I'm trying to solve the following problem:
Write $$ I = (x^2+2y^2-2,x^2+xy+y^2-2) \subset \mathbb{R}[x,y] $$ as an intersection of maximal ideals.
I have used the Gröbner basis and I found that the zeros of $I$ are equal to $(\sqrt2,0)$ and $(-\sqrt2,0)$ so I know that $$ I \subset (x-\sqrt2,y) \cap (x+\sqrt2,y). $$ I don't know how I can continue. Maybe if I could prove that $\sqrt{I} = I$ then $I =(x-\sqrt2,y) \cap (x+\sqrt2,y)$ will hold, but how I can prove $\sqrt{I} = I$ in this case?
Let $P$ be a minimal prime over $I$. Then $x^2+2y^2-2,x^2+xy+y^2-2\in P$. It follows that their difference $y^2-xy$ belongs to $P$. Since $P$ is prime we have $y\in P$ or $y-x\in P$.
In the first case $x^2-2\in P$. There are two subcases: $x-\sqrt 2\in P$ or $x+\sqrt 2\in P$. We thus get two minimal primes over $I$, the ones you already found: $(x\pm\sqrt 2,y)$.
In the second case we get $3x^2-2\in P$. Again there are two subcases: $x-\sqrt{\frac23}\in P$ or $x+\sqrt{\frac23}\in P$. We thus get another two minimal primes over $I$: $(x\pm\sqrt{\frac23},y-x)$.
It follows that $\sqrt I$ is an intersection of four prime (actually maximal) ideals.
Now let us prove that $I$ is radical, which is equivalent to $\mathbb R[x,y]/I$ reduced. In order to show this one can use, e.g., this answer. This means that we have to show that $\mathbb R[x,y]_{(x-\sqrt 2,y)}/I\mathbb R[x,y]_{(x-\sqrt 2,y)}$ (and another three rings) is a domain. It is easily seen that this quotient ring is in fact a field. Notice that $I\mathbb R[x,y]_{(x-\sqrt 2,y)}$ contains $y(y-x)$ and since $y-x\notin(x-\sqrt 2,y)$ it is invertible in the localization. Then $y\in I\mathbb R[x,y]_{(x-\sqrt 2,y)}$ and we can write $I\mathbb R[x,y]_{(x-\sqrt 2,y)}=(I,y)\mathbb R[x,y]_{(x-\sqrt 2,y)}$. But $(I,y)=(x^2-2,y)$. As above, we notice that $x+\sqrt2\notin(x-\sqrt 2,y)$ and hence it is invertible in the localization. Thus we get $I\mathbb R[x,y]_{(x-\sqrt 2,y)}=(x-\sqrt 2,y)\mathbb R[x,y]_{(x-\sqrt 2,y)}$, and the factor ring $\mathbb R[x,y]_{(x-\sqrt 2,y)}/I\mathbb R[x,y]_{(x-\sqrt 2,y)}$ is isomorphic to $\mathbb R$. We proceed the same way for the other three maximal ideals.