How do we show that an ideal of polynomials is prime I'm trying to solve this exercise:

To do so, I'm trying to prove that $(X_1^2+X_2^2+X_3^2)$ is a prime ideal. 
Suppose now $f,g\in \mathbb R[X_1,X_2,X_3]$ and $f\cdot g\in (X_1^2+X_2^2+X_3^2)$, i. e., $f\cdot g=h\cdot(X_1^2+X_2^2+X_3^2)$, with $h\in \mathbb R[X_1,X_2,X_3]$.
Suppose $f\notin (X_1^2+X_2^2+X_3^2)$, then we have to prove that $g\in (X_1^2+X_2^2+X_3^2)$...
I couldn't go further, I really need help.
Thanks a lot
 A: To show your ideal is prime it is enough to show that its generator is irreducible, for then the generator is a prime element (polynomial ring over a field is a UFD in any number of variables) and so the ideal it generates is a prime ideal. 
Now if $f = X_1^2 + X_2^2 + X_3^2$, it is enough to prove that the dehomogenized polynomial $f_\ast$ (w.r.t. the $X_3$ variable) is irreducible. We have
$$f_\ast = X_1^2 + X_2^2 + 1$$
which we may consider as a polynomial in $\left(\Bbb{R}[X_2]\right)[X_1]$. The coefficient ring is a UFD and so using Eisenstein's Criterion with the prime element $X_2^2 + 1\in \Bbb{R}[X_2]$, we get that $f_\ast$ is irreducible. Thus $f$ is irreducible.
Added for OP: Suppose that $f = gh$. We then have $f_\ast = g_\ast h_\ast$ which means wlog that $g_\ast$ is a real number since $f_\ast$ is irreducible. But now this means that $g$ is a constant times $X_3^k$ where $k$ is a non-negative integer. But this is impossible unless $k = 0$, so that $g$ is unit, thus $f$ is irreducible.
A: Maybe it is noteworthy that more generally, 
$$
F=X^2_1+X^2_2+X^2_3 \in k[X_1,X_2,X_3]
$$
is irreducible, if the characteristic of $k$ is not equal to $2$.
To see this, note that since $k[X_1,X_2]$ is a unique factorization domain, the quadratic ($!$) polynomial 
$$
F=X^2_1+X^2_2+X^2_3 \in (k[X_1,X_2])[X_3]
$$
is reducible, if and only if 
$$-(X^2_1+X^2_2)$$
is a square in $k[X_1,X_2]$.
But if $\operatorname{char}(k) \neq 2$, this is impossible (do you see why?).
Even more generally, a similar argument shows that if $\operatorname{char}(k) \neq 2$, then the quadratic form
$$
G=\sum_{i=1}^nX^2_i \in k[X_1,...,X_n]
$$
is irreducible, if $n \geq 3$ (use induction on $n$).
A: Alternative: It suffices to prove that $\mathbb{R}[x,y,z]/(x^2+y^2+z^2) \otimes_{\mathbb{R}} \mathbb{C}$ is an integral domain. But this is isomorphic to
$\mathbb{C}[x,y,z]/(x^2+y^2+z^2) \stackrel{y \mapsto iy}{\cong} \mathbb{C}[x,y,z]/((x+y)(x-y)+z^2) \cong \mathbb{C}[u,v,z]/(uv+z^2)$
which is obviously an integral domain (or use Eisenstein with the prime element $u$).
A: Dehomogenizing, $\, x^2\!+y^2\!-1\,$ is irreducible (so prime), $ $ by $\,1\!-\!y^2\, $ is nonsquare in the  UFD $\,\mathbb R[y],\,$ since the prime $\,1\!-\!y\,$ occurs to odd power (equivalently apply Eisenstein at the prime $\, 1\!-\!y).$
