The form of maximal ideal in the real polynomial ring $\mathbb R[x,y]$

Question

Every maximal ideal of the real polynomial ring $\mathbb R[x,y]$ is of the form $(x-a, y-b)$ for some $a,b \in \mathbb R$. True or false? Any suggestions?

Answer 1

Hint: How about the ideal $(x, y^2+1)$?

Answer 2

In fact the nullstellensatz allows one to deduce the answer also to the question implicit in your title, of classifying all maximal ideals of $\mathbb R[x,y]$, since there is a simple relation between maximal ideals of $\mathbb R[x,y]$ and maximal ideals of $\mathbb C[x,y]$. According to Th. 1, chap. II.4 of Mumford’s redbook of algebraic geometry, since $\mathbb R[x,y]$ is contained in $\mathbb C[x,y]$, intersecting with $\mathbb R[x,y]$ defines a surjective map from maximal ideals of $\mathbb C[x,y]$ to maximal ideals of $\mathbb R[x,y]$, whose fibers are conjugate pairs of complex ideals. Thus in addition to the maximal ideals of form (x-a,y-b) where a,b are real, the fixed points of the conjugation action, we also have the intersection of $\mathbb R[x,y]$ with maximal ideals of form (x-a,y-b) where at least one of a,b is non real. If a’,b’ are the complex conjugates of a,b, and b say is non real, this intersection ideal is generated by the quadratic (y-b)(y-b’) and the (real) linear equation for the line joining the two points (a,b) and (a’,b’). The example given by Prism is the result of intersecting $\mathbb R[x,y]$ with the complex ideal (x, y-i) or also (x,y+i). I.e. since x=0 is the line joining (0,i) and (0,-i), we get the generators (x, (y-i)(y+i)) = (x, $y^2+1$).

Answer 3

Remember $\rm M\triangleleft R~maximal\iff R/M~field$. Can you obtain $\bf C$ as a quotient?

Answer 4

This is an old question, but I would like to answer it fully for the future readers. I would extend Prism's answer and show that $(x, y^2+1)$ is indeed a maximal ideal of $\mathbb{R}[x,y]$.

We do this by exhibiting an isomorphism between $\mathbb{R}[x,y]/(x, y^2+1)$ and $\mathbb{C}$, which is a field. (Note that if $R$ is a ring and $I$ is an ideal of $R$, then $I$ is a maximal ideal of $R$ iff $R/I$ is a field).

Define $\phi: \mathbb{R}[x,y] \to \mathbb{C}$ by evaluation at $(0,i)$, or

$$f(x,y) \mapsto f(0,i)$$

It is easy to check that $\phi$ is a ring homomorphism. All that remains is to show that $\phi$ is surjective and $ker \;\phi = (x, y^2+1)$.

Any element in $\mathbb{C}$ is of the form $a+bi$ (for $a,b \in \mathbb{R}$), which is indeed the image of $\textbf{a+by}$ in $\mathbb{R}[x,y]$ under $\phi$. Hence, the map is surjective.

Further, $$ker \; \phi = \{p(x,y) \in \mathbb{R}[x,y]: p(0,i) = 0\}$$ Clearly, $$g_1(x,y)\cdot x+ g_2(x,y) \cdot (y^2 + 1) \in ker \; \phi, \; for \;g_1, g_2 \in \mathbb{R}[x,y]$$ $$\implies (x, y^2+1) \subseteq ker \; \phi$$

Now to show the other inclusion, let $p(x,y) \in ker\; \phi$. As $x$ is monic, we can divide $p(x,y)$ by $x$. This means that there exists $q(x,y)$ and $r(x,y)$ in $\mathbb{R}[x,y]$ such that: $$p = q(x,y)\cdot x + r(x,y),$$ where $r=0$ or $deg (r) < 1$ in $x$. This means that $r(x,y)$ is essentially a polynomial in $y$ only. Let us denote it by $r(y)$.

As $p \in ker\; \phi$, we have $$p(0,i) = 0 \implies r(i) = 0.$$ The polynomial in $y$ of least degree in $\mathbb{R}[x,y]$ which has $i$ as a root is $y^2+1$, so it follows that $y^2+1$ divides $r(x,y)$. Then $r(x,y)=(y^2+1)\cdot s(x,y)$ for some $s(x,y) \in \mathbb{R}[x,y]$.

Finally, notice that $$p(x,y) = q(x,y)\cdot x + s(x,y) \cdot (y^2+1)$$ $$\therefore p(x,y) \in (x, y^2+1)$$ $$\implies ker\; \phi \subseteq (x, y^2+1)$$

Thus, we have that $ker\; \phi = (x, y^2+1)$. By the first isomorphism theorem, we have $\mathbb{R}[x,y]/(x, y^2+1)$ is isomorphic to $\mathbb{C}$, so $(x, y^2+1)$ is a maximal ideal of $\mathbb{R}[x,y]$, as required. $\blacksquare$

$$$$ Note that if the ring at question was $\mathbb{C}[x,y]$, then a famous result due to Hilbert shows that the only maximal ideals of $\mathbb{C}[x, y]$ are of the form $(x-a, y-b)$ for some $a,b \in \mathbb{C}$. More generally, the result shows a bijection between points in $\mathbb{C}^m$ (geometric objects) and maximal ideals of $\mathbb{C}[x_1,\ldots , x_m]$ (algebraic objects). This is the starting point of algebraic geometry.