Studying the intersection $(X)\cap (X^{2}-Y+1)\subseteq\mathbb{R}[X,Y]$.

Question

I am trying to find the intersection of ideals $$ (X)\cap (X^{2}-Y+1)\subseteq\mathbb{R}[X,Y]. $$

This is what I have tried: $$ f\in(X^{2}-Y+1)\Rightarrow f=g\cdot (X^{2}-Y+1)\text{ for certain }g\in\mathbb{R}[X,Y], $$ $$ f\in (X)\Rightarrow 0=f(0,t)=g(0,t)\cdot (1-t)\text{ for all }t\in \mathbb{R}. $$ How could we follow?

I would also like to know if $\mathbb{R}[X,Y]/((X)\cap (X^{2}-Y+1))$ has idempotent elements different from $0,1$.

Answer 1

$(X)\cap (X^{2}-Y+1)$ is very simple to find: $X$ and $X^{2}-Y+1$ are irreducible polynomials (since they have degree one in one of the variables), and obviously $\gcd(X,X^{2}-Y+1)=1$. Then $(X)\cap (X^{2}-Y+1)=(X)(X^{2}-Y+1)$.

Now about the idempotents: if $f^2-f\in(X)\cap (X^{2}-Y+1)$, then $X\mid f(f-1)$ and $X^{2}-Y+1\mid f(f-1)$. If $X\mid f$ and $X^{2}-Y+1\mid f$, then $f=0$ in the quotient ring $\mathbb R[X,Y]/(X)\cap(X^2-Y+1)$, and similarly if $X\mid f-1$ and $X^{2}-Y+1\mid f-1$, then $f=1$. But we could have $X\mid f-1$ and $X^{2}-Y+1\mid f$ (or vice versa), so $f=1+Xg=(X^{2}-Y+1)h$. This leads to $1\in(X,X^{2}-Y+1)=(X,Y-1)$, a contradiction. In conclusion, there are no non-trivial idempotents.

Answer 2

This is not a direct answer to your question, but an answer to your final statement (which I suspect is your true motivation).

So, we want to know whether $\mathbb{R}[x,y]/I$ has any non-trivial idempotents, where

$$I=(x)\cap (x^2-y+1)$$

Well, it's well known that $\mathbb{R}[x,y]/I$ has no non-trivial idempotents if and only if

$$V(I)\subseteq\mathbb{A}^2_\mathbb{R}=\text{Spec}(\mathbb{R}[x,y])$$

is connected.

But, note that if we can show that

$$V(I)_\mathbb{C}=V(I_\mathbb{C})\subseteq\mathbb{A}^2_\mathbb{C}$$

is connected, then we're OK since there is a surjection $V(I)_\mathbb{C}\to V(I)$.

Note though that

$$V(I)_\mathbb{C}=V((x))\cup V((x^2-y+1))$$

and, moreover, $(x)$ and $(x^2-y+1)$ are prime ideals in $\mathbb{C}[x,y]$. So, certainly $V((x))$ and $V((x^2-y+1))$ are connected. Thus, $V(I)_\mathbb{C}$, being the union of these two connected spaces, will be connected if they have a point in common. But, certainly the point $(\zeta_3,0)$ (or $(x-\zeta_3,y)$ if you want to think in terms of primes) is such a point, where $\zeta_3$ is a primitive third root of unity.

Answer 3

You can use Gröbner bases. Let $I=(X)$ and $J=(X²-Y+1)$, and consider the ideal $K=tJ+(1-t)I$ in $K[t,X,Y]$. Then $I\cap J$ is the intersection of $K$ with $k[X,Y]$, and in particular you can obtain explicit generators by computing a Gröbner basis with respect to the lexicographic monomial ordering $t>X>Y$.

Mathematica gives $\{X^3-XY+X,tY-Y+X²-Y+1,tX\}$ and the term $X^3-XY+X$ is the only term not involving $t$, hence $(X)\cap(X^2-Y+1)=(X^3-XY+X)$. Note that this means $I\cap J=IJ$.