# Question about the proof that $\operatorname{Spec}\mathbb{C}[x,y]$ contains precisely $(0)$, $(x-a,y-b)$ and $(f)$ for irreductible $f$.

I am trying to show that $$\operatorname{Spec}\mathbb{C}[x,y]$$ contains exactly the ideals $$(0)$$, $$(x-a,y-b)$$ for $$a,b\in\mathbb{C}$$, and $$(f)$$ for an irreductible polynomial $$f\in\mathbb{C}[x,y]$$.

My approach (following a tip in Vakil's notes) is the following: clearly these ideals are all prime. It suffices then to prove that if $$\mathfrak{p}$$ is a prime ideal of $$\mathbb{C}[x,y]$$ which is not principal, then it is of the form $$(x-a,y-b)$$.

Since $$\mathfrak{p}$$ is not principal, it contains two polynomials $$f,g$$ without common factors. We then use Bézout in $$\mathbb{C}(x)[y]$$ to find $$p,q\in \mathbb{C}(x)[y]$$ such that $$pf+qg=1$$. Multiplying by a common denominator $$h\in\mathbb{C}[x]$$ we get $$p(x,y)f(x,y)+q(x,y)g(x,y)=h(x)$$ in $$\mathbb{C}[x,y]$$. It follows that $$h\in (f,p)\subset\mathfrak{p}$$. Since $$\mathbb{C}$$ is algebraically closed and $$\mathfrak{p}$$ is prime, there is $$a\in\mathbb{C}$$ such that $$(x-a)\in\mathfrak{p}$$. Similarly, there is $$b\in\mathbb{C}$$ such that $$(y-b)\in\mathfrak{p}$$. Finally, since $$(x-a,y-b)\subset\mathfrak{p}$$ and the former is maximal, we have equality.

There are just two points in this proof that I do not really understand. Firstly, why $$\mathfrak{p}$$ contains two polynomials without common factors? Also, why they remain relatively prime in $$\mathbb{C}(x)[y]$$?

Both appear to be obvious but I don't know how to prove them. (Two second fact seems closely related to Gauss' lemma but $$\mathbb{C}(x)[y]$$ isn't the fraction field of $$\mathbb{C}[x,y]$$.)

• $\mathfrak{p}$ contains an irreducible polynomial right? – pigeon Dec 19 '19 at 17:46

Since $$\mathfrak{p}$$ is a prime ideal, it has to contain some irreducible polynomial $$h \in \mathbb{C}[x,y]$$. For any other $$l \in \mathfrak{p}$$, either $$h \mid l$$ or $$h$$ and $$l$$ share no common factor. If $$h \mid l$$ for all $$l \in \mathfrak{p}$$ then $$\mathfrak{p}$$ is principal (generated by $$h$$). In the other case you have two polynomials with no common factor.
For the second fact, one can organize the relevant data as follows: If $$R$$ is a UFD and $$K$$ its field of fractions, and if two polynomials $$f,g \in R[x]$$ share no common factor then they share no common factor in $$K[x]$$ (converse is false, e.g. take $$R = \mathbb{Z}$$ and $$K = \mathbb{Q}$$ and $$f = 2x$$ and $$g = 2(x+1)$$).
To check the above statement, if $$f,g$$ do share a common nonunit factor $$h \in K[x]$$ (so that $$h$$ is not a constant polynomial) then there exists $$r \in R \setminus \{ 0 \}$$ with $$rf$$ and $$rg$$ sharing a common factor $$k$$ in $$R[x]$$ with $$\deg k > 0$$. That is, there exist $$a,b \in R[x]$$ with $$rf = ak$$ and $$rg = bk$$. Denoting $$c(f)$$ to be the content of $$f$$, write $$f = c(f)f_0$$, $$a = c(a)a_0, b = c(b)b_0, k = c(k)k_0$$ where $$c(f_0), c(a_0), c(b_0), c(k_0)$$ are all units in $$R$$. Then $$r c(f)$$ is an associate of $$c(a)c(k)$$ and likewise $$r c(g)$$ is an associate of $$c(b)c(k)$$, so $$f_0$$ is divisible by $$k_0$$ and $$g_0$$ is divisible by $$k_0$$. Since $$\deg k_0 > 0$$ and both $$f_0,g_0$$ are factors of $$f,g$$, $$k_0$$ is a factor of $$f,g$$, a contradiction.
• Dear hochs, thank you for your nice proof of the first fact. In regards to the second one, I understand your result but I don't know how it can be applied in my case since $\mathbb{C}(x)[y]$ is not the field of fractions of $\mathbb{C}[x,y]$. – Gabriel Dec 19 '19 at 21:36
• Hi. One does not need $\mathbb{C}(x)[y]$ to be the field of fractions of $\mathbb{C}[x,y]$. Take $R = \mathbb{C}[x]$ above, which is a UFD. One simply needs $\mathbb{C}(x)$ to be the field of fractions of $\mathbb{C}[x]$. – hochs Dec 19 '19 at 21:49