Abstract Algebra

Permutation matrix; an alternative view

Posted: February 17, 2024 in Basic Algebra, Matrices
Tags: permutation matrix

As always, in this post, we denote by $\text{GL}(n,k)$ the group of $n \times n$ invertible matrices with entries from a field $k.$ Also, $A^T$ is the transpose of a matrix $A.$

Here we defined a permutation matrix as an $n \times n$ matrix $A$ such that $Ae_i=e_{\sigma(i)}$ for some $\sigma \in S_n$ and all positive integers $i \le n,$ where $\{e_1, \cdots , e_n\}$ is the standard basis of the $\mathbb{R}$ -vector space $\mathbb{R}^n$ and $S_n$ is the group of permutations of the set $\{1, 2, \cdots , n\}.$ In other words, an $n \times n$ permutation matrix is a matrix whose columns are exactly $e_1, \cdots , e_n$ in any order. Equivalently, a permutation matrix is a matrix $A$ with this property that every row or column of $A$ has exactly one nonzero entry and that nonzero entry is $1.$ For example, for $n=2,$ we have exactly two permutation matrices: $\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}.$

In this post, we give another way to view a permutation matrix. The following problem was recently posted on the Art of Problem Solving website; here you can see the problem and my solution (post #7). I’m going to rephrase the statement of the problem and also make my solution a little bit more clear.

Problem. Show that $A \in \text{GL}(n,\mathbb{R})$ is a permutation if and only if all the entries of $A$ and $A^{-1}$ are in $\{0,1\}.$

Solution. One side is clear: if $A$ is a permutation matrix, then $A$ is invertible and all the entries of $A$ and $A^{-1}$ are in the set $\{0,1\}.$
Suppose now that $A \in \text{GL}(n,\mathbb{R})$ and all the entries of $A, A^{-1}$ are in $\{0,1\}.$ Let $I$ be the $n \times n$ identity matrix, and let $a_{ij}, b_{ij}, c_{ij}$ be, respectively, the $(i,j)$ -entry of $A,A^{-1},I.$ Choose any positive integer $i \le n.$ Since $AA^{-1}=I$ and $c_{ii}=1,$ there exists a positive integer $m \le n$ such that $a_{im}=b_{mi}=1.$ If $a_{ij}=1$ for some $j \ne m,$ then, since $A^{-1}A=I,$ we’ll get

$0=c_{mj}=b_{mi}a_{ij}+ \cdots =1 + \cdots \ne 0,$

which is a contradiction. So $a_{ij}=0$ for all $j \ne m.$ We have shown that in every row of $A,$ all the entries except for one, which is $1,$ are zero. Now, we also have that all the entries of $A^T$ and $(A^T)^{-1}$ are in $\{0,1\}$ because $(A^T)^{-1}=(A^{-1})^T.$ Therefore repeating the above argument this time for $A^T$ shows that in every column of $A,$ all the entries except for one, which is $1,$ are zero, and hence $A$ is a permutation. $\ \Box$

Left-multiplication maps as ring homomorphisms

Posted: February 14, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: abelian ring, Artin-Wedderburn theorem, center of rings, idempotent, left-multiplication, prime ring, SE, semiprime ring, semiprimitive ring, semisimple ring, upper triangular matrix

Throughout this post, $R$ is a ring with identity and $Z(R)$ is its center. We denote by $\mathcal{I}(R)$ the set of idempotents of $R,$ i.e. $r \in R$ such that $r^2=r.$ Also, $R[x]$ is the ring of polynomials in $x$ with coefficients in $R.$ Given $r \in R,$ the left-multiplication map $\ell_r: R \to R$ is defined by $\ell_r(a)=ra,$ for all $a \in R.$ Finally, ring homomorphisms in this post are not required to preserve the identity element.

Left multiplication maps $\ell_r$ are clearly right $R$ -module homomorphisms, i.e. $\ell_r(ab)=\ell_r(a)b,$ for all $a,b \in R.$ But for which $r \in R$ is the map $\ell_r$ a ring homomorphism? That’s probably not easy to answer in general, but I’m going to share some of my findings with you anyway. Before I do that, let me define a class of rings with a funny name which are related to our question.

Definition. A ring $R$ is called abelian if $\mathcal{I}(R) \subseteq Z(R),$ i.e. every idempotent of $R$ is central.

Commutative rings are clearly abelian but abelian rings are not necessarily commutative. For example, by Remark 3 in this post, every reduced ring is abelian (in fact, by Exercise 1 in that post, every reversible ring is abelian). Part vi) of the following problem gives an equivalent definition of abelian rings: a ring $R$ is abelian if and only if $\ell_r$ is a ring homomorphism for every idempotent $r$ of $R.$

Problem (Y. Sharifi). Let $R$ be a ring, and consider the set

$\mathcal{L}(R):=\{r \in R: \ \ell_r \ \text{is a ring homomorphism}\}.$

Show that

i) $\mathcal{L}(R)=\{r \in R: \ ra=rar, \ \forall a \in R\},$

ii) if $R$ is commutative, then $\mathcal{L}(R)=\mathcal{I}(R),$

iii) $\mathcal{L}(R)$ is multiplicatively closed,

iv) given $r \in \mathcal{I}(R),$ the set $S_r:=\{a \in R: \ ra=rar\}$ is a subring of $R,$

v) $Z(R) \cap \mathcal{I}(R) \subseteq \mathcal{L}(R) \subseteq \mathcal{I}(R),$

vi) $\mathcal{L}(R) = \mathcal{I}(R)$ if and only if $R$ is abelian,

vii) if $R$ is semiprime, then $\mathcal{L}(R)= Z(R) \cap \mathcal{I}(R),$

viii) if $R$ is prime, then $\mathcal{L}(R)=\{0,1\},$

ix) $\mathcal{L}(R) =R \cap \mathcal{L}(R[x])$ and if $R$ is commutative or semiprime, then $\mathcal{L}(R[x])=\mathcal{L}(R),$

x) $\mathcal{L}\left(\prod_{i \in I}R_i\right)=\prod_{i \in I}\mathcal{L}(R_i)$ for any family of rings $\{R_i\}_{i \in I}.$

Solution. i) Since $\ell_r$ is clearly additive, it is a ring homomorphism if and only if $\ell_r(ab)=\ell_r(a)\ell_r(b),$ i.e. $rab=rarb$ for all $a,b \in R.$ Choosing $b=1$ gives $ra=rar.$ Conversely, if $ra=rar,$ then $rab=rarb$ and so $\ell_r$ is a ring homomorphism.

ii) By i), $r \in \mathcal{L}(R)$ if and only if $ra=r^2a,$ for all $a \in R,$ which holds if and only if $r^2=r,$ i.e $r \in \mathcal{I}(R).$

iii) Let $r,s \in \mathcal{L}(R)$ and $a \in R.$ Then, by i), $rsar=rsa, \ sas=sa,$ and so $rsars=rsas=rsa.$ Thus $rs \in \mathcal{L}(R).$

iv) If $a,b \in S_r,$ then $rab=rarb=rarbr=rabr$ and so $ab \in S_r.$

v) Let $r \in \mathcal{L}(R).$ Then choosing $a=1$ in i) gives $r=r^2$ and so $r \in \mathcal{I}(R).$ Now, let $r \in Z(R) \cap \mathcal{I}(R),$ and $a \in R.$ Then $rar=rra=r^2a=ra$ and so $r \in \mathcal{L}(R).$

vi) If $\mathcal{I}(R) \subseteq Z(R),$ then $\mathcal{L}(R) = \mathcal{I}(R),$ by v). Suppose now that $\mathcal{L}(R) = \mathcal{I}(R)$ and $r \in \mathcal{I}(R).$ Then $1-r \in \mathcal{I}(R),$ and so, by i), $ra=rar, \ (1-r)a=(1-r)a(1-r),$ for all $a \in R.$ Thus

$a-ra=(1-r)a=(1-r)a(1-r)=a-ar-ra+rar=a-ar,$

which gives $ra=ar$ and so $r \in Z(R).$

vii) By v), we only need to show that $\mathcal{L}(R) \subseteq Z(R).$ So let $r \in \mathcal{L}(R)$ and $a,b \in R.$ If we show that $(ra-ar)b(ra-ar)=0,$ we are done because that means $(ra-ar)R(ra-ar)=(0)$ and so, since $R$ is semiprime, $ra=ar$ hence $r \in Z(R).$ Now, to show that $(ra-ar)b(ra-ar)=0,$ we have

$(ra-ar)b(ra-ar)=rabra-rabar-arbra+arbar$

and so, since by i), $rabr=rab, \ rbr=rb, \ rbar=rba,$ we get that

$(ra-ar)b(ra-ar)=raba-raba-arba+arba=0.$

viii) Clearly the inclusion $\{0,1\} \subseteq \mathcal{L}(R)$ holds for any ring $R.$ Now, suppose that $R$ is a prime ring and $r \in \mathcal{L}(R), a \in R.$ Then, by i), $ra(1-r)=0$ and so $rR(1-r)=(0),$ which gives $r=0$ or $r=1.$

ix) Let $r \in R.$ Then, by i), $r \in \mathcal{L}(R[x])$ if and only if $rf(x)=rf(x)r$ for all $f(x) \in R[x]$ if and only if $ra=rar$ for all $a \in R,$ because $x$ is central in $R[x],$ if and only if $r \in \mathcal{L}(R),$ by i).
If $R$ is commutative, then $\mathcal{L}(R)=\mathcal{I}(R)$ and so, since the polynomial ring $R[x]$ is commutative too,

$\mathcal{L}(R[x])=\mathcal{I}(R[x])=\mathcal{I}(R)=\mathcal{L}(R).$

Note that the identity $\mathcal{I}(R[x])=\mathcal{I}(R)$ is true by this post.
Now suppose that $R$ is semiprime, and $f(x)=\sum_{i=0}^nr_ix^i \in \mathcal{L}(R[x]), \ n \ge 1.$ So $f(x)a=f(x)af(x)$ for all $a \in R$ and hence, equating the coefficients of $x^{2n}$ on both sides of the equality gives $0=r_nar_n.$ Thus $r_nRr_n=(0)$ and so, since $R$ is semiprime, $r_n=0,$ contradiction. This proves that $n=0$ and hence $f(x) \in R$ giving $\mathcal{L}(R[x]) = \mathcal{L}(R).$

x) Clear, by i). $\ \Box$

Example 1. This example shows that $\mathcal{L}(R)$ is not always central. Let $C$ be a commutative domain, and let $R$ be the ring of $2 \times 2$ upper triangular matrices with entries in $C.$ Then, using the first part of the Problem, it is easy to see that

$\mathcal{L}(R)=\left \{\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix}0 & c \\ 0 & 1\end{pmatrix}, \ c \in C\right \}.$

Example 2. Let $R$ be the ring in Example 1. Then $\mathcal{L}(R[x]) \neq \mathcal{L}(R)$ because $f(x)=r+sx \in \mathcal{L}(R[x]),$ where

$r=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}, \ \ \ \ \ s=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}.$

Example 3. If $R$ is a semisimple ring, then $|\mathcal{L}(R)|=2^n$ for some positive integer $n.$

Proof. By the Artin-Wedderburn theorem, $R$ is a finite direct product of matrix rings over division rings. The result now follows from parts viii) and x) of the Problem. $\ \Box$

Semiprime rings

Posted: February 12, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: matrix ring, nilpotent ideal, prime ring, reduced ring, semiprime ideal, semiprime ring, semiprimitive ring

Throughout this post, $R$ is a ring with identity, and $M_n(R)$ is the ring of $n \times n$ matrices with entries from $R.$ Also, by ideal we always mean two-sided ideal.

Here we defined a prime ideal of $R$ as a proper ideal $P$ which satisfies this condition: if $IJ \subseteq P$ for some ideals $I,J$ of $R,$ then $I \subseteq P$ or $J \subseteq P.$ By weakening this condition, we get a set of ideals of $R$ that contains the set of prime ideals of $R;$ we call these ideals semiprime.

Definition 1. A proper ideal $Q$ of a ring $R$ is called semiprime if for any ideal $I$ of $R, \ I^2 \subseteq Q$ implies that $I \subseteq Q.$

Now, what exactly are semiprime ideals in commutative rings?

Remark 1. In a commutative ring $R,$ semiprime ideals are just radical ideals.

Proof. Since $R$ is commutative, the condition $I^2 \subseteq Q \implies I \subseteq Q$ for all ideals $I$ of $R,$ which is the definition of a semiprime ideal $Q,$ is equivalent to the condition $a^2 \in Q \implies a \in Q$ for all $a \in R,$ which itself is equivalent to $a^n \in Q \implies a \in Q$ for all $a \in R$ and all positive integers $n$ (why?). Now, the set

$\{a \in R: \ a^n \in Q, \ \text{for some positive integer} \ n\}$

is, by definition, $\sqrt{Q},$ the radical of $Q,$ which is easily seen to be an ideal containing $Q.$ If $\sqrt{Q}=Q$ or, equivalently, $\sqrt{Q} \subseteq Q,$ then we say that $Q$ is a radical ideal. So semiprime ideals of a commutative ring are just radical ideals of the ring. $\ \Box$

Remark 2. It is clear from Definition 1 that any intersection of prime ideals is semiprime. We also know from commutative ring theory that radical ideals are exactly those ideals that are intersection of some prime ideals. By Remark 1, radical ideals of a commutative ring are exactly semiprime ideals of the ring. So it is natural to ask if, in general ring theory, it is true that every semiprime ideal is an intersection of some prime ideals. It turns out that the answer is positive. In fact some authors define a semiprime ideal as an ideal which is an intersection of some prime ideals and then they show this is equivalent to Definition 1. But I did not choose this approach here because it will complicate the proof of the following Proposition hence getting into the goal of this post which is introducing semiprime rings.

By Remark 1, an ideal $Q$ in a commutative ring $R$ is semiprime if and only if $a^2 \in Q \implies a \in Q$ for all $a \in R.$ In general ring theory, we have the following similar result (compare with Proposition 1 in this post!).

Proposition. A proper ideal $Q$ of a ring $R$ is semiprime if and only if for any $a \in R, \ aRa \subseteq Q$ implies that $a \in Q.$

Proof. Suppose first that $Q$ is semiprime and $aRa \subseteq Q$ for some $a \in R.$ Let $I:=RaR.$ Then $I$ is an ideal of $R$ and $I^2=RaRaR=R(aRa)R \subseteq Q.$ Thus $I \subseteq Q,$ which gives $a \in Q.$
Conversely, suppose now that $I^2 \subseteq Q$ for some ideal $I$ of $R$ and $a \in I.$ Then

$aRa \subseteq R(aRa)R=(RaR)(RaR) \subseteq I^2 \subseteq Q$

and so $a \in Q.$ Hence $I \subseteq Q$ which proves that $Q$ is semiprime. $\ \Box$

By Remark 1, $(0)$ is a semiprime ideal of a commutative ring $R$ if and only if $\sqrt{(0)}=(0),$ i.e. if $a^n=0$ for some $a \in R$ and some positive integer $n,$ then $a=0.$ In other words, $(0)$ is a semiprime ideal of $R$ if and only if $R$ is reduced. In general ring theory, it is true that if $R$ is reduced, then $(0)$ is a semiprime ideal of $R$ (Example 1). However the converse is not true (Example 2). If $(0)$ is a semiprime ideal of a ring, then we will call the ring semiprime. So, in general ring theory, every reduced ring is semiprime but not every semiprime ring is reduced.

Definition 2. We say that a ring $R$ is semiprime if $(0)$ is a semiprime ideal of $R,$ i.e. for any ideal $I$ of $R,$ if $I^2=(0),$ then $I=(0).$

The Proposition gives the following simple yet useful test to check if a ring is semiprime or not.

Corollary. A ring $R$ is semiprime if and only if for any $a \in R, \ aRa=(0)$ implies that $a=0.$

Example 1. Every reduced ring $R$ is semiprime.

Proof. If $aRa=(0),$ for some $a \in R,$ then $a^2=0$ and so $a=0,$ because $R$ is reduced. $\ \Box$

Example 2. Every prime ring is obviously semiprime and so the matrix ring $M_n(k)$ over a field $k$ is a semiprime ring which is not reduced.

Example 3. Every semiprimitive ring $R$ is semiprime.

Proof. Suppose that $aRa=(0),$ for some $a \in R.$ Then $(ra)^2=0$ for all $r \in R$ and hence $1-ra$ is invertible. Thus $a \in J(R)=(0). \ \Box$

Example 4. Let $R$ be a ring and $S:=M_n(R).$ Then $S$ is semiprime if and only if $R$ is semiprime. In particular, if $R$ is reduced, then $S$ is semiprime.

Proof. It’s the same as the proof for prime rings (Example 4). $\ \Box$

Exercise 1. Show that a proper ideal $Q$ of a ring $R$ is semiprime if and only if the ring $R/Q$ is semiprime.

Exercise 2. Show that a proper ideal $Q$ of a ring $R$ is semiprime if and only if $I^2 \subseteq Q$ implies that $I \subseteq Q$ for any left ideal $I$ of $R.$

Exercise 3. Show that the center of a semiprime ring $R$ is reduced.
Hint. If $a^2=0$ for some central element of $R,$ then $aRa=Ra^2=(0).$

Exercise 4. Show that a direct product of rings is semiprime if and only if each ring is semiprime.

Exercise 5. Show that a ring $R$ is semiprime if and only if $R$ has no nonzero nilpotent ideal, i.e if $I^n=(0)$ for some ideal $I$ of $R$ and some positive integer $n,$ then $I=(0).$
Hint. If $n \ge 2$ is the smallest positive integer $n$ such that $I^n=0,$ then $(I^{n-1})^2=I^{2n-2}=(0).$

Note. The references for this post are Chapter 4 of T. Y. Lam’s book A First Course in Noncommutative Rings and Chapter 3 of Goodearl & Warfield’s book An Introduction to Noncommutative Noetherian Rings.

Prime rings

Posted: February 10, 2024 in Noncommutative Ring Theory Notes, Prime & Semiprime Rings
Tags: matrix ring, prime ideal, prime ring, primitive ring, reduced ring, simple ring

Throughout this post, $R$ is a ring with identity, and $M_n(R)$ is the ring of $n \times n$ matrices with entries from $R.$ Also, by ideal we always mean two-sided ideal.

In commutative ring theory, a proper ideal $P$ of a commutative ring $R$ is said to be prime if $IJ \subseteq P,$ where $I,J$ are ideals of $R,$ implies that $I \subseteq P$ or $J \subseteq P.$ The definition of a prime ideal remains the same even if $R$ is not commutative.

Definition 1. A proper ideal $P$ of a ring $R$ is called prime if for any ideals $I,J$ of $R, \ IJ \subseteq P$ implies that $I \subseteq P$ or $J \subseteq P.$

In commutative ring theory, we see that a proper ideal $P$ is prime if and only if $a,b \in R, \ ab \in P$ implies that $a \in P$ or $b \in P.$ In general ring theory, we have the following similar result.

Proposition 1. A proper ideal $P$ of a ring $R$ is prime if and only if for any $a,b \in R, \ aRb \subseteq P$ implies that $a \in P$ or $b \in P.$

Proof. Suppose first that $P$ is prime and $aRb \subseteq P$ for some $a,b \in R.$ Let $I:=RaR, \ J:=RbR.$ Then $I,J$ are ideals of $R$ and $IJ=RaRbR=R(aRb)R \subseteq P.$ Thus either $I \subseteq P,$ which gives $a \in P,$ or $J \subseteq P,$ which gives $b \in P.$
Conversely, suppose now that $IJ \subseteq P$ for some ideals $I,J$ of $R$ and $I \nsubseteq P.$ We need to show that $J \subseteq P.$ Let $a \in I \setminus P, \ b \in J.$ Then $aRb \subseteq R(aRb)R=(RaR)(RbR) \subseteq IJ \subseteq P$ and so $b \in P$ because $a \notin P.$ Hence $J \subseteq P$ which proves that $P$ is prime. $\ \Box$

In commutative ring theory, if $(0)$ is a prime ideal of a ring, the ring is called a commutative domain. In general ring theory, we still call a ring $R$ a domain if $a,b \in R, \ ab=0$ implies that $a=0$ or $b=0.$ It is true that if $R$ is a domain, then $(0)$ is a prime ideal of $R$ (Example 1). However, in general, the converse is not true (see Example 2 and Proposition 2). If $(0)$ is a prime ideal of a ring, then we will call the ring prime. So, in general ring theory, every domain is a prime ring but not every prime ring is a domain.

Definition 2. We say that a ring $R$ is prime if $(0)$ is a prime ideal of $R,$ i.e. for any ideals $I,J$ of $R,$ if $IJ=(0),$ then $I=(0)$ or $J=(0).$

Proposition 1 gives the following simple yet useful test to check if a ring is prime or not. This is useful because it’s in terms of elements not ideals, which are usually not easy to find.

Corollary. A ring $R$ is prime if and only if for any $a,b \in R, \ aRb=(0)$ implies that $a=0$ or $b=0.$

Example 1. Every domain $R$ is prime.

Proof. Suppose that $aRb=(0)$ for some $a,b \in R.$ Then $ab=0$ and so either $a=0$ or $b=0,$ because $R$ is a domain. $\ \Box$

Example 2. Every simple ring $R$ is prime. In particular, every matrix ring $M_n(k)$ over a field $k$ is prime.

Proof. The only ideals of $R$ are $(0), R.$ So $IJ=(0),$ where $I,J$ are ideals of $R,$ implies that $I=(0)$ or $J=(0). \ \Box$

So not every prime ring is a domain. The following characterizes all prime rings that are domain.

Proposition 2. A ring is a domain if and only if it’s both prime and reduced.

Proof. See Remark 2 in this post. $\ \Box$

The following example generalizes Example 2.

Example 3. Every left primitive ring is prime.

Proof. See Fact 2 in this post. $\ \Box$

Example 4. Let $R$ be a ring and $S:=M_n(R).$ Then $S$ is prime if and only if $R$ is prime. In particular, if $R$ is a domain, then $S$ is prime.

Proof. It is a well-known fact that ideals of $S$ are in the form $M_n(I),$ where $I$ is any ideal of $R.$ Now, suppose that $R$ is prime and $KL=(0)$ for some ideals $K,L$ of $S.$ We have $K=M_n(I), \ L=M_n(J)$ for some ideals $I,J$ of $R.$ So $M_n(I)M_n(J)=(0)$ and hence $IJ=(0),$ which gives $I=(0)$ or $J=(0),$ because $R$ is prime. Therefore $K=(0)$ or $L=(0)$ and so $S$ is prime. The proof for the converse is similar. $\ \Box$

Exercise 1. Show that a proper ideal $P$ of a ring $R$ is prime if and only if the ring $R/P$ is prime.

Exercise 2. Show that a proper ideal $P$ of a ring $R$ is prime if and only if $IJ \subseteq P$ implies that $I \subseteq P$ or $J \subseteq P,$ for any left ideals $I,J$ of $R.$

Exercise 3. Show that the center of a prime ring $R$ is a commutative domain.
Hint. If $ab=0$ for some central elements of $R,$ then $aRb=Rab=(0).$

Exercise 4. Show that a direct product of two or more rings can never be a prime ring.
Hint. If $R=R_1 \times R_2,$ and $a=(1,0), \ b=(0,1),$ then $aRb=(0).$

Exercise 5. Show that every maximal ideal of a ring is prime.

Note. The reference for this post is the first few pages of Chapter 4 of T. Y. Lam’s book A First Course in Noncommutative Rings.

Every nil, multiplicatively closed set of matrices is nilpotent

Posted: January 26, 2024 in Basic Algebra, Matrices, Rings
Tags: finite dimensional algebras, matrix algebras, nil, nilpotent, nilpotent left ideals

Throughout this post, $k$ is a field and $M_n(k)$ is the ring of $n \times n$ matrices with entries in $k.$

We have already seen the concepts of nil and nilpotency in a ring in this blog several times, but let’s see it here again. Let $R$ be a ring, and $S$ a subset of $R.$ We say that $S$ is nil if every element of $S$ is nilpotent, i.e. for every $s \in S$ there exists a positive integer $n$ such that $s^n=0.$ We say that $S$ is nilpotent if there exists a positive integer $n$ such that $S^n=(0),$ i.e. $s_1s_2 \cdots s_n=0$ for any $s_1, s_2, \cdots , s_n \in S.$ It is clear that $S$ is nilpotent if and only if $S^n=(0)$ for all positive integers $n$ that are large enough.

If $S$ is nilpotent, then $S$ is clearly nil too but, as the following example shows, the converse is not true.

Example. Let $\{e_{11},e_{12},e_{21},e_{22}\}$ be the standard $k$ -basis for $M_2(k),$ and let $S:=\{e_{12},e_{21}\} \subset M_2(k).$ Then $e_{12}^2=e_{21}^2=0$ and so $S$ is nil. But $S$ is not nilpotent because $(e_{12}e_{21})^n=e_{11}^n=e_{11} \ne 0,$ for any positive integer $n.$

Notice that the set $S$ in the above Example is not multiplicatively closed because $e_{12}e_{21}=e_{11} \notin S.$ We are now going to show that if $S \subset M_n(k)$ is both nil and multiplicatively closed, then $S$ is nilpotent. In fact, we show that $S^n=(0).$ But first a couple of useful lemmas.

Lemma 1. Any finite sum of nilpotent left ideals of a ring $R$ is nilpotent.

Proof. i) By induction, we only need to prove that for two ideals. So let $I,J$ be two nilpotent left ideals of $R.$ So $I^m=J^n=(0)$ for some positive integers $m,n.$ We claim that $(I+J)^{m+n-1}=(0).$ So we need to show that if $a_i \in I, b_i \in J, \ 1 \le i \le m+n-1,$ then

$(a_1+b_1) \cdots (a_{m+n-1}+b_{m+n-1})=0.$

The product on the left-hand side is a sum of terms of the form $c_1c_2 \cdots c_{m+n-1},$ where for each $i,$ either $c_i \in I$ or $c_i \in J.$ So we just need to show that $c_1c_2 \cdots c_{m+n-1}=0.$ Since there are $m+n-1$ of $c_i,$ either at least $m$ of $c_i$ are in $I$ or at least $n$ of $c_i$ are in $J.$ We will assume that at least $m$ of $c_i$ are in $I$ since the argument for the other case is similar. So we can write

$c_1c_2 \cdots c_{m+n-1}=u_1v_1u_2v_2\cdots u_mv_mu_{m+1},$

where each $v_i$ is in $I$ and each $u_i$ is in $I \cup J.$ Notice that there might be nothing before $v_1$ or after $v_m$ or between $v_i$ and $v_{i+1}$ for some $i,$ but that won’t cause any problem for our argument. Now, $u_iv_i \in I$ for all $i$ because $v_i \in I$ and $I$ is a left ideal. We also have that $I^m=(0).$ Thus

$c_1c_2 \cdots c_{m+n-1}=(u_1v_1)(u_2v_2)\cdots (u_mv_m)u_{m+1} \in I^mu_{m+1}=(0),$

and so $c_1c_2 \cdots c_{m+n-1}=0. \ \Box$

Next is to generalize this basic fact that if $A \in M_n(k)$ is nilpotent, then $A^n=0.$

Lemma 2. If $S$ is a nilpotent subset of $M_n(k),$ then $S^n=(0).$

Proof. We will assume $S \ne (0),$ since there is nothing to prove in this case. Let $k^n$ be the $k$ -vector space of all $n \times 1$ vectors with entries in $k$ and let

$V_j:=\{v \in k^n: \ S^jv=(0)\}, \ \ \ \ j \ge 1.$

So $V_j$ is the annihilator of $S^j$ in $k^n.$ Clearly each $V_j$ is a $k$ -vector subspace of $k^n$ and $V_1 \subseteq V_2 \subseteq \cdots.$ Since $S$ is nilpotent, there exists the smallest positive integer $m \ge 2$ such that $S^m=(0).$ So $V_m=k^n,$ and $V_{m-1} \ne k^n.$

Claim. $V_j \ne V_{j+1}$ for all $1 \le j \le m-1.$

Proof. By contradiction. Let $j \le m-1$ be the largest positive integer such that $V_j=V_{j+1}.$ That means if $S^{j+1}v=(0),$ for some $v \in k^n,$ then $S^jv=(0).$ Notice that in fact we have $j \le m-2$ because $V_{m-1} \ne V_m=k^n.$ But then $S^{j+2}v=S^{j+1}(Sv)=0, \ v \in k^n,$ implies that $S^{j+1}v=S^j(Sv)=(0)$ and so $V_{j+1}=V_{j+2},$ contradicting the maximality of $j.$ This completes the proof of the Claim.

So, by the Claim, $V_1 \subset V_2 \subset \cdots \subset V_{m-1} \subset V_m=k^n.$ Therefor

$\dim_k V_1 < \dim_k V_2 < \cdots < \dim_k V_{m-1} < \dim_k V_m=n,$

and so $m \le n,$ which gives $S^n=(0)$ because $S^m=(0). \ \Box$

We are now ready to prove the main result.

Theorem. Let $S$ be a nil, multiplicatively closed subset of $M_n(k).$ Then $S^n=(0).$

Proof. Since there is nothing to prove if $S=(0),$ we will assume that $S \ne (0).$ Let $V$ be the $k$ -vector subspace of $M_n(k)$ generated by $S.$ We are done if we show that $V^n=(0)$ because $S \subseteq V.$ Notice that $V$ is also a ring because $S$ is multiplicatively closed.
The proof of $V^n=(0)$ is by induction over $\dim_k V.$ If $\dim_k V=1,$ then $V=ks$ for some $s \in S$ and so $V^n=ks^n=(0).$
Suppose now that $\dim_k V > 1.$ Let $s \in S$ and define the map $f: V \to Vs$ by $f(v)=vs,$ for all $v \in V.$ Clearly $f$ is an onto $k$ -linear map. Also, $\ker f \ne (0)$ because if $s=0,$ then $\ker V=V$ and if $s \ne 0,$ then $0 \ne s^{m-1} \in \ker f,$ where $m$ is the smallest positive integer such that $s^m=0.$ So, as $k$ -vector spaces, $V/\ker f \cong Vs$ and hence $\dim_k Vs=\dim_k V-\dim_k \ker f < \dim_k V.$ Therefore since $Vs$ is the $k$ -vector space generated by $Ss \subseteq S$ and $Ss$ is a nil multiplicatively closed subset of $M_n(k),$ because $S$ is so, we have $(Vs)^n=(0),$ by our induction hypothesis. So we have shown that $(Vs)^n=0$ for all $s \in S.$ Now, $V=\sum_{j=1}^m Vs_j$ for some positive integer $m$ and so $s_j \in S,$ because $\dim_k V < \infty.$ Each $Vs_j$ is clearly a left ideal of $V$ and we just showed that they are all nilpotent. Thus, by Lemma 1, $V$ is nilpotent and so, by Lemma 2, $V^n=(0). \ \Box$

The Theorem in fact holds in any finite-dimensional $k$ -algebra not just $M_n(k).$ To prove that, we need the following simple yet important lemma.

Lemma 3 (Cayley). Every $n$ -dimensional $k$ -algebra with identity $R$ can be embedded into $M_n(k).$

Proof. Since $\dim_k R=n,$ the ring of matrices $M_n(k)$ is just $\text{End}_k(R),$ the ring of $k$ -linear maps $R \to R.$ Define the map $f: R \to \text{End}_k(R)$ by $f(r)(x)=rx$ for all $r,x \in R,$ and see that $f$ is an injective ring homomorphism. $\ \Box$

Corollary. Let $R$ be an $n$ -dimensional $k$ -algebra with identity, and let $S$ be a nil, multiplicatively closed subset of $R.$ Then $S^n=(0).$

Proof. By Lemma 3, $S$ is a subset of $M_n(k)$ and so, by the Theorem, $S^n=(0). \ \Box$

Note. The reference for this post is Chapter 1, Section 3, of the book Polynomial Identities in Ring Theory by Louis Rowen.

Exercise. Try to fully understand the proof of Lemma 3.

Finite sum of reciprocals of squares of integers modulo primes of the form 6k + 1

Posted: January 17, 2024 in Basic Number Theory

Let’s begin this post with an extension of the concept of divisibility of integers by a prime number to divisibility of positive rational numbers by a prime number.

Definition. Let $p$ be a prime number, and $r=\frac{a}{b},$ where $a,b$ are positive integers and $p \nmid b.$ Then we say that $p \mid r$ or, equivalently, $r \equiv 0 \mod p$ if $p \mid a.$

The following problem was posted on the Art of Problem Solving website a couple of years ago and remained unsolved until I saw it a couple of months ago! haha …

Problem. Let $k$ be a positive integer such that $p=6k+1$ is a prime number. Show that

$\displaystyle \sum_{i=1}^{2k}\frac{1}{i^2}\equiv 0 \mod p \iff \sum_{i=1}^{k}\frac{1}{i^2} \equiv 0 \mod p.$

Solution. I wrote the problem here as it was posted on the AoPS website but I think a much better problem is to prove that

$\displaystyle 5\sum_{i=1}^{2k}\frac{1}{i^2}-\sum_{i=1}^k\frac{1}{i^2} \equiv 0 \mod p, \ \ \ \ \ \ \ \ \ \ (*)$

which obviously solves the problem posted on AoPS too.

Proof of $(*).$ For a positive integer $i < p,$ we’ll write $i^{-1}$ for the inverse of $i$ modulo $p.$ Also, for simplicity, we’ll write $\equiv$ instead of $\equiv \mod p.$ It is clear that $\frac{1}{i} \equiv i^{-1}.$ Now

$\displaystyle \sum_{i=1}^{p-1}\frac{1}{i^2} \equiv \sum_{i=1}^{p-1}(i^{-1})^2=\sum_{i=1}^{p-1}i^2=\frac{p(p-1)(2p-1)}{6}=k(2p-1)p \equiv 0$

and so

$\displaystyle 2\sum_{i=1}^{3k}\frac{1}{i^2}=2\sum_{i=1}^{\frac{p-1}{2}}\frac{1}{i^2} \equiv \sum_{i=1}^{p-1}\frac{1}{i^2} \equiv 0,$

which gives

$\displaystyle \sum_{i=1}^{3k}\frac{1}{i^2} \equiv 0. \ \ \ \ \ \ \ \ \ \ \ (1)$

On the other hand,

$\displaystyle \sum_{i=1}^{3k}\frac{1}{i^2}=\sum_{i=1}^{2k}\frac{1}{i^2}+\sum_{i=2k+1}^{3k}\frac{1}{i^2}=\sum_{i=1}^{2k}\frac{1}{i^2}+\sum_{i=1}^{k}\frac{1}{(3k+1-i)^2} \equiv \sum_{i=1}^{2k}\frac{1}{i^2} + \sum_{i=1}^{k}\frac{1}{(2^{-1}(2i-1))^2}$

$\displaystyle \equiv \sum_{i=1}^{2k}\frac{1}{i^2} + 4\sum_{i=1}^{k}\frac{1}{(2i-1)^2} =\sum_{i=1}^{2k}\frac{1}{i^2} + 4\left(\sum_{i=1}^{2k}\frac{1}{i^2}-\sum_{i=1}^k\frac{1}{(2i)^2}\right)$

$\displaystyle =5\sum_{i=1}^{2k}\frac{1}{i^2}-\sum_{i=1}^k\frac{1}{i^2}. \ \ \ \ \ \ \ \ \ \ \ (2)$

The equivalence $(*)$ now follows from $(1),(2). \ \Box$

Rings satisfying (xy)^n=x^ny^n, for all n, may not be commutative

Posted: January 14, 2024 in Basic Algebra, Rings

In this post, we took a look at groups $G$ satisfying $(xy)^n=x^ny^n$ for some integer $n$ and all $x,y \in G.$ We showed that if $n=2,$ then $G$ is abelian, but for $n > 2,$ the group $G$ may or may not be abelian. We can now ask the same question for rings. If $R$ is a commutative ring, then obviously $(xy)^n=x^ny^n$ for all $x,y \in R$ and all positive integers $n.$

Question. Are there noncommutative rings $R$ with this property that $(xy)^n=x^ny^n$ for all $x,y \in R$ and all positive integers $n$ ?

Answer. No if $R$ has $1,$ the multiplicative identity, and yes if $R$ doesn’t have $1.$

Proof. Suppose first that $R$ has $1,$ and $(xy)^2=x^2y^2$ for all $x,y \in R.$ Then $((x+1)y)^2=(x+1)^2y^2$ gives

$(xy)^2+y^2+xy^2+yxy=x^2y^2+2xy^2+y^2$

and hence

$xy^2=yxy, \ \ \ \ \ \ \ \ \ (*)$

for all $x,y \in R.$ Now, in $(*),$ change $y$ to $y+1$ to get

$xy^2+2xy+x=yxy+yx+xy+x,$

which together with $(*),$ gives $xy=yx$ and so $R$ is commutative.

Now, we give an example of a noncommutative ring $R$ (without $1$ ) with this property that $(xy)^n=x^ny^n$ for all $x,y \in R$ and all positive integers $n.$ Let $R$ be the subring of $M_2(\mathbb{C}),$ the ring of $2 \times 2$ matrices with complex entries, defined as follows

$R:=\left \{\begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}: \ \ a,b \in \mathbb{C}\right\}.$

See that $R$ is a noncommutative ring. Now, let $n$ be a positive integer and $x,y \in R.$ So

$x=\begin{pmatrix} a & 0 \\ b & 0 \end{pmatrix}, \ \ \ \ \ \ y=\begin{pmatrix} c & 0 \\ d & 0 \end{pmatrix},$

for some $a,b,c,d \in \mathbb{C}.$ Then

$xy=\begin{pmatrix} ac & 0 \\ bc & 0 \end{pmatrix}.$

An easy induction over $n$ shows that

$x^n=\begin{pmatrix} a^n & 0 \\ a^{n-1}b & 0 \end{pmatrix}$

and so

$(xy)^n=x^ny^n=\begin{pmatrix} a^nc^n & 0 \\ a^{n-1}bc^n & 0 \end{pmatrix}. \ \Box$

Integro-differential rings

Posted: December 30, 2023 in Derivation, Noncommutative Ring Theory Notes
Tags: Derivation, differential ring, integration, integro-differential ring

Throughout, $R$ is a ring with identity.

Definition 1. We say that $(R,\delta)$ is a differential ring if $\delta$ is a derivation of $R.$

If $(R,\delta)$ is a differential ring, then $C:=\ker \delta$ is a subring of $R$ called the ring of constants of $\delta.$ Also, $\delta$ is $C$ -linear.

Definition 2. Let $(R,\delta)$ be a differential ring, and let $C:=\ker \delta.$ An integration on a differential ring $(R,\delta)$ is a $C$ -linear map $\int : R \to R$ such that $\delta \int (a)=a$ for all $a \in R.$ We then say that $(R,\delta, \int)$ is an integro-differential ring.

When exactly does a differential ring have an integration? The following Theorem answers this question.

Theorem. Let $(R,\delta)$ be a differential ring, and let $C$ be the ring of constants of $\delta.$ Then $(R,\delta)$ has an integration if and only if $\delta$ is surjective and there exists a $C$ -module $D$ such that $R=C \oplus D,$ as $C$ -modules.

Proof. Suppose first that $(R,\delta,\int)$ is an integro-differential ring. Then clearly $\delta$ is surjective because, by definition, $a= \delta(\int (a))$ for all $a \in R.$ Now let $D:=\int (R)=\{\int(a): \ a \in R\}.$ Since, by definition, $\int$ is $C$ -linear, $D$ is a $C$ -module. Also, for all $a \in R,$

$\delta(a-\int \delta (a))=\delta(a)-\delta \int \delta (a)=\delta(a)-\delta(a)=0$

and hence $a - \int \delta(a) \in C,$ which then gives $R=C+D,$ because $\int \delta (a) \in D.$ To show that the sum $C+D$ is direct, we just need to show that $C \cap D=(0).$ Well, if $c \in C \cap D,$ then $c=\int (a)$ for some $a \in R$ and hence $0=\delta(c)=\delta \int (a)=a,$ which gives $c=\int (0)=0.$

Suppose now that $\delta$ is surjective and $R=C \oplus D,$ for some $C$ -module $D.$ Let $a \in R.$ Then $\delta(b)=a$ for some $b \in R.$ We can also write $b=c+d,$ for some $c \in C, \ d \in D,$ which then gives $a=\delta(b)=\delta(d).$ So there exists $d \in D$ such that $\delta(d)=a.$ This $d$ is unique because if $\delta(d')=a$ for some other $d' \in D,$ then $\delta(d-d')=\delta(d)-\delta(d')=a-a=0,$ and so $d-d' \in C \cap D=(0).$ So we have shown that for every $a \in R,$ there exists a unique $d_a \in D$ such that $\delta(d_a)=a.$ Define the map $\int : R \to R$ by $\int(a)=d_a.$ Then $\delta \int (a)=\delta(d_a)=a.$ So in order to complete the proof, we just need to show that $\int$ is $C$ -linear. Let $a,b \in R, c \in C.$ Then, since $\delta$ is $C$ -linear, we have

$\delta(cd_a+d_b)=c\delta(d_a)+\delta(d_b)=ca+b,$

and so $\int (ca+b) =d_{ca+b}=cd_a+d_b=c\int(a)+\int(b). \ \Box$

Note. The reference for this post is the first few pages of this paper. Happy New Year and my 500th post!

Ideals of a polynomial ring R[x] which are of the form I[x] for some ideal I of R

Posted: December 26, 2023 in Basic Algebra, Rings
Tags: finitely generated, maximal ideals, nilradical, polynomial ring, prime ideal, radical of ideals

Let $R$ be a commutative ring with identity, and let $I$ be an ideal of $R.$ Let $R[x]$ be the ring of polynomials over $R,$ and let $I[x]$ be the set of all polynomials in $R[x]$ with all coefficients in $I.$ So an element of $I[x]$ has the form $r_0+r_1x + \cdots + r_nx^n, \ n \ge 0,$ where $r_i \in I$ for all $i.$ It is quite easy to see that $I[x]$ is an ideal of $R[x].$ Now, you may ask: how are algebraic properties of $I$ and $I[x]$ related? In this post, we look into some of those properties and relations.

Note. Regarding parts v), vi) of the following Problem, for the definitions of radical of an ideal and the nilradical of a commutative ring, see this post.

Problem. Let $R$ be a commutative ring with identity, and let $I$ be an ideal of $R.$ Let $R[x]$ be the ring of polynomials over $R.$ Show that

i) $I$ is a finitely generated ideal of $R$ if and only if $I[x]$ is a finitely generated ideal of $R[x],$

ii) $R[x]/I[x] \cong (R/I)[x],$

iii) $I$ is a prime ideal of $R$ if and only if $I[x]$ is a prime ideal of $R[x],$

iv) $I[x]$ is never a maximal ideal of $R[x],$

v) $\sqrt{I[x]}=\sqrt{I}[x],$

vi) if $I$ is the nilradical of $R,$ then $I[x]$ is the nilradical of $R[x].$

Solution. i) If $I$ is generated by $r_1, \cdots , r_n,$ i.e. $I=\sum_{i=1}^nRr_i,$ then it’s clear that $I[x]$ is also generated by $r_1, \cdots , r_n,$ i.e. $I[x]=\sum_{i=1}^nR[x]r_i.$ Conversely, if $I[x]$ is generated by $p_1(x), \cdots , p_n(x),$ then $I$ is generated by $r_1, \cdots , r_n,$ where $r_i$ is the constant term of $p_i(x).$

ii) Define the map $f: A[x] \to (A/I)[x]$ by

$f(r_nx^n+ \cdots +r_1x+r_0)=(r_n+I)x^n+ \cdots + (r_1+I)x+r_0+I, \ \ \ r_i \in R, \ n \ge 0.$

See that $f$ is an onto ring homomorphism and $\ker f=I[x].$

iii) $I$ is a prime ideal of $R$ if and only if $R/I$ is a domain if and only if $(R/I)[x]$ is a domain if and only if $R[x]/I[x]$ is a domain, by ii), if and only if $I[x]$ is a prime ideal of $R[x].$

iv) Suppose $I[x]$ is a maximal ideal of $R[x]$ for some ideal $I$ of $R.$ Then $R[x]/I[x]$ is a field and so, by ii), $(R/I)[x]$ is a field, which is nonsense because a polynomial ring can never be a field (because, for example, $x$ has no inverse in there). Here’s an easier proof. If $I=R,$ then $I[x]=R[x],$ which is not maximal. If $I \ne R,$ then $I[x]+\langle x \rangle$ is a proper ideal of $R[x]$ which contains $I[x]$ properly, because $x \notin I[x].$

v) For any $p(x)=\sum_{i=0}^nr_ix^i \in R[x],$ let $\overline{p(x)}:=\sum_{i=1}^n(r_i+I)x^i \in (R/I)[x].$ By ii), the function $\tilde{f}: R[x]/I[x] \to (R/I)[x]$ defined by $\tilde{f}(p(x)+I[x])=\overline{p(x)}$ is a ring isomorphism. So $p(x) \in \sqrt{I}$ if and only if $(p(x))^m \in I[x]$ for some positive integer $m$ if and only if $p(x)+I[x]$ is nilpotent in $R[x]/I[x]$ if and only if $\overline{p(x)}$ is nilpotent in $(R/I)[x]$ if and only if $r_i+I$ is nilpotent in $R/I$ for all $i,$ by the third part of the Problem in this post, if and only if $r_i^{k_i} \in I$ for some positive integers $k_i$ if and only if $r_i \in \sqrt{I}.$

vi) In v), choose $I=(0). \ \Box$

Exercise. By the second part of the above Problem, $R[x]/I[x]$ is a polynomial ring. Suppose now that $R$ is a commutative ring with identity, and $J$ is an ideal of the polynomial ring $R[x].$ Suppose also that $R[x]/J$ is a polynomial ring. Does that imply $J=I[x],$ for some ideal $I$ of $R$ ?
Hint. No. For example, choose $R$ to be the polynomial ring $\mathbb{R}[t]$ and $J$ to be the ideal $\langle x\rangle.$

Euclidean domains with a special Euclidean function

Posted: December 22, 2023 in Basic Algebra, Rings
Tags: Euclidean domains, Euclidean functions, Fields, group of units, polynomial ring

In this post we are going to show that if $R$ is a Euclidean domain with a Euclidean function $\phi$ that satisfies the inequality $\phi(a+b) \le \max \{\phi(a),\phi(b)\},$ whenever $a,b,a+b \in R \setminus \{0\},$ then $R$ is either a field or a polynomial ring over a field. This result was posted as a problem on the Art of Problem Solving website recently; you can see the problem and my solution here. I’m now going to give a slightly improved version of that solution here.

Let’s begin with the definition of Euclidean domains.

Definition. A Euclidean domain is a commutative domain with identity $R$ such that there exists a function $\phi: R \setminus \{0\} \to \mathbb{Z}_{\ge 0}$ with the following properties.

i) If $a,b \in R, \ b \ne 0,$ then $a=sb+r$ for some $r,s \in R$ with either $r=0$ or $\phi(r) < \phi(b).$

ii) if $a,b \in R \setminus \{0\},$ then $\phi(a) \le \phi(ab).$

The function $\phi$ is called a Euclidean function for $R.$

Not all ring theorists are happy with having the condition ii) in the definition of a Euclidean domain since it can be shown that i) implies the existence of a function $R \setminus \{0\} \to \mathbb{Z}_{\ge 0}$ that satisfies both i), ii). But in this post, we are going to stay away from that argument and keep ii).

There are many examples of Euclidean domains but, for a reason you’re going to see in a moment, we are only interested in two of them in this post.

Example 1. Any field $F$ is a Euclidean domain. Just choose any nonnegative integer $n$ and define $\phi(a)=n,$ for all $0 \ne a \in F.$ Then the condition ii) clearly holds and if $0 \ne b \in R,$ then $a=(ab^{-1})b+0,$ so the condition i) is also satisfied.

Example 2. The ring of polynomials $F[x]$ over a field $F$ is a Euclidean domain. Here we define the Euclidean function by $\phi(p(x))=\deg p(x),$ for all $0 \ne p(x) \in F[x].$ Then the condition ii) clearly holds because

$\phi(p(x)q(x))=\deg (p(x)q(x))=\deg p(x)+\deg q(x) \ge \deg p(x)=\phi(p(x)).$

The condition i) is just the familiar process of dividing two polynomials.

Remark. Both Euclidean functions in Examples 1,2 satisfy the inequality $\phi(a+b) \le \max \{\phi(a),\phi(b)\},$ whenever $a,b,a+b \in R \setminus \{0\}.$ Interestingly, there are no other Euclidean domains whose Euclidean functions satisfy the said inequality. The goal in this post, as mentioned in the first paragraph of this post, is to prove that result.

Notation. From now on, $R$ is a Euclidean domain with a Euclidean function $\phi,$ and $U(R)$ is the multiplicative group of units of $R.$

Next is to find $U(R).$ This is a well-known result about Euclidean domains but since it is important and also for the sake of completeness, I’m going to prove it here.

Lemma. Let $a,b \in R \setminus \{0\}.$ Then

1) $\phi(1) \le \phi(a),$ i.e. $\phi(1)$ is the minimum value of $\phi,$

2) $\phi(a)=\phi(ab)$ if and only if $b$ is a unit. In particular,

$U(R)=\{a \in R \setminus \{0\}: \ \phi(a)=\phi(1)\}.$

Proof. 1) $\phi(a)=\phi(1 \cdot a) \ge \phi(1).$

2) If $b$ is a unit, then $\phi(ab) \ge \phi(a)=\phi(abb^{-1}) \ge \phi(ab)$ and so $\phi(a)=\phi(ab).$ Suppose now that $\phi(a)=\phi(ab).$ If $Ra=Rab,$ then $b$ is a unit. Otherwise, there exists $c \in Ra \setminus Rab$ and so $c=sab+r$ for some $r,s \in R$ with either $r=0$ or $\phi(r) < \phi(ab).$ If $r=0,$ then $c \in Rab,$ which is not true. Therefore $\phi(r) < \phi(ab)=\phi(a),$ which is not possible because $r=c-sab \in Ra$ and so $r=da$ for some $d \in R,$ which then gives the contradiction $\phi(r)=\phi(da) \ge \phi(a). \ \Box$

We are now ready to prove the result mentioned in the above Remark.

Theorem. Let $F:=U(R) \cup \{0\},$ and suppose that

$\phi(a+b) \le \max\{\phi(a),\phi(b)\},$

whenever $a,b,a+b \in R \setminus \{0\}.$ Then

1) $\phi(a+b)=\phi(a),$ for all $a,b \in R$ with $a \ne 0, \ a+b \ne 0, \ b \in F,$

2) $F$ is a subfield of $R,$

3) either $R = F$ or $R$ is a polynomial ring over $F.$

Proof. 1) That is clear if $b=0.$ Suppose now that $b \in U(R).$ Then by the Lemma

$\phi(a+b) \le \max \{\phi(a),\phi(b)\}=\max \{\phi(a), \phi(1)\}=\phi(a).$

Similarly,

$\phi(a)=\phi(a+b-b) \le \max \{\phi(a+b),\phi(-b)\}=\max \{\phi(a+b),\phi(1)\}=\phi(a+b).$

2) The only thing that we need to prove is that if $a,b \in F,$ then $a+b \in F.$ Well, that’s clear if $a=0$ or $a+b=0.$ If $a, a+b$ are nonzero, then, by 1), $\phi(a+b)=\phi(a)=\phi(1),$ and so $a+b \in F.$

3) Choose $x \in R \setminus F$ such that $\phi(x)$ is a minimal element of $\{\phi(a): \ a \in R \setminus F\}.$ Notice that $x$ is not algebraic over the field $F$ because then either $x=0$ or $x$ would be a unit, contradicting $x \notin F.$ So we are done if we show that $R=F[x].$ Let $a \in R.$ We prove by induction over $\phi(a)$ that $a \in F[x].$ By the first part of the Lemma, $\phi(1)$ is the minimum value of $\phi.$ If $\phi(a)=\phi(1),$ then $a \in F \subset F[x],$ and we are done. Suppose now that $a \notin F.$ We have $a=sx+r$ for some $r,s \in R$ with either $r=0$ or $\phi(r) < \phi(x).$ If $\phi(r) < \phi(x),$ then $r \in F,$ by the minimality condition on $\phi(x).$ So either way, $r \in F.$ So if $s \in F,$ then $a \in F[x]$ and we are done. Thus we may assume that $s \notin F.$ Then, by the Lemma and the first part of the Theorem, $\phi(s) < \phi(sx)=\phi(a)$ and so, by induction, $s \in F[x].$ Therefore $a=sx+r \in F[x]$ and that completes the proof of the Theorem. Merry Christmas! $\ \Box$