2 On base fields

We begin with the following remark.

From Remark 2.1(iii), we continue with the case where the algebra is simply the field itself, that is, $\mathcal {A} = F$ . A quick counterexample is the polynomial $f(x) = x^2$ . Over any field F of characteristic different from $2$ , we observe that $f(1) = 1 = f(-1)$ , even though $1 \ne -1$ . Hence, the evaluation map $\mathrm {eva}_{f, F}$ fails to be injective in this case. When the field F is algebraically closed, the story sharpens and a stronger statement holds, as follows.

Proof. By Remark 2.1, it suffices to show that if the evaluation map $ \mathrm {eva}_{f, F} $ is injective, then f must have degree $1$ in one variable.

We begin with the base case where f is a polynomial in one variable. Observe that any polynomial f with degree greater than $1$ can be factored into linear factors (possibly repeated). If it has a single repeated factor, say $f(x) = a(x - b)^n$ , where $a,b\in F$ , $a \ne 0$ and $n\geq 2$ , then distinct inputs symmetrically placed around b will map to the same nonzero value, that is, it is easy to find two distinct values of x that map to the same nonzero value under f. If it splits into several distinct roots, say

$$ \begin{align*}f(x)=a(x-b_1)^{m_1}(x-b_2)^{m_2}\cdots(x-b_k)^{m_k},\end{align*} $$

where $a \ne 0$ , $m_i \ge 1$ , $b_i \ne b_j$ for $i \ne j$ , and $\sum _{i=1}^k m_i = n$ , then already $f(b_i) = f(b_j) = 0$ for $i \ne j$ , even though $b_i \ne b_j$ . In either scenario, we always find two different elements sent to the same result, so injectivity fails unless f has degree $1$ .

Lastly, we consider the case in which f is a polynomial in at least two variables. By Remark 2.1, if the evaluation map $ \mathrm {eva}_{f,F} $ is injective, then $ f $ must be a nonconstant polynomial. This implies that there exists an index $ i \in \{1, 2, \ldots , m\} $ such that the degree of $ x_i $ in $ f $ is at least 1. We can write $ f = \sum _{j=0}^k g_j x_i^j $ as a polynomial in $ x_i $ , keeping the other variables $ x_1, x_2, \ldots , x_{i-1}, x_{i+1}, \ldots , x_m $ fixed, where $ g_j \in F[x_1, x_2, \ldots , x_{i-1}, x_{i+1}, \ldots , x_m] $ and $ k $ is the degree of $ f $ with respect to $ x_i $ . Since $ \mathrm {eva}_{f,F} $ is injective, the polynomial $ f $ is still nonconstant in $ x_i $ . Because $ F $ is algebraically closed, we know that $ f $ , as a polynomial in $ x_i $ , must have a root in $ F $ . Since the other variables $ x_1, x_2, \ldots , x_{i-1}, x_{i+1}, \ldots , x_m $ are fixed, it follows that $ f $ has infinitely many distinct roots in $ F^m $ . Therefore, the evaluation map $ \mathrm {eva}_{f,F} $ cannot be injective, which leads to a contradiction, as promised.

Proof. First, let us consider the case where $ F $ is a finite field. In this case, $ F^m $ is a finite set. If the evaluation map $ \mathrm {eva}_{f,F} $ were injective, the number of elements in $ F^m $ would have to be less than or equal to the number of elements in $ F $ , which implies $ m = 1 $ . This leads to a contradiction with the assumption that $m\geq 2$ , so there is no polynomial whose evaluation map is injective in this case.

We now turn to the case where $ F = \mathbb {R} $ . Assume, for contradiction, that the evaluation map $\mathrm {eva}_{f,\mathbb {R}} : \mathbb {R}^m \to \mathbb {R}$ is injective. Restrict it to the unit sphere $S^{m-1} \subseteq \mathbb {R}^m$ and denote

$$ \begin{align*} g := \mathrm{eva}_{f,\mathbb{R}}|_{S^{m-1}} : S^{m-1} \to \mathbb{R}. \end{align*} $$

Since f is a polynomial, it follows that g is continuous and since $\mathrm {eva}_{f,\mathbb {R}}$ is injective, so is g. The domain $S^{m-1}$ is compact and connected; hence, the image $K := g(S^{m-1})$ is also compact and connected in $\mathbb {R}$ . Thus, K must be either a single point or a closed interval $[a,b]$ .

If K is a point, then g is constant, which contradicts injectivity (see Remark 2.1). If $K=[a,b]$ , then $g : S^{m-1} \to [a,b]$ is a continuous bijection from a compact space to a Hausdorff space; hence, a homeomorphism. Pick $y \in (a,b)$ and let $x = g^{-1}(y)$ . Then,

$$ \begin{align*} S^{m-1}\setminus \{x\} \cong [a,b]\setminus \{y\}. \end{align*} $$

However, for $m\geq 2$ , the set $S^{m-1}\setminus \{x\}$ is connected (indeed, homeomorphic to $\mathbb {R}^{m-1}$ ), whereas $[a,b]\setminus \{y\}$ is disconnected. This contradiction shows that no such polynomial can exist.

3 Algebras containing idempotent or nilpotent elements

Theorem 2.2 leads to the following theorem, which addresses the injectivity of the evaluation map on algebras over algebraically closed fields.

Proof. We begin with two observations.

If $ \mathcal {A} $ is unital and the evaluation map $ \mathrm {eva}_{f,\mathcal {A}} $ is injective, then the evaluation map $ \mathrm {eva}_{f,F} $ is also injective. Indeed, let $ \alpha = (\alpha _1, \ldots , \alpha _m) $ and $ \beta = (\beta _1, \ldots , \beta _m) $ be elements in $ F^m $ such that $ f(\alpha ) = f(\beta ) $ . Evaluate $ f $ at the corresponding scalars to obtain

$$ \begin{align*} f(\alpha_1 \mathbf{1}_{\mathcal{A}}, \ldots, \alpha_m \mathbf{1}_{\mathcal{A}}) = f(\alpha) \mathbf{1}_{\mathcal{A}} = f(\beta) \mathbf{1}_{\mathcal{A}} = f(\beta_1 \mathbf{1}_{\mathcal{A}}, \ldots, \beta_m \mathbf{1}_{\mathcal{A}}). \end{align*} $$

Since we assumed the evaluation map $ \mathrm {eva}_{f,\mathcal {A}} $ is injective, we must have

$$ \begin{align*} (\alpha_1 \mathbf{1}_{\mathcal{A}}, \ldots, \alpha_m \mathbf{1}_{\mathcal{A}}) = (\beta_1 \mathbf{1}_{\mathcal{A}}, \ldots, \beta_m \mathbf{1}_{\mathcal{A}}), \end{align*} $$

which implies $ \alpha _i = \beta _i $ for all $ i $ . Thus, $ \alpha = \beta $ , showing that $ \mathrm {eva}_{f,F} $ is injective.

If $ \mathcal {A} $ is nonunital and $ \mathcal {A} $ contains a nonzero idempotent element $ e $ , then the corner subalgebra $ e \mathcal {A} e = \{ eae \mid a \in \mathcal {A} \} $ forms a unital subalgebra with identity element  $ e $ . If $ \mathrm {eva}_{f,\mathcal {A}} $ is injective, then the restriction $ \mathrm {eva}_{f,e\mathcal {A}e} $ is also injective. Indeed, if $ {\alpha , \beta \in (e\mathcal {A}e)^m} $ satisfy $ f(\alpha ) = f(\beta ) $ , then injectivity of $ \mathrm {eva}_{f,\mathcal {A}} $ implies $ \alpha = \beta $ . This means that the restriction $ \mathrm {eva}_{f,e\mathcal {A}e} $ is also injective. However, $ e\mathcal {A}e $ is a unital algebra and the previous observation has already shown how injectivity behaves in that setting.

By combining these observations with Theorems 2.2, 2.3 and Remark 2.4, the proof is completed.

We now adopt a structured approach, drawing inspiration from the ideas found in [Reference Buşe, O’Regan and Saierli5, Lemma 4.2]. Our version, however, includes modifications adapted to the specific context of our study, which we outline in detail as follows.

As far as we can tell, the collection of unital associative algebras admitting a nonzero nilpotent of index two seems to be rather broad. Matrix algebras offer a concrete instance of this phenomenon. This leads us naturally to the focus of the next section.

4 Matrix algebras over fields

As shown in Proposition 2.2, when $ F $ is an algebraically closed field and $ f \in F[x] $ is nonconstant, a necessary and sufficient condition for the evaluation map $ \mathrm {eva}_{f,F} $ to be injective is that $ f $ has degree one. Since $ F $ can be naturally identified with $ \mathrm {M}_1(F) $ , this observation suggests that a similar criterion may hold for matrix algebras of higher dimension. With this in mind, we establish the following theorem as a step toward extending the result to arbitrary fields. We use the standard notation $ \mathrm {M}_n(F) $ to denote the algebra of $ n \times n $ matrices with entries in a field $ F $ .

We pose the following open problem, which arises from Theorem 4.1.

Remark 4.3. To illustrate Problem 4.2, consider the polynomial $f(x)=x^4+2x+c \in \mathbb {Q}[x]$ , where $\mathbb {Q}$ is the field of rational numbers. Let

$$ \begin{align*}A=\begin{pmatrix}0&\tfrac{1}{2}\\[2pt] 1&-1\end{pmatrix}\oplus 0_{n-2} \quad\text{and}\quad B=\begin{pmatrix}0&-\tfrac{3}{2}\\[2pt] 1&1\end{pmatrix}\oplus 0_{n-2},\end{align*} $$

where $\oplus $ denotes the block direct sum. A straightforward computation shows that $f(A)=\tfrac 14(3+4c)\mathrm {I}_2 \oplus c\mathrm {I}_{n-2} = f(B),$ while $A\neq B$ . Hence, the evaluation map $\mathrm {eva}_{f,\mathrm {M}_n(\mathbb {Q})}$ fails to be injective. This example is worth emphasising. Indeed, writing $f(x)-f(0)=x^4+2x = x(x^3+2),$ we see that the decomposition involves $m=1$ and $d=3$ , with $h(x)=x^3+2$ irreducible in $\mathbb {Q}[x]$ . Thus, the noninjectivity already appears in the case $n=2<d$ , exactly the range not covered by Theorem 4.1.

By the way, the evaluation map $\mathrm {eva}_{f,\mathbb {Q}}$ is injective. Indeed, suppose that there exist distinct rationals $x,y\in \mathbb {Q}$ with $f(x)=f(y)$ . Then,

$$ \begin{align*} f(x)-f(y)=(x-y)((x+y)(x^2+y^2)+2)=0, \end{align*} $$

so necessarily,

(4.1) $$ \begin{align} (x+y)(x^2+y^2)=-2. \end{align} $$

Write $x={a}/{u}$ and $y={b}/{u}$ with integers $a,b,u$ such that the greatest common divisor of $a,b,u$ is $1$ . Clearing denominators in (4.1) gives

(4.2) $$ \begin{align} (a+b)(a^2+b^2)=-2u^3. \end{align} $$

A parity check shows that a and b cannot be of opposite parity, for otherwise, $(a+b)(a^2+b^2)$ would be odd, which contradicts the right-hand side. Thus, both $a,b$ are either even or odd. Since the greatest common divisor of $a,b,u$ is $1$ , it follows that a and b must both be odd. In this case, $a+b \equiv 2 \pmod {4}$ and $a^2+b^2 \equiv 2 \pmod {4},$ so the $2$ -adic valuation $v_2(a+b)=v_2(a^2+b^2)=1$ . Hence, $v_2((a+b)(a^2+b^2))=2$ . However, from (4.2), $v_2(-2u^3)=1+3v_2(u),$ which can never equal $2$ . This contradiction shows that no such pair $(x,y)$ exists, as claimed.

From Remark 4.3, we can also formulate another open problem.

Following Theorem 4.1 again, when $ d \in \{1,2\} $ , this issue does not arise, since the condition $ n \geq d $ is always satisfied for $ n \geq 2 $ . A fundamental example is any algebraically closed field, where every nonconstant polynomial splits into linear factors; hence, all irreducible polynomials are of degree one. Using Theorem 4.1, we arrive at the following conclusion regarding matrix algebras over algebraically closed fields: if $ F $ is an algebraically closed field and $ f \in F[x] $ is a polynomial of degree greater than one, then for every integer $ n \geq 1 $ , the evaluation map $ \mathrm {eva}_{f,\mathrm {M}_n(F)} $ is not injective. Interestingly, this statement coincides exactly with a special case of Theorem 3.1, which can be derived directly from Proposition 2.2 without invoking Theorem 4.1.

Another classical example where all irreducible polynomials have degree at most two is the field of real numbers $ \mathbb {R} $ . More broadly, it is known that real closed fields are precisely those in which every irreducible polynomial over the field has degree either one or two. Therefore, for any real closed field, Theorem 4.1 applies fully and we may deduce the following corollary.

It is easy to see that the conclusion of Corollary 4.5 does not necessarily hold when $ n = 1 $ . For instance, the polynomial $ f = x^3 $ serves as a concrete counterexample: the corresponding evaluation map $ \mathrm {eva}_{f,F} $ is injective, even though $ f $ has degree 3.

Drawing upon Theorems 3.1 and 4.5, we are led naturally to the following corollary.

Returning to Theorem 4.1, we note that when $ n < d $ , difficulties arise due to the limitation on matrix size. However, this issue can be circumvented by working within the framework of the finitary matrix algebra. For clarity, we briefly recall its construction. Let $ F $ be a field. For each positive integer $ n $ , there is a natural embedding of $ \mathrm {M}_n(F) $ into $ \mathrm {M}_{n+1}(F) $ that sends a matrix $ A = (a_{i,j}) $ to the matrix $ A' $ , obtained by appending a row and a column of zeros, that is, $ A' = A \oplus 0 $ where $0\in \mathrm {M}_1(F)$ . This yields an ascending chain of matrix algebras:

$$ \begin{align*} \mathrm{M}_1(F) \subseteq \mathrm{M}_2(F) \subseteq \cdots \subseteq \mathrm{M}_n(F) \subseteq \cdots. \end{align*} $$

The union of this chain, denoted $ \mathrm {M}_\infty (F) $ , is called the finitary matrix algebra over $ F $ :

$$ \begin{align*} \mathrm{M}_\infty(F) = \bigcup_{n \geq 1} \mathrm{M}_n(F). \end{align*} $$

Each element of $ \mathrm {M}_\infty (F) $ is a countably infinite matrix with only finitely many nonzero entries. In particular, every such matrix can be written in the form $ A \oplus 0 $ for some $ A \in \mathrm {M}_n(F) $ , with $ n \geq 1 $ , and where $ 0 $ denotes the infinite zero matrix. Without loss of generality, we may always assume $ n \geq 2 $ . With this setup in place, we now establish the following result concerning finitary matrix algebras. Note that $\mathrm {M}_\infty (F)$ is nonunital, so we restrict our attention to polynomials whose constant term is zero. Note that this result follows immediately from Theorem 3.1 when the field $ F $ is algebraically closed and, more generally, it can be derived from Theorem 4.1.

In what follows, we continue our exploration of injective maps between algebras, particularly those that are isomorphic to matrix algebras. Before proceeding, we require the following technical lemma. While it may seem straightforward, we include it here for clarity and to highlight its specific role in the proof.

In light of Lemma 4.8, our results at this stage may apply to any algebra $ \mathcal {A} $ that is isomorphic to $ \mathrm {M}_n(\mathcal {A}) $ for some integer $ n \geq 2 $ (or weaker forms such as direct products of matrix algebras). This condition defines a remarkably broad class of algebras. For example, the Wedderburn–Artin theorem tells us that any finite-dimensional semisimple algebra over a field is isomorphic to a finite direct product of matrix algebras over division rings. Additional examples arise from functional analysis: consider the algebra $ B(X) $ of all bounded linear operators on a Banach space $ X $ . If $ X \cong X \oplus X $ , then $ B(X) \cong \mathrm {M}_2(B(X)) $ . In particular, infinite-dimensional Hilbert spaces $ H $ satisfy $ H \cong H \oplus H $ , so this property holds for most Banach spaces. In fact, the first known infinite-dimensional Banach space not isomorphic to its square was only constructed in 1960 [Reference Bessaga and Pełczyński4]. A similar phenomenon occurs in algebra: if $ X $ is a free module of infinite rank over a ring $ R $ , then its endomorphism ring $ \mathrm {End}_R(X) $ is isomorphic to $ \mathrm {M}_2(\mathrm {End}_R(X)) $ , since $ X \cong X \oplus X $ in this setting as well.

We conclude this section with a result concerning functions of several variables that are naturally associated with polynomials in one variable over a field $ F $ . The motivation arises from observing that $ \mathrm {M}_n(F) $ , the algebra of $ n \times n $ matrices over $ F $ , is isomorphic to the $ F $ -vector space $ F^{n^2} $ via the linear map

$$ \begin{align*} A = (a_{i,j}) \mapsto \varphi(A) = (a_{1,1}, a_{1,2}, \ldots, a_{1,n}, a_{2,1}, \ldots, a_{2,n}, \ldots, a_{n,1}, \ldots, a_{n,n}). \end{align*} $$

However, it is important to emphasise that this map is not a ring homomorphism, since a ring homomorphism from $ \mathrm {M}_n(F) $ to F would send all nilpotent matrices to zero, which is clearly not the case here. Thus, the assumption that $ \varphi $ is an algebra homomorphism in Lemma 4.8 plays a crucial role.

Now, let $ f \in F[x] $ be a polynomial and let $ n\geq 2 $ be an integer. We proceed by recalling the earlier remark that $ \mathrm {M}_n(F) \cong F^{n^2} $ as vector spaces. Define a map

$$ \begin{align*} \overline{f} : F^{n^2} \xrightarrow{\theta} \mathrm{M}_n(F) \xrightarrow{\mathrm{eva}_{f, \mathrm{M}_n(F)}} \mathrm{M}_n(F) \xrightarrow{\varphi} F^{n^2}, \end{align*} $$

where the map $ \theta $ is defined by arranging the entries of $ F^{n^2} $ into an $ n \times n $ matrix: given $ (a_1, \ldots , a_{n^2}) \in F^{n^2} $ , we set $ \theta (a_1, \ldots , a_{n^2}) = (b_{i,j}) $ , where $ b_{i,j} = a_{(i-1)n + j} $ . It is straightforward to check that $ \theta $ is bijective and in fact the inverse of $ \varphi $ , that is, $ \theta = \varphi ^{-1} $ .

It follows that if $ \overline {f} $ is injective, then so is the evaluation map $ \mathrm {eva}_{f, \mathrm {M}_n(F)} $ , due to the bijectivity of both $ \theta $ and $ \varphi $ . Now, invoking Theorem 4.1, we see that if $ n $ is sufficiently large, then the injectivity of $ \mathrm {eva}_{f, \mathrm {M}_n(F)} $ implies that $ f $ must be linear. Similarly, even when $ n $ is not large, the same conclusion holds under the assumption that $ F $ is algebraically closed or real closed, as established in Proposition 2.2 and Corollary 4.5.

This discussion leads us to the following theorem.

As is well known, every complex polynomial is in fact an entire function on the complex plane. This naturally leads us to ask what can be said about the injectivity of entire functions. Recall that in complex analysis, an entire function in one variable is a complex-valued function that is holomorphic on all of $\mathbb {C}$ . Moreover, any entire function $f:\mathbb {C}\to \mathbb {C}$ admits a power series expansion $ f(z) = \sum _{i=0}^{\infty } a_i z^i,$ which converges everywhere on $\mathbb {C}$ .

It is a classical fact that every injective entire function must necessarily be of the form $f(z) = az + b$ with $a \neq 0$ and $a, b \in \mathbb {C}$ (see [Reference Remmert11, Ch. 10, Section 2, Theorem, page 311] for details). Since the radius of convergence of f is infinite, its defining power series $ f(z) = \sum _{i=0}^\infty a_i z^i$ remains valid when substituting a matrix A over $\mathbb {C}$ , leading to the natural definition $ f(A) = \sum _{i=0}^\infty a_i A^i,$ as discussed in [Reference Frommer and Simoncini7, Theorem 1]. Following a similar line of reasoning, one can extend this framework to the setting of unital associative Banach algebras. More precisely, if $\mathcal {A}$ is such an algebra and $f:\mathbb {C}\to \mathbb {C}$ is entire, then the series $ \sum _{i=0}^\infty a_i \alpha ^i$ converges for every element $\alpha \in \mathcal {A}$ . Indeed, the root test shows absolute convergence, and completeness of $\mathcal {A}$ ensures that the limit lies in $\mathcal {A}$ .

With this in hand, we arrive at the following conclusion: if $\mathcal {A}$ is a unital associative Banach algebra over $\mathbb {C}$ and the evaluation map $\operatorname {eva}_{f,\mathcal {A}}$ associated with an entire function f is injective, then necessarily f must be affine linear, that is, of the form $f(z) = az + b$ with $a \neq 0$ .

In parallel with the complex field $ \mathbb {C} $ , attention has also been given to the study of entire functions over $ \mathbb {C}_p $ , the field of $ p $ -adic complex numbers. Both $ \mathbb {C} $ and $ \mathbb {C}_p $ are complete, algebraically closed fields of characteristic zero, each endowed with a nontrivial absolute value which is Archimedean in the case of $ \mathbb {C} $ and non-Archimedean in the case of $ \mathbb {C}_p $ . According to the classical result of Ostrowski [Reference Grillet8, Theorem 5.5], up to isomorphism preserving absolute values, $ \mathbb {R} $ and $ \mathbb {C} $ are the only complete fields with respect to an Archimedean absolute value. Consequently, $ \mathbb {C} $ is, up to isomorphism, the only complete algebraically closed field with a nontrivial Archimedean absolute value. In contrast, the landscape in the non-Archimedean case is significantly richer and more intricate (for a detailed discussion, see [Reference Barría Comicheo, Shamseddine, Escassut, Perez-Garcia and Shamseddine3, Section 9]). Hence, we anticipate that this work may also be of interest to researchers in non-Archimedean analysis, as we believe that our results concerning entire functions continue to hold in the non-Archimedean setting.