Groebner basis of a maximal ideal

Question

is it a true for a maximal ideal $I=\langle x-a,\,y-b\rangle$ the vector space $\mathbb{C}[x,y]/I$ always has the dimension one?

I thought we would have a Groebner basis $G$ of the same form as $I$ and $\mathbb{N}^2\setminus deg(G)=(0,0)$ therefore the dimension of $\mathbb{C}[x,y]/I$ is equal to one.

However I am not very sure about this..

Answer 1

In excruciating detail: if $I = \langle x-a, y-b \rangle$, then the $\mathbb{C}$-algebra homomorphism $\mathbb{C}[x,y] \to \mathbb{C}$ taking $x$ to $a$ and $y$ to $b$ (i.e., evaluation of a polynomial at $(a,b)$) factors through $I$, so defines a $\mathbb{C}$-algebra homomorphism $\mathbb{C}[x,y]/I \to \mathbb{C}$ which is surjective because (the class of) $1$ goes to $1$, and injective because any element of $\mathbb{C}[x,y]$ which vanishes at $(a,b)$ belongs to $I$ (indeed, we can assume $(a,b)=(0,0)$ by translating, and then a polynomial vanishes at $(0,0)$ iff it has no constant term, in which case it is clear that it can be written as $x g(x) + y h(x,y)$, so it belongs to $I = \langle x,y\rangle$). So this defines an isomorphism of rings between $\mathbb{C}[x,y]/I$ and $\mathbb{C}$, and, in particular, the former has dimension $1$.

Yes, you can also do this with Gröbner bases if you want: $x-a$ and $y-b$ are already a Gröbner basis for $I$ (for any monomial order!), so this allows us to compute the dimension as you state. (But using Gröbner bases in this context is akin to cracking nuts with a sledgehammer.)

What is not so obvious, on the other hand, is (as you assumed implicitly) that every maximal ideal of $\mathbb{C}[x,y]$ is of the form $\langle x-a, y-b \rangle$: in fact, this is more or less equivalent to Hilbert's Nullstellensatz. If we were to replace $\mathbb{C}$ by $\mathbb{R}$, this would be false: $\langle x^2+1, y\rangle$ is maximal in $\mathbb{R}[x,y]$ and the quotient has dimension $2$.