We already know that the ring of polynomials of one variable over a field is a PID. Some common examples are $\mathbb{Q}[x]$, $\mathbb{R}[x]$, $\mathbb{C}[x]$ and $\mathbb{F}_p[x]$ (with $p$ a prime number). Prime ideals of such rings, therefore, can be generated by a single element $f(x)$, where $f(x)$ is an irreducible polynomial over the field of interest. In the rings $\mathbb{R}[x]$ and $\mathbb{C}[x]$, we can write down the explicit form of $f(x)$. In $\mathbb{R}[x]$, the irreducible polynomials are precisely the ones of degree one and the ones of degree two that have no real roots. In $\mathbb{C}[x]$, a polynomial is irreducible if and only if its degree is one.
However, the same story does not happen if the base ring is not a field, such as $\mathbb{Z}$ and $\mathbb{C}[x]$. For example, $(2,x)$ is a prime ideal of $\mathbb{Z}[x]$ that is not principal. Similarly, $(x,y)$ is prime and non-principal in $\mathbb{C}[x,y]$ (Recall that ($\mathbb{C}[x,y]$ is defined as $\mathbb{C}[x][y]$). Hence, there are more prime ideals than just ideals generated by an irreducible polynomial. The question now is: could we determine all of the prime ideals in $\mathbb{Z}[x]$ or $\mathbb{C}[x,y]$? The answer is yes: in fact, we could do this for an arbitrary ring of polynomials over a PID. First, let us recall the necessary definitions and properties of a PID.
1. BACKGROUND
Throughout this section, we assume that $R$ is an integral domain.
Definition 1.1.
a) Two elements $a,b \in R$ is called associated if $a \mid b$ and $b \mid a$. In this case, we also say that $a$ is associated with $b$ (or $b$ is associated with $a$).
b) A non-zero and non-unit element $p \in R$ is called irreducible if it has no proper divisors (i.e. divisors that are neither units nor elements linked with $p$).
c) An element $p \in R$ is called prime if for any $a,b \in R$, we have $p \mid ab \Rightarrow p \mid a \vee p \mid b$.
Since the association is an equivalence relation, all the elements associated together are the same. This relation helps us reduce troubles when working with associated irreducible elements.
Definition 1.2. Let $I$ be an ideal of $R$.
a) $I$ is called prime if for any $a,b \in R$, we have $ab \in I \Rightarrow a \in I \vee b \in I$.
b) $I$ is called maximal if there does not exist an intermediate proper ideal $Q$ between $I$ and $R$, i.e. $I \subsetneq Q \subsetneq R$.
In an arbitrary integral domain, primality implies irreducibility but not vice versa. However, in a PID the converse statement is true. The same event occurs for the primality and maximality of ideals.
Proposition 1.3. Let $R$ be a PID and $p \in R$. The following statements are equivalent:
a) $p$ is irreducible in $R$,
b) $p$ is prime in $R$,
c) $(p)$ is a prime ideal of $R$.
d) $(p)$ is a maximal ideal of $R$.
Proposition 1.4. Let $I \triangleleft R$ be a proper ideal. Then $I$ is prime (resp. maximal) if and only if $R/I$ is an integral domain (resp. a field).
Proof: We have:
$\bullet$ $I$ is prime $\Leftrightarrow \forall x,y \in R: (xy \in I \Rightarrow x \in I \vee y \in I)$
$\Leftrightarrow \forall x,y \in R: (\overline{xy}=\overline{0} \Rightarrow \overline{x}=\overline{0} \vee \overline{y} \vee \overline{0})$
$\Leftrightarrow R/I$ is an integral domain.
$\bullet$ $I$ is maximal $\Leftrightarrow$ there exists no ideal $Q \triangleleft R$ with $I \subsetneq Q \subsetneq R$
$\Leftrightarrow$ the ring $R/I$ has only two trivial ideals $(0)=I/I$ and $R/I$
$\Leftrightarrow$ $R/I$ is a field. $\square$
Next, we will describe all the prime ideals in a polynomial ring over a field. We also give two examples demonstrating that the description might be incomplete if the base ring is not a field.
Proposition 1.5. Consider the ring of polynomials $R[x]$. Then $R[x]$ is a PID if and only if $R$ is a field. If this happens, any prime ideal of $R[x]$ is of the form $(f(x))$ for some irreducible $f(x) \in R[x]$.
Example 1.6.
a) $I=(2,x) \triangleleft \mathbb{Z}[x]$ is prime but not principal.
Proof: Primality: Suppose that $f(x)g(x) \in I$, i.e. $f(x)g(x)$ can be written as $2p(x)+xq(x)$ for some $p(x),q(x) \in \mathbb{Z}[x]$. Note that every $h(x) \in I$ can be rewritten as
$$h(x)=a_0+a_1x+...+a_nx^n,$$
where $a_0$ is an even integer. Therefore the constant term of $f(x)g(x)$ is even, which implies that the constant term of $f(x)$ or $g(x)$ is even. Thus $f(x) \in I$ or $g(x) \in I$.
Principality: Suppose that $I=(f(x))$ with $f(x) \in \mathbb{Z}[x]$. Since $2 \in (f(x))$, we must have $2=f(x)g(x)$ where $g(x) \in \mathbb{Z}[x]$. Taking degree two sides we get $0=\deg f(x)+\deg g(x)$, which implies $\deg f(x)=0$. But $\deg f(x)=0$ means $f(x)$ is a constant divides $2$, therefore $f(x)=1$ or $f(x)=2$, both of which are impossible. Hence $I$ is not principal.
b) $I=(x,y) \triangleleft \mathbb{R}[x,y]$ is a prime ideal generated by at least two elements.
Proof: The proof for this ideal can be done similarly to part a). Here we present another approach by investigating the quotient ring $\mathbb{R}[x,y]/I$.
Rewrite $\mathbb{R}[x,y]=\mathbb{R}[x][y]$. Consider the projection map
$$\iota: \mathbb{R}[x,y] \to (\mathbb{R}[x]/(x))[y] \cong \mathbb{R}[y] \to \mathbb{R}[y]/(y) \cong \mathbb{R}$$
transforming $f(x,y)$ into $\overline{f}(x,y)$ modulo $(x)$ and then into $\widetilde{f}(x,y)$ modulo $(y)$. We will show that $\ker \iota=(x,y)$. Indeed,
$$\widetilde{f}(x,y)=0 \Leftrightarrow x \mid \overline{f}(x,y) \Leftrightarrow \overline{a_0}=\overline{0}$$
$$\Leftrightarrow x \mid a_0(x) \Leftrightarrow f(x,y)=a_1(x)y+...+a_n(x)y^n \in (x,y).$$
By Noether's first isomorphism theorem, we have $\mathbb{R}[x,y]/(x,y) \cong \mathbb{R}$, which is a field. Thus $(x,y)$ is a maximal ideal (and therefore prime) of $\mathbb{R}[x,y]$. $\square$
Using the same technique, we can also prove that $(2,x)$ is a maximal ideal of $\mathbb{Z}[x]$ (part a) of the above example). More generally, all the prime ideals in $\mathbb{Z}[x]$ of the form $(p,f(x))$, when $p$ is a prime number and $f(x)$ is irreducible in $(\mathbb{Z}/p\mathbb{Z})[x]$, are maximal ideals. Summing up the above results, we see that there are at least three types of prime ideals in $\mathbb{Z}[x]$:
- $(0)$,
- $(f(x))$, where $f(x)$ is an irreducible polynomial of $R[x]$,
- $(p,f(x))$, where $p$ is a prime number and $f(x)$ is irreducible in $(\mathbb{Z}/p\mathbb{Z})[x]$.
(Note that the case $f(x)=p$ is included in the second case because a prime element of $p$ can be considered as an irreducible polynomial in $R[x]$)
A natural question arises: Are these all the prime ideals in $\mathbb{Z}[x]$? Moreover, can we generalize the result for $R[x]$ if $R$ is a PID? We will find the answer now in the next section.
2. THE MAIN THEOREM
Before giving the statement and the proof of the main result, we state and prove two necessary lemmas as follows:
Lemma 2.1. Let $\mathfrak{p}$ be a non-zero prime ideal of $R[x]$ where $R$ is a UFD. If $\mathfrak{p}$ does not consist of non-zero constants, then it contains an irreducible polynomial $f$ of minimal degree.
Proof: The set of all the degrees of the polynomials in $\mathfrak{p}$ is a bounded set in $\mathbb{N}$, therefore it contains a polynomial $f' \in \mathfrak{p}$ whose degree is minimal. By unique factorization, we can write $f'=sf$ where $s$ is the $\gcd$ of the coefficients of $f$. Since $\mathfrak{p}$ does not have any non-zero constants, $s$ could not lie in $\mathfrak{p}$, hence $f \in \mathfrak{p}$. Now $f$ is irreducible in $R[x]$; otherwise we can factor $f=gh$ where $g,h \in R[x]$ are non-constant and have the degree less than $f$, contradicts the minimality of $\deg f$ in $\mathfrak{p}$.
Lemma 2.2. Let $\mathfrak{p}$ be prime ideal in a ring $R$ and $\phi: R \to S$ be a surjective ring homomorphism such that $\ker \phi \subseteq \mathfrak{p}$. Then $\phi (\mathfrak{p})$ is prime in $S$.
Proof: Suppose $ab \in \phi (\mathfrak{p})$. Since $\phi$ is surjective, we can write $a=\phi(x)$, $b=\phi(y)$ and $ab=\phi(z)$ for some $x,y,z \in R$. Furthermore, we can choose $z$ such that $z \in \mathfrak{p}$. Note that
$$\phi(xy-z)=\phi(x)\phi(y)-\phi(z)=ab-ab=0.$$
This implies $xy-z \in \ker \phi \subseteq \mathfrak{p}$. Because $z \in \mathfrak{p}$, we get $xy \in \mathfrak{p}$. By the primality of $\mathfrak{p}$, we get $x \in \mathfrak{p}$ or $y \in \mathfrak{p}$. Thus $a=\phi(x) \in \phi(\mathfrak{p})$ or $b=\phi(y) \in \phi (\mathfrak{p})$. $\square$
Now back to the main problem. Suppose $\mathfrak{p}$ is an ideal of $R[x]$ where $R$ is a PID. To "play" with $\mathfrak{p}$, we can restrict $\mathfrak{p}$ to an ideal of $R$, from which we can determine the form of the ideal thanks to the PID property of $R$. For this reason, we consider the inclusion map $\iota: R \to R[x]$, which is also an injective ring homomorphism. Then $\iota^{-1}(\mathfrak{p})=\mathfrak{p} \cap R$ is a prime ideal of $R$. Since $R$ is a PID, we have $\iota^{-1}(\mathfrak{p})=(0)$ or $\iota^{-1}(\mathfrak{p})=(p)$ for some prime $p \in R$.
First, consider the case $\iota^{-1}(\mathfrak{p})=(0)$. If $\mathfrak{p}=(0)$ there is nothing to prove, so we assume that $\mathfrak{p} \neq (0)$. It is true that $\mathfrak{p}$ does not consist of non-zero constant polynomials: if there is such a constant called $r_0$, then $r_0 \in \iota^{-1}(\rho)=(0)$ which implies $r_0=0$. Applying Lemma 2.1, there is an irreducible polynomial $f(x) \in R[x]$ lying in $\mathfrak{p}$. Now $\mathfrak{p} \supseteq (f(x))$, we only need to prove the converse inclusion. To do this, we need to "extend" the coefficient ring $R$ to its fraction field $F$.
Pick an $g(x) \in R[x]$ and view $f(x)$ and $g(x)$ as elements of $F[x]$. Let $I=(f(x),g(x))$ be an ideal of $F[x]$, then $I$ can be rewritten as $I=(q(x))$ with $q(x) \in F[x]$ irreducible, which implies $q(x) \mid f(x)$. Since $f(x)$ is irreducible in $F[x]$ due to the Gauss' lemma, $q(x)$ is associated with either $1$ or $f(x)$ (but not both of them). However, the former case is impossible. Indeed, if this happens there exists $a(x),b(x) \in F[x]$ such that $a(x)f(x)+b(x)g(x)=1$. Moreover, we can eliminate the denominators of coefficients in $a(x)$ and $b(x)$ by multiplying them by $c_1,c_2 \in R$, respectively, and get
$$c_1c_2a(x)f(x)+c_1c_2b(x)g(x)=c_1c_2.$$
Since the left-hand side lies in $\mathfrak{p}$, so does the right-hand side $c_1c_2$. Applying the primality of $\mathfrak{p}$, we get $c_1 \in \mathfrak{p}$ or $c_2 \in \mathfrak{p}$, which contradicts the fact that $\mathfrak{p}$ does not contain constants in $R$. Now, the association between $q(x)$ and $f(x)$ follows that $I=(f(x))$ and thus $f(x) \mid g(x)$ in $F[x]$. According to Gauss' lemma (again), we have $f(x) \mid g(x)$ in $R[x]$. We conclude that $\mathfrak{p}=(f(x))$.
Next, assume that $\iota^{-1}(\mathfrak{p})$ is generated by a prime $p \in R$. Observe that $(p)$ is a maximal ideal because every prime ideal of a PID is maximal. It follows that $F:= R/(p)$ is a field. This time, we will embed the coefficients from $R$ to $F$ by the natural projection $\pi: R \to F$. Recall that $\pi$ induces the ring homomorphism $\widetilde{\pi}: R[x] \to F[x]$ defined by taking modulo $(p)$ of the coefficients $a_i$. It is not difficult to see that
$$g(x) \in \ker \widetilde{\pi} \Leftrightarrow \pi(a_i)=0, \forall i \Leftrightarrow a_i \in (p), \forall i \Leftrightarrow g(x) \in (p),$$
which follows that $\ker \widetilde{\pi}=(p) \subseteq \mathfrak{p}$. By Lemma 2.2 $\widetilde{\pi}(\mathfrak{p})$ is a prime ideal $F[x]$. Hence $\widetilde{\pi}(\mathfrak{p})=(0)$ or $\widetilde{\pi}(\mathfrak{p})=(\overline{f}(x))$ where $\overline{f}(x)$ is monic and irreducible over $F$. If $\widetilde{\pi}(\mathfrak{p})=(0)$ then $\mathfrak{p}=(p)$ due to the argument that
$$g(x) \in \mathfrak{p} \Rightarrow \widetilde{\pi}(g(x))=0 \Rightarrow g(x) \in \ker \widetilde{\pi}=(p).$$
Now suppose $\widetilde{\pi}(\mathfrak{p})=(\overline{f}(x))$. We can lift $\overline{f}(x)$ to a monic $f(x) \in R[x]$. $f(x)$ must be irreducible in $R[x]$; otherwise,
$$f(x)=a(x)b(x) \Rightarrow \overline{f}(x)=\widetilde{\pi}(\overline{f}(x))=\widetilde{\pi}(a(x))\widetilde{\pi}(b(x)),$$
which claims that $\overline{f}(x)$ is reducible, a contradiction. Now we can verify that
$$\widetilde{\pi}^{-1}((\overline{f}(x)))=\widetilde{\pi}^{-1}(\widetilde{\pi}(\mathfrak{p}))=\mathfrak{p}+(p)=\mathfrak{p}.$$
This implies $f(x) \in \widetilde{\pi}^{-1}((\overline{f}(x)))=\mathfrak{p}$ and $p \in (p) \subseteq \mathfrak{p}$ and then $(p,f(x)) \subseteq \mathfrak{p}$. Now take an element $g(x) \in \mathfrak{p}=\widetilde{\pi}^{-1}((\overline{f}(x)))$, we have
$$\overline{g}(x)=\overline{f}(x)\overline{q}(x) \Rightarrow \widetilde{\pi}(g(x))=\widetilde{\pi}(f(x)g(x))$$
$$\Rightarrow g(x)-f(x)q(x) \in \ker \widetilde{\pi}=(p) \Rightarrow g(x) \in (p,f(x)).$$
We conclude that $\mathfrak{p}=(p,f(x))$.
Now let us sum up and state the main result of the article.
Theorem 2.3. Let $R$ be a PID and $I$ be a prime ideal of $R[x]$. Then $I$ is of one of the following forms:
- $(0)$,
- $(f(x))$, where $f(x)$ is a non-constant irreducible polynomial of $R[x]$,
- $(p)$, where $p \in R$ is prime,
- $(p,f(x))$, where $p$ is prime in $R$ and $f(x)$ is a non-constant irreducible in $(R/(p))[x]$.
Since maximal ideals are prime, the possible maximal ideals of $R$ lie in the above cases. To continue, we will check which types of prime ideals are maximal and which types are not.
Theorem 2.4. Consider the ideals of the ring $R[x]$ mentioned in Theorem 2.3. The only prime ideals of the fourth type are maximal.
Proof: Obviously, $(0)$ is not maximal. Take randomly a prime $p \in R$ and a non-constant irreducible $f(x) \in R[x]$, we need to show that $(p)$ and $(f(x))$ are not maximal of $R[x]$. Indeed, for the case $(p)$, we have the isomorphic rings
$$R[x]/(p) \cong (R/(p))[x],$$
with the LHR is not a field since it is a polynomial ring over a field. For the ideal $(f(x))$, it is obviously that
$$(f(x)) \subsetneq (p,f(x)),$$
since the ideal $(f(x))$ only contains polynomials with the degree $\geq \deg f(x) \geq 1$, therefore does not contains $p$.
Now eliminate the irreducibility of $f(x)$ in $R[x]$ and suppose that $\widetilde{f}(x) \pmod{(p)}$ is irreducible in $(R/(p))[x]$, we have
$$R[x]/(p,f(x)) \cong (R/(p))[x]/(\widetilde{f}(x)).$$
Since $(R/(p))[x]$ is a PID, $(R/(p))[x]/(\widetilde{f}(x))$ is a field. Thus $(p,f(x))$ is a maximal ideal. This completes the proof. $\square$
Finally, we will give some examples of prime ideals and maximal ideals over several popular polynomial rings.
3. EXAMPLES
Example 3.1. Given two ideals $I=(5,x^3+2x-3)$, $J=(4,x^2+x+1,x^2+x-1)$ in the ring $\mathbb{Z}[x]$. We will check which ideal of these two is maximal and which one is prime.
Firstly, since $I$ is an ideal of the fourth type mentioned in Theorem 2.3, we will check if the polynomial $x^3+2x-3 \pmod{5}$ is irreducible. The answer is no since $x^3+2x-3=(x-1)(x^2+x+3)$. Therefore $I$ is neither prime nor maximal. For the ideal $J$, note that
$$J=(x^2+x+1,x^2+x-1)=(2,x^2+x+1)$$
since $4=2(x^2+x+1)-2(x^2+x-1)$. The polynomial $x^2+x+1$ is irreducible modulo 2 since it is of degree 2 and does not have roots modulo 2, thus $J$ is a maximal (and also prime) ideal.
Example 3.2. Consider the ring $R=\mathbb{C}[x,y]$. The prime ideals of $R$ are precisely $(0)$, $(f(x,y))$ where $f(x,y)$ is a irreducible polynomial and $(x-a,y-b)$ where $a,b \in \mathbb{C}$. Furthermore, only prime ideals of the third type are maximal.
Proof: From Theorem 2.3 and 2.4 we just need to show that ideals of the fourth type are exactly $(x-a,y-b)$ for some $a,b \in \mathbb{C}$. Note that $\mathbb{C}[x,y]$ can be written as $\mathbb{C}[x][y]$, hence a prime $p \in \mathbb{C}[x]$ must be of the form $x-a$ and an irreducible polynomial in $\mathbb{C}[x,y]/(x-a) \cong \mathbb{C}[y]$ must be of the form $y-b$. Thus $(x-a,y-b)$ is a maximal ideal. $\square$
References
1. https://math.colorado.edu/~kearnes/Teaching/Courses/F20/HW/calg2p7.pdf
