Lemma 1.1. Let $A$ be a domain, and $P_1=(p_1),P_2=(p_2)$ be two non-zero principal prime ideals of $A$, such that $P_1\ne P_2$, then $P_1\not\subset P_2$.
Proof. Assume that $P_1\subsetneq P_2$, or equivalently $(p_1)\subsetneq (p_2)$, then there exists $a\in A$ such that $p_1=ap_2$. It can be seen that $(p)$ is non-zero prime ideal iff $p$ is a non-zero prime element. Then either $a$ or $p_2$ are invertible, but $p_2$ cannot be invertible because it is a prime element. From this, $a$ is invertible and hence $(p_1)=(p_2)$, a contradiction. This finishes our proof. (Q.E.D)
Lemma 1.2. Let $A$ be an U.F.D, then $P$ is a non-zero prime ideal of $A$. Then $ht(P)=1$ iff $P$ is principal.
Proof. Let $a\ne 0$ in $P$, then we can factorize $a=\prod_{i=1}^np_i^{a_i}$, where $p_i$ are non-zero prime elements in $A$. Because $P$ is prime, at leats one of $p_i$ must lie in $P$, say $p$. We have $(0)\subsetneq (p)\subset P$ is a chain in $P$. Hence, $ht(P)=1$ implies that $P=(p)$, or $P$ is principal.
Conversely, if $P=(p)$ is principal prime ideal, and if there exists a prime ideal $Q$ such that $(0)\subsetneq Q\subsetneq (p)$. Take any $a\ne 0$ in $Q$, we can factorize $a=\prod_{i=1}^np_i^{a_i}$, where $p_i$ are non-zero prime elements of $A$. Similarly, because of the primality of $Q$, there exists $p_1\in Q$, or $(0)\subsetneq (p_1)\subsetneq (p)$ is a chain in $A$. But it is impossible due to the first lemma. This implies $ht(P)=1$. (Q.E.D)
Theorem 1.3. Let $A$ be a domain with dimension 1, then $A[x]$ has dimension 2 or 3. If $A$ is a PID, then $A[x]$ has dimension 2.
Proof. First, let $P$ be a prime ideal in $A$, and $f(x)$ is irreducible in $A[x]$, then $A[x]/(P,f(x))$ is al field. This implies $\dim A\ge 2$. Let $0=M_0\subsetneq M_1\subsetneq...\subsetneq M_n$ be any chain of prime ideals in $A[x]$ with $M_n$ is maximal. Let $j$ be a smallest integer such that $M_j\cap A=P\ne (0)$, then $P$ is a maximal ideal of $A$, for $\dim A=1$. And any prime ideal of $A[x]$ lies above $P$ will contain $PA[x]$. From the correspondence between $\{P\in Spec(A[x])|PA[x]\subset P\}$ and $\{\overline{P}\in Spec(A[x]/PA[x])\}$, we can see $A[x]/PA[x]\cong (A/P)[x]$, which is a field. Hence, prime ideals strictly contains $PA[x]$ will be maximal. Or $n\le j+1$. We now find an upper bound for $j$.
For an integer $i$ from 0 to $j-1$, we can see $M_i\cap A=0$. Let us localize $A[x]$ at $S=A\setminus \{0\}$, then $(A[x])_S=K[x]$, where $K$ is the fraction field of $A$. From the $1:1$ correspondence between $\{P\in Spec(A[x])|P\cap S=\emptyset\}$ and $\{P\in Spec(K[x])\}$, we can see $j-1\le 1$. Hence, $n\le 3$, or $\dim A\le 3$.
In the case $A$ is principal, then $A[x]$ is an U.F.D. Look at the first paragraph, the chain is maximal if we start with $M_j=PA[x]$. And from $P=(p)$, we can see $PA[x]=(p)$, where $p\in A$, is principal and is of height 1, by Lemma 1.2. Hence, $j=1$ in this case, and $PA[x]$ is not maximal ideal, for $A[x]/PA[x]\cong A/P[x]$, which is not a field. By the correspondence theorem, any ideal strictly contains $PA[x]$ is maximal. Therefore, $\dim A[x]=2$ in this case. (Q.E.D)
Via this, one can see if $k$ is a field, then $k[x]$ is a PID, and hence, $k[x,y]$ is of dimension 2. It is natural to ask whether the maximal ideal of $k[x,y]$ is generated by just one element? It is impossible, at least for the case $f(x,y)$ is not irreducible. Since the quotient ring $k[x,y]/(f(x,y))$ is not an integral domain. How about the case $f(x,y)$ is irreducible?
Lemma 1.4. Let $k$ be a field, and $f(x,y)\in k[x,y]$ is an irreducible polynomial, then the ideal generated by $f(x,y)$ is not maximal.
Proof. If $(f(x,y))$ is maximal, and $(a,b)$ is a point of $f(x,y)=0$ in $\overline{k}^2$. Hence, $f(x,y)\in (x-a,y-b)=M$, which is maximal in $\overline{k}[x,y]$. Hence, $M\cap k[x,y]$ is an ideal of $k[x,y]$ containing $f(x,y)$. This implies $M\cap k[x,y]=(f(x,y))$. Let $h(x)\in k[x]$ be the minimum polynomial of $a$, then $h(x)$ is irreducible in $k[x,y]$ (for the quotient $k[x,y]/(h(x))\cong (k[x]/(h(x))[y]$ is a domain). And $h(x)\in M$. From this, there exists $g(x,y)\in k[x,y]$ such that $h(x) = f(x,y)g(x,y)$, this implies $g(x,y)$ is invertible, or $(h(x))=(f(x,y))$. However $k[x,y]/(h(x))\cong (k[x]/(h(x))[y]$ is a domain, and never a field. This contradiction implies that $(f(x,y))$ is not maximal. (Q.E.D)
From this, one can see any maximal ideal in $k[x,y]$ is generated by at least two elements. And as an application of the lemma, one gets
Theorem 1.5. Let $f(x,y)$ be an irreducible polynomial of $k[x,y]$, then the ring $R=k[x,y]/(f(x,y))$ is a domain of dimension 1.
Proof. The fact that $R$ is a domain is obvious. From the previous lemma, (f(x,y)) is not a maximal ideal, and it is of height 1, by Lemma 1.2. Because $\dim k[x,y]=2$, any prime ideal that strictly contains $f(x,y)$ will be maximal, and their existence is ensured, for any ideal is contained in a maximal ideal. Via the correspondence between $\{\overline{P}\in Spec(R)\}$ and $\{P\in Spec(k[x,y])|f(x,y)\in P\}$, one can see $R$ is of dimension 1. (Q.E.D)
Now, we can see the coordinate ring of an irreducible curve is of dimension 1, and Noetherian. For any non-zero prime ideal of $R$, we will prove that its intersection with $k[x]$ is also non-zero. By this, one can see $k[x]$ plays a role as $\mathbb{Z}$ in algebraic number theory, where the intersection between any non-zero prime ideal of a number ring and $\mathbb{Z}$ is non-zero prime ideal of $\mathbb{Z}$. And we look up from $p$, find characterizations of primes lies over it. The same occurs for $k[x,y]$. The following lemma will be necessary
Lemma 1.6. Let $M$ be a maximal ideal of $k[x,y]$, then $M\cap k[x]\ne 0$.
Proof. Assume that $M\cap k[x]=0$. Let $A=k[x]$, and $K$ the quotient field of $A$. Via the correspondence between $\{P\in Spec(A[y])|P\cap A = 0\}$, and $\{Q\in Spec(K[z])\}$, we can see non-zero ideal $M$ with $M\cap A=0$ is corresponding to a maximal ideal of $K[z]$, and hence, $ht(M) = 1$. Because $A[y]$ is an UFD, by Lemma 1.2, this implies $M$ is principal. Lemma 1.4 says that $M$ cannot be a maximal ideal. This contradiction finishes our proof. (Q.E.D)
2. Factorization in $C_f$. In this section, $k$ is algebraically closed. We recall that $C_f$ is the coordinate ring of an irreducible curve $f(x,y)=0$. From Lemma 1.6, if we take any non-zero prime ideal in $C_f$, then its intersection with $k[x]$ is non-zero prime ideal of $k[x]$, and hence, is of the form $(x-a)$, where $a\in k$. We are now interested in prime ideals of $C_f$ that lies above $(x-a)$. Such ideals must contain $(x-a)C_f$. Consider the quotient $C_f/(x-a)C_f\cong k[x,y]/(f(x,y),x-a)\cong k[y]/(f(a,y))$, we can see non-zero prime ideal containing $(x-a)C_f$ is $1:1$ corresponding to ideal of $k[y]$ that contain $f(a,y)$. We can see also $f(a,y)$ is invertible in $C_f$ iff $(x-a)$ is invertible in $C_f$. In this case $(x-a)C_f=C_f.$
Otherwise, we can factorize $f(a,y)=\prod_{i=1}^n(y-b_i)^{e_i}$, then via the isomorphism, prime ideals that strictly contains $(x-a)C_f$ is of the form $(x-a,y-b_i)$. Note that in the case of algebraic number theory, as Kummer-Dedekind pointed out, the determination of prime ideals lying over $p$ is done by considering the reduction modulo $p$. In the case of algebraic curves, it is done by reduction modulo $k[y]$.
(There is another way to prove there exists only finitely many maximal prime ideals that lies over $(x-a)C_f$. Note that $C_f$ is a Noetherian domain of dimension 1, and $C_f/(x-a)\cong k[y]/(f(a,y))$ is Noetherian domain of dimension 0, and hence, an Artinian domain, whose has finitely many maximal ideals).
Let us denote $M_i=(x-a,y-b_i)$. We now prove that $\prod_{i=1}^nM_i^{e_i}\subset (x-a)C_f$. And the equality holds iff all $(x-a,y-b_i)$ are smooth points. First, via the isomorphism, we can write $f(x,y)=(x-a)g(x,y)+\prod_{i=1}^n(y-b_i)^{e_i}$ (for $f(a,y)=\prod_{i=1}^n(y-b_i)^{e_i}$). And $M_i=(x-a)C_f+(y-b_i)C_f$ implies $M_i^{e_i}\subset (x-a, (y-b_i)^{e_i})$. Hence, $\prod_{i=1}^nM_i^{e_i}\subset (x-a)C_f +f(a,y)C_f$. Besides, in $C_f,f(x,y)=0$, or $f(a,y)\in (x-a)C_f$. This implies $\prod_{i=1}^nM_i^{e_i}\subset (x-a)C_f$. Then, we next prove that all $(a,b_i)$ are smooth points iff the equality occurs.
Taking the partial derivatives, one have
$$\frac{df}{dx}=g(x,y) + (x-a)\frac{dg}{dx}$$
$$\frac{df}{dy}=(x-a)\frac{dg}{dy}+\sum_{i=1}^n(e_i(y-b_i)^{e_i-1}\prod_{j\ne i}(y-b_j)^{e_j})$$
$(a,b_i)$ is a smooth point iff $g(a,b_i)\ne 0$, or $e_i=1$. In the case $e_i=1$, then the localization $\prod_{i=1}^nM_i^{e_i}= M_i(C_f)_{M_i}$, and this implies $((x-a)C_f)_{M_i}= M_i(C_f)_{M_i}$. If $g(a,b_i)\ne 0$, then $g(x,y)\notin M_i$, and hence, it is invertible in the local ring $(C_f)_{M_i}$, and the localization of $M_i$ yields $((x-a)C_f)_{M_i}=(M_i(C_f)_{M_i})^{e_i}=(\prod_{i=1}^n M_i^{e_i})_{M_i}$. And hence, in any case, we always have $\prod_{i=1}^n(M_i)^{e_i}$ is equals to $(x-a)C_f$ at any localization of maximal ideals. This implies they are equal in $C_f$. And we conclude this post by a theorem
Theorem 2.1. Let $(C): f(x,y) = 0$ be an irreducible algebraic curve over $k[x,y]$ where $k$ is algebraically closed. Then the coordinate ring $C_f$ is a Dedekind domain iff $(C)$ has no singular point.
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