Determine if $\mathbb R[X,Y]/I$ is a field

Question

Let $I=\langle Y+X^2-1,XY-2Y^2+2Y \rangle$, decide whether if $\mathbb R[X,Y]/I$ is a field.

What I've tried to do was the following:

If $\mathbb R[X,Y]/I$ is a field, then $I$ is a maximal which also implies $I$ is a prime. As I couldn't prove $\mathbb R[X,Y]/I$ is a field, I've tried to show it isn't. First let me call $$p=Y+X^2-1, \space q=XY-2Y^2+2Y$$ I've noticed that $$p=Y+(X-1)(X+1),q=Y(X-2(Y-1))$$

So I took $g=Y,h=X-2(Y-1)$, then $gh=q \in I$ but $g,h \not \in I$.

I have the feeling that my solution is correct but I don't know how to formally prove that $g$ and $h$ are not in $I$. Would it be correct to argue: any element $s \in I$ is of grade at least 2 but $deg(g)=deg(h)=1$?

I would appreciate if someone could take a look at my answer and make the necessary corrections, thanks in advance.

Answer 1

If $Y\in I$, then $Y=(Y+X^2-1)f(X,Y)+(XY-2Y^2+2Y)g(X,Y)$. It follows that $Y\mid (Y+X^2-1)f(X,Y)$ and since $\gcd(Y,Y+X^2-1)=1$ we must have $Y\mid f(X,Y)$ hence $f(X,Y)=Yf_1(X,Y)$. Thus we get $1=(Y+X^2-1)f_1(X,Y)+(X-2Y+2)g(X,Y)$. Now set $X=0$ and $Y=1$ and reach a contradiction.

If $X-2Y+2\in I$, then $X-2Y+2=(Y+X^2-1)f(X,Y)+(XY-2Y^2+2Y)g(X,Y)$. It follows that $X-2Y+2\mid (Y+X^2-1)f(X,Y)$ and since $\gcd(X-2Y+2,Y+X^2-1)=1$ (why?) we must have $X-2Y+2\mid f(X,Y)$ hence $f(X,Y)=(X-2Y+2)f_1(X,Y)$. Thus we get $1=(Y+X^2-1)f_1(X,Y)+Yg(X,Y)$. Now set $X=1$ and $Y=0$ and reach a contradiction.

Answer 2

Yes, you are right.

The elements of $I$ are all of the form $f\cdot p\ +\ g\cdot q$, and unless all higher terms cancel each other somehow, it has indeed degree $\ge 2$. This 'unless' part might be clarified e.g. by verifying that $p$ and $q$ are coprime in $\Bbb R[X,Y]$.

For another approach, consider first the ideal $I_1:=\langle Y+X^2-1 \rangle$. Taking the quotient by it means that we force $\ Y+X^2-1\ :=\ 0\ $ (in the quotient ring), i.e. that $Y=1-X^2$. So, formally we could just keep $X$ and substitute $Y=1-X^2$ for $Y$.

To make it precise, consider the homomorphism $\varphi_1:\Bbb R[X,Y]\to\Bbb R[X]$ which sends (evaluates) $Y\mapsto 1-X^2$ and fixes the whole $\Bbb R[X]$. Check that $\ker\varphi_1=I_1$, so that $\Bbb R[X,Y]/I_1\cong\Bbb R[X]$.

Finally, using $\varphi_1$, unless I miscalculated, we will get $\Bbb R[X,Y]/I\cong\Bbb R[X]/\langle X(1-X)(1+X)(1+2X)\rangle$ which seems to have lots of zero divisors.

Answer 3

Note that $I \subseteq \langle X+1, Y \rangle \cap \langle X-1, Y \rangle$, so it cannot be a maximal ideal since it is contained in two distinct maximal ideals.

To show this, note that $q \in \langle Y \rangle $, while $p = Y + (X-1)(X+1) \in \langle X+1, Y \rangle \cap \langle X-1, Y \rangle$.

Answer 4

Consider $f(x,y)=y$ and $g(x,y)=x-2y+2$.

Then $f\cdot g = 0 \in \Bbb R[X,Y]/I$

Answer 5

Fix $I=\langle y+x^2-1,xy-2y^2+2y \rangle=\langle f_1,f_2\rangle$. I will find a Gröbner basis for $I$. We first cancel leading terms $$\Delta_{1,2}=yf_1-x f_2=(2x+1)(y^2-y)=2xy^2-2xy+y^2-y$$

Performing polynomial division, $$\Delta_{1,2}=(2y-2)f_2+(4y^3-7y^2+3y)$$

Set $f_3=4y^3-7y^2+3y$, note $f_3\in I$. Then $$\Delta_{1,3}=4y^3 f_1-x^2f_3=7x^2y^2-3x^2y+4y^4-4y^3$$

and again $$\Delta_{1,3}=(7y^2-3y)f_1+(y+1) f_3$$

Now finally $$\Delta_{2,3}=(7y-3)f_2+(2y-2)f_3$$

This means that $(x^2+y-1,xy+2y(1-y),y(y-1)(4y-3))$ is Gröbner base for you ideal. Thus, to decide if a polynomial is in $I$ it suffices you divide this polynomial by such base. Observe that since $x^2,xy,y^3$ are not divisible by each other, this is a reduced Gröbner basis. Moreover, $x^2$ divides no monomial term in $f_2,f_3$, $xy$ divides no monomial term in $f_1,f_3$ and $y^3$ divides no monomial term in $f_1,f_2$, so that is is the unique reduced Gröbner basis of $I$ with respect to the lexicographical monomial ordering $x>y$. Since $y$ is its own remainder upon division, one gets that $y\notin I$, and similarly $x+2(y-1)$ is its own remainder upon division, so this is not in $I$ either. You can conclude your quotient is not even a domain.