Finite rings in which sum and product of all non-zero elements are non-zero

Posted: March 29, 2019 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem. Let $R$ be a finite ring with $1$ and suppose that $R$ satisfies the following properties

i) $\displaystyle \sum_{r \in R} r \ne 0$

ii) $\displaystyle \prod_{0 \ne r \in R} r \ne 0.$

Show that either $R \cong \mathbb{Z}_2$ or $R \cong \mathbb{Z}_4.$

Solution. First notice that both $\mathbb{Z}_2$ and $\mathbb{Z}_2$ satisfy i), ii). Since $\mathbb{Z}_2$ is the only ring with two elements, we may assume that $|R| > 2.$ Let $n:=|R|.$ First we show that $n$ is even. Suppose, to the contrary, that $n$ is odd. Then the group $(R,+)$ would have no element of order $2,$ i.e. $r \ne -r$ for all $0 \ne r \in R,$ and thus

$\displaystyle \sum_{r \in R} r = \sum (r+(-r))=0,$

Now let $m:=\text{char}(A).$ We show that $m=p^k$ for some prime $p$ and $k \in \{1,2\}.$ To see that, let $m=\prod_{i=1}^{\ell}p_i^{k_i}$ be the prime factorization of $m.$ If $\ell > 1,$ then

$\displaystyle \prod_{i=1}^{\ell} p_i^{k_i}1_R=m1_R=0,$

contradicting the property ii), because $p_i^{k_i}1_R$ are non-zero distinct elements of $R.$ So $\ell=1,$ i.e. $m=p^k$ for some prime number $p$ and positive integer $k.$ Now if $k > 2,$ then we can write

$0=m1_A=(p1_R)(p^{k-1}1_R)$

and again, by the minimal property of $m,$ the elements $p1_R$ and $p^{k-1}1_R$ are non-zero and distinct and that contradicts the property ii). So $k \le 2.$
We now show that in fact $m=4.$ We just proved that $m=p^k$ for some prime $p$ and $k \in \{1,2\}.$ Suppose that $q \ne p$ is a prime divisor of $n.$ Then there exists $r \in (R,+)$ that has order $q.$ But then $qa=ma=0$ which gives the false result $r=0,$ because $q,m$ are coprime. So $n$ is a power of $p.$ But we have already showed that $n$ is even. So $p=2$ and thus either $m=2$ or $m=4.$ If $m=2,$ then $R$ would be a vector space over $\mathbb{Z}_2,$ i.e. $R=\sum_{i=1}^s \mathbb{Z}_2r_i$ for some integer $s \ge 2,$ because $n > 2,$ and $r_i \in R.$ But then

$\displaystyle \sum_{r \in R}r =2^{s-1}\sum_{i=1}^s r_i=0,$

contradicting the property ii). So $m=4$ and hence $R$ contains a subring $R_0 \cong \mathbb{Z}_4.$
Finally, we show that $R=R_0.$ Suppose, to the contrary, that $R \neq R_0$ and let $r \in R \setminus R_0.$ We have $(21_R)(2r)=0,$ because $\text{char}(R)=4,$ and so, since $21_R \ne 0,$ we must have either $21_R=2r$ or $2r=0,$ by property ii). If $21_R=2r,$ then $(21_R)(r-1_R)=0,$ which gives $r=1_R \in R_0,$ by property ii), and that’s a contradiction. So $2r=0,$ i.e. we have shown that $2r=0$ for any element $r \in R \setminus R_0.$ But if $r \in R \setminus R_0,$ then $r+1_R \in R \setminus R_0$ too, because $1_R \in R_0,$ and so $2r=2(r+1_R)=0,$ which gives $21_R=0,$ contradiction. So $R=R_0 \cong \mathbb{Z}_4. \ \Box$

Rings with no proper left ideals

Posted: November 16, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules

Suppose that $R \ne (0)$ is a ring with no proper left ideals. If $R$ has $1,$ then  $R$ is a division ring. To see this, let $0 \ne x \in R.$ Then $Rx=R$ and so $yx=1$ for some $y \in R.$ Since $y \ne 0,$ we have $Ry=R$ and hence $zy=1$ for some $z \in R.$ Then $x=zyx=z$ and so $yx=xy=1$ proving that $R$ is a division ring.

But what if $R$ doesn’t have $1 ?$ The following problem answers this question.

Problem. Let $R \ne (0)$ be a ring, which may or may not have $1.$ Show that if $R$ has no proper left ideals, then either $R$ is a division ring or $R^2=(0)$ and $|R|=p$ for some prime number $p.$

Solution. Let

$I:=\{r \in R: \ \ Rr=(0)\}.$

Then $I$ is a left ideal of $R$ because it’s clearly a subgroup of $(R,+)$ and, for $s \in R$ and $r \in I,$ we have $Rsr \subseteq Rr =(0)$ and so $Rsr=(0),$ i.e. $sr \in I.$ So either $I=(0)$ or $I=R.$

Case 1: $I=R.$ That means $sr=0$ for all $r,s \in R$ or, equivalently, $R^2=(0).$ Thus every subgroup of $(R,+)$ is a left (in fact, two-sided) ideal of $R.$ Hence $(R,+)$ has no proper subgroup (because $R$ has no proper left ideals) and thus $|R|=p$ for some prime $p.$

Case 2: $I=(0).$ Choose $0 \ne r \in R.$ So $r \notin I$ and hence $Rr=R$ because $Rr$ is clearly a left ideal of $R.$ Thus there exists $e \in R$ such that $er=r.$ Now

$\text{ann}(r):=\{s \in R: \ sr=0\},$

the left-annihilator of $r$ in $R$, is obviously a left ideal of $R$ and we can’t have $\text{ann}(r)=R$ because then $Rr=(0).$ So $\text{ann}(r)=(0).$ Since

$(re-r)r=rer-r^2=r^2-r^2=0,$

we have $re-r \in \text{ann}(r)=(0).$ Thus $re=er=r.$ Let

$J=\{x \in R: \ \ xe=x\}.$

Clearly $J$ is a left ideal of $R$ and $0 \ne r \in J.$ Thus $J=R.$ So $xe=x$ for all $x \in R.$ Now let $0 \ne r'$ be any element of $R.$ Then, by what we just proved, $r'e=r'.$ On the other hand, by the same argument we used for $r,$ we find $e' \in R$ such that $r'e'=e'r'=r'.$ Thus $r'(e-e')=0,$ i.e. $r' \in \text{ann}(e-e').$
So $\text{ann}(e-e') \ne (0)$ and hence $\text{ann}(e-e')=R,$ i.e. $R(e-e')=(0)$ and thus $e-e' \in I=(0).$
So $e=e'$ and hence $r'e=er'=r'$ for all $r' \in R.$ Thus $e=1_R$ proving that $R$ is a division ring. $\Box$

Remark. The same result given in the above problem holds if $R$ has no proper right ideals.

Example. Let $p$ be a prime number. The ring

$\displaystyle R:= \left \{\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}: \ \ \ a \in \mathbb{Z}/p\mathbb{Z}\right\} \subset M_2(\mathbb{Z}/p\mathbb{Z})$

is not a division ring and it has no proper left (or right) ideals.

In finite commutative rings of odd order, number of idempotents divides number of units

Posted: October 20, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules
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All rings in this post are assumed to have the multiplicative identity $1.$

In this post, we showed that the ring $R:= \mathbb{Z}/n\mathbb{Z}$ has $2^m$ idempotents, where $m$ is the number of prime divisors of $n.$ Clearly $m$ is also the number of prime (= maximal) ideals of $R$ (recall that, in general, in commutative Artinian rings, prime ideals are maximal). Let $\varphi$ be the Euler’s totient function. The number of units of $R$ is $\varphi(n).$ If $n=\prod_{i=1}^m p_i^{n_i}$ is the prime factorization of $n,$ then $\varphi(n)=\prod_{i=}^m(p_i^{n_i}-p_i^{n_i-1}).$ If $n$ is odd, then $p_i^{n_i}-p_i^{n_i-1}$ is even for all $i$ and hence $2^m$ divides $\varphi(n).$ So if $n$ is odd, then the number of idempotents of $R$ divides the number of units of $R.$ As we are going to show now, this is a property of any finite commutative ring of odd order.

Recall that a commutative ring $R$ is called semilocal if the number of maximal ideals of $R$ is finite and it is called local if it has only one maximal ideal.

Example 1. If $p$ is a prime and $k \ge 1$ is an integer, then $\mathbb{Z}/p^k\mathbb{Z}$ is a local ring with the unique maximal ideal $p\mathbb{Z}/p^k\mathbb{Z}.$ If $n \ge 2$ is an integer, then $\mathbb{Z}/n\mathbb{Z}$ is a semilocal ring (what are its maximal ideals?).

Example 2. Generalizing the above example, Artinian rings are semilocal. This is easy to see; let $R$ be an Artinian ring and let $S$ be the set of all finite intersections of maximal ideals of $R.$ Since $R$ is Artinian, $S$ has a minimal element $I:=\bigcap_{i=1}^m M_i.$ Let $M$ be any maximal ideal of $R.$ Since $I \bigcap M \in S$ and $I$ is a minimal element of $S,$ we must have $M_1M_2 \cdots M_m \subseteq I \subseteq M.$ So $M_i \subseteq M$ for some $i$ and hence $M_i=M.$ Therefore $M_1, \cdots , M_m$ are all the maximal ideals of $R.$

Problem 1. Show that if $R$ is a local ring, then $0,1$ are the only idempotents of $R.$

Solution. Let $M$ be the maximal ideal of $R$ and let $e$ be an idempotent of $R.$ Then $e(1-e)=0 \in M.$ Thus either $e \in M$ or $1-e \in M.$ If $e \in M,$ then $1-e \notin M$ because otherwise $1=e+(1-e) \in M,$ which is false. So $1-e$ is a unit because there’s no other maximal ideal to contain $1-e.$ So $(1-e)r=1$ for some $r \in R$ and thus $e=e(1-e)r=0.$ Similarly, if $1-e \in M,$ then $e \notin M$ and thus $e$ is a unit. So $er=1$ for some $r \in R$ implying that $1-e=(1-e)er=0$ and hence $e=1. \ \Box$

Problem 2 Show that the number of idempotents of a commutative Arinian ring $R$ is $2^m,$ where $m$ is the number of maximal ideals of $R.$

Solution. Since $R$ is Artinian, it has only finitely many maximal ideals, say $M_1, \cdots , M_m$ (see Example 2). The Jacobson radical of $R$ is nilpotent, hence there exists an integer $k \ge 1$ such that

$(0)=\left(\bigcap_{i=1}^m M_i \right)^k=\prod_{i=1}^m M_i^k=\bigcap_{i=1}^m M_i^k.$

Thus, by the Chinese remainder theorem for commutative rings, $R \cong \prod_{i=1}^m R/M_i^k.$ Since each $R/M_i^k$ is a local ring, with the unique maximal ideal $M_i/M_i^k,$ it has only two idempotents, by Problem 1, and so $R$ has exactly $2^m$ idempotents. $\Box$

Problem 3. Let $R$ be a finite commutative ring. Show that if $|R|,$ the number of elements of $R,$ is odd, then the number of idempotents of $R$ divides the number of units of $R.$

Solution. Since $R$ is finite, it is Artinian. Let $\{M_1, \cdots , M_m\}$ be the set of maximal ideals of $R.$ By problem 2, the umber of idempotents of $R$ is $2^m$ and $R \cong \prod_{i=1}^m R/M_i^k$ for some integer $k \ge 1.$
Since $|R|$ is odd, each $|M_i|$ is odd too because $(M_i,+)$ is a subgroup of $(R,+)$ and so $|M_i|$ divides $|R|.$ Also, units in a local ring are exactly those elements of the ring which are not in the maximal ideal. So the number of units of each $R/M_i^k$ is $|R/M_i^k|-|M_i/M_i^k|,$ which is an even number because both $|R|$ and $|M_i|$ are odd. So the number of units of $R,$ which is the product of the number of units of $R/M_i^k, \ 1 \le i \le m,$ is divisible by $2^m,$ which is the number of idempotents of $R. \ \Box$

Remark 1. The result given in Problem 3 is not always true if the number of elements of the ring is even. For example, $\mathbb{Z}/2\mathbb{Z}$ has one unit and two idempotents. However, the result is true in $\mathbb{Z}/2^n\mathbb{Z}, \ n \ge 2,$ which has $\varphi(2^n)=2^n-2^{n-1}$ units and two idempotents.
Can we find all even integers $n$ for which the result given in Problem 3 is true in $\mathbb{Z}/n\mathbb{Z}$? Probably not because this question is equivalent to finding all integers $n$ such that $2^m \mid \varphi(n),$ where $m$ is the number of prime divisors of $n,$ and that is not an easy thing to do.

Remark 2. The result given in Problem 3 is not necessarily true in noncommutative rings with an odd number of elements. For example, consider $R:=M_2(\mathbb{F}_3),$ the ring of $2\times 2$ matrices with entries from the field of order three. Then $R$ has $(3^2-3)(3^2-1)=48$ units (see Problem 3 in this post!) but, according to my calculations, $R$ has $14$ idempotents and $14$ does not divide $48.$

Fermat’s last theorem for finite rings

Posted: October 3, 2018 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem. Let $R$ be a finite ring with $1$ and let $R^*=R \setminus \{0\}.$ Show that for every integer $n \ge 1$ there exist $x,y,z \in R^*$ such that $x^n+y^n=z^n$ if and only if $R$ is not a division ring.

Solution. Suppose first that $R$ is a division ring. Then, since $R$ is a finite ring, $R$ is a finite field, by the Wedderburn’s little theorem. Let $|R|=q.$ Then $x^{q-1}=1$ for all $x \in R^*$ and so $x^{q-1}+y^{q-1}=z^{q-1}$ has no solution in $R^*.$
Conversely, suppose that $R$ is not a division ring (equivalently, a field because $R$ is finite). So $|R| > 2$ and hence the equation $x+y=z$ has solutions in $R^*$ (just choose $y=1, \ z \ne 0,1$ and $x=z-1$).
Let $J(R)$ be the Jacobson radical of $R.$ Since $R$ is finite, it is Artinian and so $J(R)$ is nilpotent.
So if $J(R) \neq (0),$ then there exists $a \in R^*$ such that $a^2=0$ and so $a^n=0$ for all $n \ge 2.$ Therefore the equation $x^n+y^n=z^n, \ n \ge 2,$ has a solution $x=a, y=z=1$ in $R^*.$
If $J(R)=(0),$ then by the Artin-Wedderburn’s theorem,

$\displaystyle R=\prod_{i=1}^k M_{n_i}(F_i),$

for some finite fields $F_i.$ Since $R$ is not a field, we have either $n_i >1$ for some $i$ or $n_i=1$ for all $i$ and $k \ge 2.$ If $n_i > 1$ for some $i,$ then $M_{n_i}(F_i),$ and hence $R,$ will have a non-zero nilpotent element and we are done. If $n_i=1$ for all $i$ and $k \ge 2,$ then

$x=(1,0,0, \cdots ,0), \ y = (0,1,0, \cdots , 0), \ z = (1,1,0, \cdots , 0)$

will satisfy $x^n+y^n=z^n. \ \Box$

Throughout this post, $U(R)$ and $J(R)$ are the group of units and the Jacobson radical of a ring $R.$ Assuming that $U(R)$ is finite and $|U(R)|$ is odd, we will show that $|U(R)|=\prod_{i=1}^k (2^{n_i}-1)$ for some positive integers $k, n_1, \ldots , n_k.$ Let’s start with a nice little problem.

Problem 1. Prove that if $U(R)$ is finite, then $J(R)$ is finite too and $|U(R)|=|J(R)||U(R/J(R)|.$

Solution. Let $J:=J(R)$ and define the map $f: U(R) \to U(R/J))$ by $f(x) = x + J, \ x \in U(R).$ This map is clearly a well-defined group homomorphism. To prove that $f$ is surjective, suppose that $x + J \in U(R/J).$ Then $1-xy \in J,$ for some $y \in R,$ and hence $xy = 1-(1-xy) \in U(R)$ implying that $x \in U(R).$ So $f$ is surjective and thus $U(R)/\ker f \cong U(R/J).$
Now, $\ker f = \{1-x : \ \ x \in J \}$ is a subgroup of $U(R)$ and $|\ker f|=|J|.$ Thus $J$ is finite and $|U(R)|=|\ker f||U(R/J)|=|J||U(R/J)|. \Box$

Problem 2. Let $p$ be a prime number and suppose that $U(R)$ is finite and $pR=(0).$ Prove that if $p \nmid |U(R)|,$ then $J(R)=(0).$

Solution. Suppose that $J(R) \neq (0)$ and $0 \neq x \in J(R).$ Then, considering $J(R)$ as an additive group, $H:=\{ix: \ 0 \leq i \leq p-1 \}$ is a subgroup of $J(R)$ and so $p=|H| \mid |J(R)|.$ But then $p \mid |U(R)|,$ by Problem 1, and that’s a contradiction! $\Box$

There is also a direct, and maybe easier, way to solve Problem 2: suppose that there exists $0 \neq x \in J(R).$ On $U(R),$ define the relation $\sim$ as follows: $y \sim z$ if and only if $y-z = nx$ for some integer $n.$ Then $\sim$ is an equivalence relation and the equivalence class of $y \in U(R)$ is $[y]=\{y+ix: \ 0 \leq i \leq p-1 \}.$ Note that $[y] \subseteq U(R)$ because $x \in J(R)$ and $y \in U(R).$ So if $k$ is the number of equivalence classes, then $|U(R)|=k|[y]|=kp,$ contradiction!

Problem 3. Prove that if $F$ is a finite field, then $|U(M_n(F))|=\prod_{i=1}^n(|F|^n - |F|^{i-1}).$ In particular, if $|U(M_n(F))|$ is odd,  then $n=1$ and $|F|$ is a power of $2.$

Solution. The group $U(M_n(F))= \text{GL}(n,F)$ is isomorphic to the group of invertible linear maps $F^n \to F^n.$ Also, there is a one-to-one correspondence between the set of invertible linear maps $F^n \to F^n$ and the set of (ordered) bases of $F^n.$ So $|U(M_n(F))|$ is equal to the number of bases of $F^n.$ Now, to construct a basis for $F^n,$ we choose any non-zero element $v_1 \in F^n.$ There are $|F|^n-1$ different ways to choose $v_1.$ Now, to choose $v_2,$ we need to make sure that $v_1,v_2$ are not linearly dependent, i.e. $v_2 \notin Fv_1 \cong F.$ So there are $|F|^n-|F|$ possible ways to choose $v_2.$ Again, we need to choose $v_3$ somehow that $v_1,v_2,v_3$ are not linearly dependent, i.e. $v_3 \notin Fv_1+Fv_2 \cong F^2.$ So there are $|F|^n-|F|^2$ possible ways to choose $v_3.$ If we continue this process, we will get the formula given in the problem. $\Box$

Problem 4. Suppose that $U(R)$ is finite and $|U(R)|$ is odd. Prove that $|U(R)|=\prod_{i=1}^k (2^{n_i}-1)$ for some positive integers $k, n_1, \ldots , n_k.$

Solution. If $1 \neq -1$ in $R,$ then $\{1,-1\}$ would be a subgroup of order 2 in $U(R)$ and this is not possible because $|U(R)|$ is odd. So $1=-1.$ Hence $2R=(0)$ and $\mathbb{Z}/2\mathbb{Z} \cong \{0,1\} \subseteq R.$ Let $S$ be the ring generated by $\{0,1\}$ and $U(R).$ Obviously $S$ is finite, $2S=(0)$ and $U(S)=U(R).$ We also have $J(S)=(0),$ by Problem 2. So $S$ is a finite semisimple ring and hence $S \cong \prod_{i=1}^k M_{m_i}(F_i)$ for some positive integers $k, m_1, \ldots , m_k$ and some finite fields $F_1, \ldots , F_k,$ by the Artin-Wedderburn theorem and Wedderburn’s little theorem. Therefore $|U(R)|=|U(S)|=\prod_{i=1}^k |U(M_{m_i}(F_i))|.$ The result now follows from the second part of Problem 3. $\Box$

Example. It’s clear from Problem 4 that there’s no ring with only $5$ units. We can prove that directly without using the result given in Problem 4, as follows. Suppose, to the contrary, that there exists a ring $R$ such that $|U(R)|=5.$ Then $U(R)$ is cyclic and so it has a generator $g.$ If $2 \ne 0$ in $R,$ then $\{1,-1\}$ would be a subgroup of order $2$ in $U(R)$ and that is not possible because $|U(R)|$ is odd. So $2=0.$ Now see that

$(1+g^2+g^3)(1+g+g^4)=1$

and so $1+g^2+g^3 \in U(R).$ Thus $1+g^2+g^3=g^n$ for some integer $0 \le n \le 4.$ Clearly we can’t have $n=0,2,3$ because the order of $g$ is $5.$ If $n=1,$ then $1+g^2+g^3=g$ gives $g^2+g^4+1=g^3$ and hence $g^4=g^3+g^2+1=g$ implying that  $g^3=1,$ which is false. Similarly, the case $n=4$ is impossible. So $|U(R)| \ne 5.$

Maximal and prime ideals of a polynomial ring over a PID (2)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
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See part (1) here! Again, we will assume that $R$ is a PID and $x$ is a varibale over $x.$ In this post, we will take a look at the maximal ideals of $R[x].$ Let $I$ be a maximal ideal of $R[x].$ By Problem 2, if $I \cap R \neq (0),$ then $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is irreducible modulo $p.$ If $I \cap R =(0),$ then $I=\langle f(x) \rangle$ for some irreducible element $f(x) \in R[x].$ Before investigating maximal ideals of $R[x]$ in more details, let’s give an example of a PID $R$ which is not a field but $R[x]$ has a maximal ideal $I$ which is principal. We will see in Problem 3 that this situation may happen only when the number of prime elements of $R$ is finite.

Example 1. Let $F$ be a filed and put $R=F[[t]],$ the formal power series in the variable $t$ over $F.$ Let $x$ be a variable over $R.$ Then $I:=\langle xt - 1 \rangle$ is a maximal ideal of $R[x].$

Proof. See that $R[x]/I \cong F[[t,t^{-1}]]$ and that $F[[t,t^{-1}]]$ is the field of fractions of $R.$ Thus $R[x]/I$ is a field and so $I$ is a maximal ideal of $R[x]. \ \Box$

Problem 3. Prove that if $R$ has infinitely many prime elements, then an ideal $I$ of $R[x]$ is maximal if and only if $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is irreducible modulo $p.$

Solution. We have already proved one direction of the problem in Problem 1. For the other direction, let $I$ be a maximal ideal of $R[x].$ By the first case in the solution of Problem 2 and the second part of Problem 1, we  only need to show that $I \cap R \neq (0).$ So suppose to the contrary that $I \cap R=(0).$ Then, by the second case in the solution of Problem 2, $I=\langle f(x) \rangle$ for some $f(x) \in R[x].$ We also know that $R[x]/I$ is a field because $I$ is a maximal ideal of $R[x].$ Since $R$ has infinitely many prime elements, we can choose a prime $p \in R$ such that $p$ does not divide the leading coefficient of $f(x).$ Now, consider the natural ring homomorphism $\psi : R[x] \to R[x]/I.$ Since $I \cap R=(0),$ $\psi(p) \neq 0$ and so $\psi(p)$ is invertible in $R[x]/I.$ Therefore $pg(x)-1 \in \ker \psi = I$ for some $g(x) \in R[x].$ Hence $pg(x)-1=h(x)f(x)$ for some $h(x) \in R[x].$ If $p \mid h(x),$ then we will have $p \mid 1$ which is non-sense. So $h(x)=pu(x) + v(x)$ for some $u(x),v(x) \in R[x]$ where $p$ does not divide the leading coefficient of $v(x).$ Now $pg(x) - 1 =h(x)f(x)$ gives us $p(g(x)-u(x)f(x)) - 1 =v(x)f(x)$ and so the leading coefficient of $v(x)f(x)$ is divisible by $p.$ Hence the leading coefficient of $f(x)$ must be divisible by $p,$ contradiction! $\Box$

Example 2. The ring of integers $\mathbb{Z}$ is a PID and it has infinitely many prime elements. So, by Problem 3, an ideal $I$ of $\mathbb{Z}[x]$ is maximal if and only if $I=\langle p, f(x) \rangle$ for some prime $p \in \mathbb{Z}$ and some $f(x)$ which is irreducible modulo $p.$ By Problem 2, the prime ideals of $\mathbb{Z}[x]$ are the union of the following sets:
1) all maximal ideals
2) all ideals of the form $\langle p \rangle,$ where $p \in \mathbb{Z}$ is a prime
3) all ideals of the form $\langle f(x) \rangle,$ where $f(x)$ is irreducible in $\mathbb{Z}[x].$

Maximal and prime ideals of a polynomial ring over a PID (1)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
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We know that if $R$ is a field and if $x$ is a variable over $R,$ then $R[x]$ is a PID and a non-zero ideal $I$ of $R[x]$ is maximal if and only if $I$ is prime if and only if $I$ is generated by an irreducible element of $R[x].$ If $R$ is a PID which is not a field, then $R[x]$ could have prime ideals which are not maximal. For example, in $\mathbb{Z}[x]$ the ideal $\langle 2 \rangle$ is prime but not maximal. In this two-part post, we will find prime and maximal ideals of $R[x]$ when $R$ is a PID.

Notation. Throughout this post, $R$ is a PID and $R[x]$ is the polynomial ring in the variable $x$ over $R.$ Given a prime element $p \in R,$ we will denote by $\phi_p$ the natural ring homomorphism $R[x] \to R[x]/pR[x].$

Definition Let $p$ be a prime element of $R.$ An element $f(x) \in R[x]$ is called irreducible modulo $p$ if $\phi_p(f(x))$ is irreducible in $R[x]/pR[x].$ Let $\gamma : R \to R/pR$ be the natural ring homomorphism. Then, since $R[x]/pR[x] \cong (R/pR)[x],$ an element $f(x)=\sum_{i=0}^n a_ix^i \in R[x]$ is irreducible modulo $p$ if and only if $\sum_{i=0}^n \gamma(a_i)x^i$ is irreducible in $(R/pR)[x].$ Note that $R/pR$ is a field because $R$ is a PID.

Problem 1. Prove that if $p \in R$ is prime and if $f(x) \in R[x]$ is irreducible modulo $p,$ then $I:=\langle p, f(x) \rangle$ is a maximal ideal of $R[x].$ If $f =0,$ then $I$ is a prime but not a maximal ideal of $R[x].$

Solution. Clearly $I/pR[x]=\phi_p(I)=\langle \phi_p(f(x)) \rangle.$ So $\phi_p(I)$ is a maximal ideal of $R[x]/pR[x]$ because $\phi_p(f(x))$ is irreducible in $R[x]/pR[x] \cong (R/pR)[x]$ and $R/pR$ is a field. So $I$ is a maximal ideal of $R[x].$ If $f =0,$ then $I=\langle p \rangle=pR[x]$ and so $R[x]/I \cong (R/pR)[x]$ is a domain which implies that $I$ is prime. Finally, $I= \langle p \rangle$ is not maximal because, for example, $I \subset \langle p,x \rangle \ \Box$

Problem 2. Prove that a non-zero ideal $I$ of $R[x]$ is prime if and only if either $I= \langle f(x) \rangle$ for some irreducible element $f(x) \in R[x]$ or $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is either zero or irreducible modulo $p.$

Solution. If $f(x) \in R[x]$ is irreducible, then $\langle f(x) \rangle$ is a prime ideal of $R[x]$ because $R[x]$ is a UFD. If $f(x)=0$ or $f(x)$ is irreducible modulo a prime $p \in R,$ then $I=\langle p, f(x) \rangle$ is a prime ideal of $R[x]$ by Problem 1.
Conversely, suppose that $I$ is a non-zero prime ideal of $R[x].$ We consider two cases.
Case 1. $I \cap R \neq (0)$ : Let $0 \neq r \in I \cap R.$ Then $r$ is clearly not a unit because then $I$ wouldn’t be a proper ideal of $R[x].$ So, since $r \in I$ and $I$ is a prime ideal of $R[x],$ there exists a prime divisor $p$ of $r$ such that $p \in I.$  So $pR[x] \subseteq I$ and hence $\phi_p(I)=I/pR[x]$ is a prime ideal of $R[x]/pR[x] \cong (R/pR)[x].$ Thus we have two possibilities. The first possibility is that $\phi_p(I)=(0),$ which gives us $I \subseteq \ker \phi_p = pR[x]$ and therefore $I=pR[x]=\langle p \rangle.$ The second possibility is that $\phi_p(I)=\langle \phi_p(f(x)) \rangle= \phi_p(\langle f(x) \rangle)$ for some irreducible element $\phi_p(f(x)) \in R[x]/pR[x],$ which gives us $I=\langle p, f(x) \rangle$ because $\ker \phi_p =pR[x].$
Case 2. $I \cap R = (0)$ : Let $Q$ be the field of fractions of $R$ and put $J:=IQ[x].$ Then $J$ is a non-zero prime ideal of $Q[x]$ because $I$ is a prime ideal of $R[x].$ Note that $J=\{g(x)/r : \ g(x) \in I, \ 0 \neq r \in R \}.$ So, since $Q[x]$ is a PID, $J=q(x)Q[x]$ for some irreducible element $q(x) \in Q[x].$ Obviously, we can write $q(x)=\alpha f(x),$ where $\alpha \in Q$ and $f(x) \in R[x]$ is irreducible and the gcd of the coefficients of $f(x)$ is one. Thus $J = f(x)Q[x]$ and, since $f(x) \in J,$ we have $f(x) = g(x)/r$ for some $0 \neq r \in R$ and $g(x) \in I.$ But then $rf(x)=g(x) \in I$ and so $f(x) \in I$ because $I$ is prime and $I \cap R = (0).$ Hence $\langle f(x) \rangle \subseteq I.$ We will be done if we prove that $I \subseteq \langle f(x) \rangle.$ To prove this, let $h(x) \in I \subseteq J=f(x)Q[x].$ So $h(x)=f(x)q_0(x)$ for some $q_0(x) \in Q[x].$ Therefore, since the gcd of the coefficients of $f(x)$ is one, we must have $q_0(x) \in R[x]$ by Gauss’s lemma. Hence $h(x) \in \langle f(x) \rangle$ and the solution is complete. $\Box$

See the next part here!