\to \mathbb F_9$ such that $f(p)=p(i)$ for all $p\in \mathbb Z[X]$. This is very easy to see that this is in fact a homomorphism of rings.
It is clear that $f(3)=0$ and $f(X^2+1)=0$. Since $f$ is a map of rings, it follows that the whole ideal $I$ generated by $3$ and by $X^2+1$ in $\mathbb Z[X]$ is contained in the kernel of $f$.
Now obviously $f(1)=1\neq0$, so $1\in \mathbb Z[X]$ is not contained in the kernel of $f$ and, a fortiori, it is also not contained in the ideal $I$.
A: Reformulating Barry's answer in a possibly more down to Earth way: 
All the polynomials $p(x)$ in your ideal have the property that $p(i)$ is divisible by three in the ring of Gaussian integers. It is easy to show that polynomials with that property form an ideal in $\mathbb{Z}[x]$. It is also easy to give an example of a polynomial that does not have this property, so $I$ is proper.
