$\mathbb{Q}[x,y]/\langle x-1,y-3 \rangle \cong \mathbb{Q}$

Question

We want to proove that $\mathbb{Q}[x,y]/\langle x-1,y-3 \rangle \cong \mathbb{Q}$.

Answer: We define $\varphi:\mathbb{Q}[x,y] \longrightarrow \mathbb{Q},f(x,y) \mapsto\varphi\big(f(x,y)\big)=f(1,3).$ So, $\varphi$ is an epimorphism.

Now, we have to find $\ker\varphi$ in order to apply the fundamental isomorphism theorem.

We will show that $\ker\varphi=\langle x-1,y-3 \rangle$.

$ \subseteq; \ $ $x-1,x-3\in \ker\varphi \implies \langle x-1,y-3 \rangle \subseteq \ker\varphi $

$\supseteq;\ \ $Let $f(x,y) \in \ker\varphi$.

$f(x,y)=a_{00}+x(a_{10}+a_{11}y+a_{21}xy+a_{12}y^2+...)+y(a_{01}+a_{02}y+...) \in \mathbb{Q}[x,y] $

We put $\hat{x}-1=x$ and $\hat{y}-3=y$. So $f(\hat{x}-1,\hat{y}-3)\in \ker\varphi \implies f(0,0)=0=a_{00}$

I stuck there... Any help?

Answer 1

It is easier if you notice that a basis for $\Bbb Q [x,y]$ is given by the monomials

$$\{ (x-1)^m (y-3)^n \mid m,n \in \Bbb N \} .$$

If $f = \sum \limits _{m,n \ge 0} f_{mn} (x-1)^m (y-3)^n$ and $\varphi (f) = 0$, this means that $f_{00} = 0$, so $f \in \langle x-1,y-3 \rangle$.

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In general, if $K$ is a commutative field and $a_1, \dots, a_n \in K$, then

$$\{ (x_1 - a_1)^{k_1} \dots (x_n - a_n)^{k_n} \mid k_1, \dots k_n \in \Bbb N\}$$

is a basis of $K[x_1, \dots, x_n]$.

This can be easily seen by noticing that the map defined on the "usual" basis monomials by

$$F(x_1 ^{k_1} \dots x_n ^{k_n}) = (x_1 - a_1)^{k_1} \dots (x_n - a_n)^{k_n}$$

and extended by linearity has an inverse given by

$$G \big( (x_1 - a_1)^{k_1} \dots (x_n - a_n)^{k_n} \big) = x_1 ^{k_1} \dots x_n ^{k_n}$$

extended by linearity.

Checking this is really easy because you just apply the definitions of $F$ and $G$:

$$ (G \circ F) (x_1 ^{k_1} \dots x_n ^{k_n}) = G \big( (x_1 - a_1)^{k_1} \dots (x_n - a_n)^{k_n} \big) = x_1 ^{k_1} \dots x_n ^{k_n}$$

and similarly for $F \circ G$.

This shows that $F$ is an isomorphism of vector spaces, and as such it takes bases to bases. Since the subset

$$\{ x_1 ^{k_1} \dots x_n ^{k_n} \mid k_1, \dots, k_n \in \Bbb N\}$$

is a basis in the space of polynomials by construction, so will be the subset

$$\{ (x_1 - a_1)^{k_1} \dots (x_n - a_n)^{k_n} \mid k_1, \dots k_n \in \Bbb N\}$$

as its image through $F$.

Answer 2

$f(x+1,y+3)(0,0)=f(1,3)=0$. You can write $f(x+1,y+3)$ is a polynomial $\sum a_{ij}x^iy^j$ where $a_{00}=0$, $f(x,y)=f(x+1,y+3)(x-1,y-3)=\sum a_{ij}(x-1)^i(y-3)^j, a_{00}=0$.