Proving Maximal Ideal of $\mathbb{C}[X,Y]$

[evinda]

Hello! (Wave)

How can I show that the ideal $I=<X-1,Y+X^2-1>$ is a maximal ideal of $\mathbb{C}[X,Y]$ ? (Thinking)

 

[Siron]

Perhaps you can use the equivalence:
$$\frac{\mathbb{C}[X,Y]}{(X-1,Y+X^2-1) }\ \mbox{is a field} \ \Leftrightarrow (X-1,Y+X^2-1) \ \mbox{is a maximal ideal in} \ \mathbb{C}[X,Y]$$

That means if you can show that $\frac{\mathbb{C}[X,Y]}{(X-1,Y+X^2-1)}$ is isomorphic to a field, you're done.

 

[Euge]

Adding to Siron's point, consider what the elements of $\Bbb C_1 := \Bbb C[X,Y]/<X-1,Y+X^2-1>$ look like. You can view $\Bbb C_1$ as the set of polynomials in $\Bbb C[X,Y]$ subject to the relations $X -1 = 0$ and $Y + X^2 - 1 = 0$. Those relations are equivalent to $X = 1$ and $Y = 0$. Thus, $\Bbb C_1$ can be identified with the set of all constant polynomials over $\Bbb C$, i.e., $\Bbb C_1$ can be identified with $\Bbb C$.

To be more rigorous, define a mapping $\phi: \Bbb C[X,Y] \to \Bbb C$ by setting $\phi(f(x,y)) = f(1,0)$. Show that $\phi$ is a $\Bbb C$-linear map. The kernel $\text{ker}(\phi)$ contains $I$ since $X - 1$ and $Y + X^2 - 1$ map to zero under $\phi$. On the other hand, if $f\in \text{ker}(\phi)$, then since $X - 1$ and $Y$ belong to $I$ ($Y\in I$ since $Y = (Y + X^2 - 1) + (1 - X)(Y - 1)$),

$$ f(X,Y) = \sum_{1\, \le i + j\, \le \deg(f)} \frac{\partial^{i+j}(f)}{\partial X^i\, \partial Y^j}(1,0)\, (X - 1)^i Y^j \in I.$$

Thus $\text{ker}(\phi) = I$. Now you just have to show that $\phi$ is surjective. Then you can claim that $\Bbb C_1$ is isomorphic to $\Bbb C$. Since $\Bbb C$ is a field, it follows from Siron's comment that $I$ is maximal in $\Bbb C[X,Y]$.

 

[evinda]

Could you explain me further the definition of:
$$\mathbb{C}[X,Y]/<X-1, Y+X^2-1>$$
? (Thinking)

 

[Euge]

Could you explain me further the definition of:
$$\mathbb{C}[X,Y]/<X-1, Y+X^2-1>$$
? (Thinking)

It's the quotient of $\Bbb C[X,Y]$ by the ideal $<X-1, Y+X^2-1>$. It consists of all cosets $p(X,Y) + <X - 1, Y + X^2 - 1>$ where $p(X,Y) \in \Bbb C[X,Y]$.

 

[evinda]

It's the quotient of $\Bbb C[X,Y]$ by the ideal $<X-1, Y+X^2-1>$. It consists of all cosets $p(X,Y) + <X - 1, Y + X^2 - 1>$ where $p(X,Y) \in \Bbb C[X,Y]$.

A ok.. (Nod) How could I continue, in order to show that $\Bbb C[X,Y]/<X-1, Y+X^2-1>$ is isomorphic to $\mathbb{C}$ ? (Thinking)

 

[Euge]

A ok.. (Nod) How could I continue, in order to show that $\Bbb C[X,Y]/<X-1, Y+X^2-1>$ is isomorphic to $\mathbb{C}$ ? (Thinking)

Please refer back to post #3. :D

 

[evinda]

You can view $\Bbb C_1$ as the set of polynomials in $\Bbb C[X,Y]$ subject to the relations $X -1 = 0$ and $Y + X^2 - 1 = 0$. Those relations are equivalent to $X = 1$ and $Y = 0$. Thus, $\Bbb C_1$ can be identified with the set of all constant polynomials over $\Bbb C$, i.e., $\Bbb C_1$ can be identified with $\Bbb C$.

Could you explain me why we view $\Bbb C_1$ as the set of polynomials in $\Bbb C[X,Y]$ subject to the relations $X -1 = 0$ and $Y + X^2 - 1 = 0$? (Worried)

 

[Euge]

Could you explain me why we view $\Bbb C_1$ as the set of polynomials in $\Bbb C[X,Y]$ subject to the relations $X -1 = 0$ and $Y + X^2 - 1 = 0$? (Worried)

Let $\overline{g(X,Y)} \in \Bbb C_1$ denote the equivalence class of a complex polynomial $g(X,Y)$. Given $f(X,Y) = \sum_{i,j} a_{ij} X^i Y^j \in \Bbb C[X,Y]$, we have $\overline{f(X,Y)} = \sum_{i,j} a_{ij}\overline{X}^i \overline{Y}^j$. In addition, $\overline{X - 1} = 0$ and $\overline{Y + X^2 - 1} = 0$, i.e., $\overline{X} - \overline{1} = 0$ and $\overline{Y} + \overline{X}^2 - \overline{1} = 0$. If we associate $X$ with $\overline{X}$ and $Y$ with $\overline{Y}$, then we may view $\Bbb C_1$ as the set polynomials subject to the relations $X - 1 = 0$ and $Y + X^2 - 1 = 0$.

 

[evinda]

Let $\overline{g(X,Y)} \in \Bbb C_1$ denote the equivalence class of a complex polynomial $g(X,Y)$. Given $f(X,Y) = \sum_{i,j} a_{ij} X^i Y^j \in \Bbb C[X,Y]$, we have $\overline{f(X,Y)} = \sum_{i,j} a_{ij}\overline{X}^i \overline{Y}^j$. In addition, $\overline{X - 1} = 0$ and $\overline{Y + X^2 - 1} = 0$, i.e., $\overline{X} - \overline{1} = 0$ and $\overline{Y} + \overline{X}^2 - \overline{1} = 0$. If we associate $X$ with $\overline{X}$ and $Y$ with $\overline{Y}$, then we may view $\Bbb C_1$ as the set polynomials subject to the relations $X - 1 = 0$ and $Y + X^2 - 1 = 0$.

Is this the only way to show that $\mathbb{C}[X,Y]/<X-1,Y+X^2-1>$ is isomorphic to $\mathbb{C}$ ? (Thinking)

 

[Euge]

Is this the only way to show that $\mathbb{C}[X,Y]/<X-1,Y+X^2-1>$ is isomorphic to $\mathbb{C}$ ? (Thinking)

No, follow the steps in the second paragraph of post #3.

 

[evinda]

Could we do it like that?

Since $\langle x-1,y+x^2-1\rangle =\langle x-1, y\rangle$ we have to show that $\mathbb{C}/\langle x-1,y\rangle$ is isomorphic to $\mathbb{C}$.

We define the homomorphism $\phi(z)=z, \forall z\in \mathbb{C} , \phi(x)=1, \phi(y)=0$.

A polynomial in $\mathbb{C}$ is of the form $$\sum a_{mn} x^m y^n=\sum a_{mn} (x-1+1)^my^n=\sum a_{mn}\binom{m}{k} (x-1)^ky^n$$

Let $p(x)=q(x)+\sum a_{mn} \binom{m}{k} (x-1)^ky^n$
$$\phi(p(x))=q(x)\Rightarrow \phi(p(x))=0 \Leftrightarrow x-1 \mid p(x) \text{ and } y\mid p(x)$$ $$\Rightarrow ker \phi=\langle x-1,y\rangle$$

From the theorem :

Let $R,S$ rings and $\phi :R\to S$ homomorphism. There exists an isomirphism between the rings $R/ker \phi $ and $S$.

we have that $\mathbb{C}[x,y]/\langle x-1, y+x^2-1\rangle $ is isomorphic to $\mathbb{C}$.

Is it right? Or have I done something wrong? (Thinking)

 

[caffeinemachine]

Could we do it like that?

Since $\langle x-1,y+x^2-1\rangle =\langle x-1, y\rangle$ we have to show that $\mathbb{C}/\langle x-1,y\rangle$ is isomorphic to $\mathbb{C}$.

We define the homomorphism $\phi(z)=z, \forall z\in \mathbb{C} , \phi(x)=1, \phi(y)=0$.

A polynomial in $\mathbb{C}$ is of the form $$\sum a_{mn} x^m y^n=\sum a_{mn} (x-1+1)^my^n=\sum a_{mn}\binom{m}{k} (x-1)^ky^n$$ I think you are using Binomial theorem. In that case shouldn't there be one more summation appearing in the last term? I may have missed something. [/color]

Let $p(x)=q(x)+\sum a_{mn} \binom{m}{k} (x-1)^ky^n$
$$\phi(p(x))=q(x)\Rightarrow \phi(p(x))=0 \Leftrightarrow x-1 \mid p(x) \text{ and } y\mid p(x)$$ $$\Rightarrow ker \phi=\langle x-1,y\rangle$$

From the theorem :

Let $R,S$ rings and $\phi :R\to S$ homomorphism. There exists an isomirphism between the rings $R/ker \phi $ and $S$.

we have that $\mathbb{C}[x,y]/\langle x-1, y+x^2-1\rangle $ is isomorphic to $\mathbb{C}$.

Is it right? Or have I done something wrong? (Thinking)

Now. Here's another way of doing it.

Define homomorpsim $\phi:\mathbf C[x,y]\to \mathbf C[y]$ which sends $x$ to $1$, $y$ to $y$ ans preserves each element of $\mathbf C$. (There is a unique homomorphism with this property.)

Since $\phi$ is surjective, by the First Isomorphism Theorem, we have

$$\mathbf C[x,y]/\langle x-1\rangle\cong \mathbf C[y]$$.

Now define $I=\langle x-1,y\rangle$.

Note that $I$ contains $\ker\phi$ and thus by the lattice theorem we have

$$\mathbf C[x,y]/I\cong C[y]/\phi(I)$$

Now $\phi(I)=\langle y\rangle$. Can you finish?