$\DeclareMathOperator{\im}{im}$

1.3 Algebra Homomorphisms

As with any algebraic structure (like vector spaces, groups, rings) one needs to define and study maps between algebras which ‘preserve the structure’.

Definition 1.22.

Let $A$ and $B$ be $K$-algebras. A map $ϕ:A→B$ is a K-algebra homomorphism (or homomorphism of $K$-algebras) if
  1. (i)

    $ϕ$ is a $K$-linear map of vector spaces,

     
  2. (ii)

    $ϕ(ab)=ϕ(a)ϕ(b)$ for all $a,b∈A$,

     
  3. (iii)

    $ϕ$(1$A$) = 1$B$.

     
The map $ϕ:A→B$ is a $K$-algebra isomorphism if it is a $K$-algebra homomorphism and is in addition bijective. If so, then the $K$-algebras $A$ and $B$ are said to be isomorphic, and one writes $A$≅$B$. Note that the inverse of an algebra isomorphism is also an algebra isomorphism, see Exercise 1.14.

Remark 1.23.

  1. (1)

    To check condition (ii) of Definition 1.22, it suffices to take for $a,b$ any two elements in some fixed basis. Then it follows for arbitrary elements of $A$ as long as $ϕ$ is $K$-linear.

     
  2. (2)

    Note that the definition of an algebra homomorphism requires more than just being a homomorphism of the underlying rings. Indeed, a ring homomorphism between $K$-algebras is in general not a $K$-algebra homomorphism.

    For instance, consider the complex numbers $ℂ$ as a $ℂ$-algebra. Let $ϕ :ℂ→ ℂ$, $ϕ (z)=\bar {z}$ be the complex conjugation map. By the usual rules for complex conjugation $ϕ$ satisfies axioms (ii) and (iii) from Definition 1.22, that is, $ϕ$ is a ring homomorphism. But $ϕ$ does not satisfy axiom (i), since for example \begin{aligned}ϕ(i i) = ϕ(i^2) = ϕ(-1) = -1 ≠ 1 = i(-i)= i ϕ(i). \end{aligned} So $ϕ$ is not a $ℂ$-algebra homomorphism. However, if one considers $ℂ$ as an algebra over $ℝ$, then complex conjugation is an $ℝ$-algebra homomorphism. In fact, for $r∈ ℝ$ and $z∈ ℂ$ we have \begin{aligned}ϕ(rz) = \overline{rz} = \bar{r}\bar{z} = r\bar{z} = rϕ(z). \end{aligned}
     

We list a few examples, some of which will occur frequently. For each of these, we recommend checking that the axioms of Definition 1.22 are indeed satisfied.

Example 1.24.

Let $K$ be a field.
  1. (1)

    Let $Q$ be the one-loop quiver with one vertex $v$ and one arrow $α$ such that $s(α)=v=t(α)$. As pointed out in Example 1.13, the path algebra $KQ$ has a basis consisting of 1, $α,α^2$, …. The multiplication is given by $α^i⋅α^j=α^{i+j}$. From this we can see that the polynomial algebra $K[X]$ and $KQ$ are isomorphic, via the homomorphism defined by sending $∑_iλ_iX^i$ to $∑_iλ_iα^i$. That is, we substitute $α$ into the polynomial $∑_iλ_iX^i$.

     
  2. (2)
    Let $A$ be a $K$-algebra. For every element $a∈A$ we consider the ‘evaluation map’ \begin{aligned}φ_a : K[X] → A\mbox{ , ~}\sum_i λ_iX^i ↦ \sum_i λ_i a^i. \end{aligned} This is a homomorphism of $K$-algebras.
     
  3. (3)

    Let $A$ be a $K$-algebra and $I⊂A$ a proper two-sided ideal, with factor algebra $A/I$. Then the canonical map $π:A→A/I,a↦a+I$ is a surjective $K$-algebra homomorphism.

     
  4. (4)
    Let $A=A_1×…×A_n$ be a direct product of $K$-algebras. Then for each $i$ ∈{1, …, $n$} the projection map \begin{aligned} π_i : A → A_i\mbox{ , ~}(a_1,…,a_n)↦ a_i \end{aligned} is a surjective $K$-algebra homomorphism. However, note that the embeddings \begin{aligned}ι_i:A_i→ A\mbox{ , ~}a_i↦ (0,…,0,a_i,0,…,0) \end{aligned} are not $K$-algebra homomorphisms when $n$ ≥ 2 since the identity element $1_{A_i}$ is not mapped to the identity element $1_A=(1_{A_1},… ,1_{A_n})$.
     
  5. (5)
    Let $A=T_n(K)$ be the algebra of upper triangular matrices. Denote by $B=K×…×K$ the direct product of $n$ copies of $K$. Define $ϕ:A→B$ by setting \begin{aligned}ϕ\begin{pmatrix} a_{11} & & & \\ & ⋱ & & ∗ \\ 0 & & ⋱ & \\ & & & a_{nn} \end{pmatrix} = (a_{11},…,a_{nn}).\end{aligned} Then $ϕ$ is a homomorphism of $K$-algebras.
     
  6. (6)
    Consider the matrix algebra $M_n(K)$ where $n$ ≥ 1. Then the opposite algebra $M_n(K)$op (as defined in Definition 1.6) is isomorphic to the algebra $M_n(K)$. In fact, consider the map given by transposition of matrices \begin{aligned}τ:M_n(K)→ M_n(K)^{op}\mbox{ , ~} m↦ m^t.\end{aligned} Clearly, this is a $K$-linear map, and it is bijective (since $τ^2$ is the identity). Moreover, for all matrices $m$1, $m$2 ∈ $M_n(K)$ we have $(m_1m_2)^t=m_2^tm_1^t$ by a standard result from linear algebra, that is, \begin{aligned} τ(m_1m_2) = (m_1m_2)^t = m_2^tm_1^t = τ(m_1)∗τ(m_2).\end{aligned} Finally, $τ$ maps the identity matrix to itself, hence $τ$ is an algebra homomorphism.
     
  7. (7)

    When writing linear transformations of a finite-dimensional vector space as matrices with respect to a fixed basis, one basically proves that the algebra of linear transformations is isomorphic to the algebra of square matrices. We recall the proof, partly as a reminder, but also since we will later need a generalization.

    Suppose $V$ is an $n$-dimensional vector space over the field $K$. Then the $K$-algebras End$K$$(V$ ) and $M_n(K)$ are isomorphic.

    In fact, we fix a $K$-basis of $V$ . Suppose $α$ is a linear transformation of $V$ , let $M(α)$ be the matrix of $α$ with respect to the fixed basis. Then define a map \begin{aligned}ψ: \mathrm{End}_K(V) → M_n(K) , \ \ ψ(α):= M(α).\end{aligned} From linear algebra it is known that $ψ$ is $K$-linear and that it preserves the multiplication, that is, $M(β)M(α)=M(β∘α)$. The map $ψ$ is also injective. Suppose $M(α)$=0, then by definition $α$ maps the fixed basis to zero, but then $α$=0. The map $ψ$ is surjective, because every $n×n$-matrix defines a linear transformation of $V$ .
     
  8. (8)
    We consider the algebra $T$2$(K)$ of upper triangular 2 × 2-matrices. This algebra is of dimension 3 and has a basis of matrix units $E$11, $E$12 and $E$22. Their products can easily be computed (for instance using the formula in Exercise 1.10), and they are collected in the multiplication table below. Let us now compare the algebra $T$2$(K)$ with the path algebra $KQ$ for the quiver $1 \xleftarrow {α} 2$ which has appeared already in Example 1.13. The multiplication tables for these two algebras are given as follows
    $\begin{array}{c|c|c|c}\cdot&E_{11}&E_{12}&E_{22} \\ \hline E_{11}&E_{11}&E_{12}&0 \\ \hline E_{12}&0&0&E_{12} \\ \hline E_{22}&0&0&E_{22}\end{array}\qquad\begin{array}{c|c|c|c}\cdot&e_{1}&\alpha&e_{2} \\ \hline e_{1}&e_{1}&\alpha&0 \\ \hline \alpha&0&0&\alpha \\ \hline e_{2}&0&0&e_{2}\end{array}$

    From this it easily follows that the assignment $E$11↦$e$1, $E$12↦$α$, and $E$22↦$e$2 defines a $K$-algebra isomorphism $T$2$(K)$ → $KQ$.

     

Remark 1.25.

The last example can be generalized. For every $n∈ ℕ$ the $K$-algebra $T_n(K)$ of upper triangular $n×n$-matrices is isomorphic to the path algebra of the quiver \begin{aligned}1⟵ 2 ⟵ … ⟵ n-1 ⟵ n \end{aligned} See Exercise 1.18.

In general, homomorphisms and isomorphisms are very important when comparing different algebras. Exactly as for rings we have an isomorphism theorem for algebras. Note that the kernel and the image of an algebra homomorphism are just the kernel and the image of the underlying $K$-linear map.

Theorem 1.26 (Isomorphism Theorem for Algebras).

Let K be a field and let A and B be K-algebras. Suppose ϕ : A  B is a K-algebra homomorphism. Then the kernel $\ker(ϕ)$ is a two-sided ideal of A, and the image $\im(ϕ)$ is a subalgebra of B. Moreover:
  1. (a)
    If I is an ideal of A and I $⊆\ker(ϕ)$ then we have a surjective algebra homomorphism \begin{aligned}\bar{ϕ}: A/I→ \im(ϕ), \ \ \bar{ϕ}(a+I) = ϕ(a). \end{aligned}
     
  2. (b)
    The map $\bar {ϕ }$ is injective if and only if I = $\ker(ϕ)$. Hence ϕ induces an isomorphism \begin{aligned}Ψ:A/\mathrm{ker}(ϕ)→ \im(ϕ)\mathit{\mbox{ , ~}} a+ \mathrm{ker}(ϕ) ↦ ϕ(a). \end{aligned}
     

Proof.

From linear algebra we know that the kernel $\ker(ϕ)$ = {$a∈A$ | $ϕ(a)$=0} is a subspace of $A$, and $\ker(ϕ)$ ≠ $A$ since $ϕ$(1$A$) = 1$B$ (see Definition 1.22). If $a∈A$ and $x∈\ker(ϕ)$ then we have \begin{aligned}ϕ(ax) = ϕ(a)ϕ(x) = ϕ(a)⋅ 0 = 0 = 0⋅ ϕ(a) = ϕ(x)ϕ(a) = ϕ(xa),\end{aligned} that is, $ax∈\ker(ϕ)$ and $xa∈\ker(ϕ)$ and $\ker(ϕ)$ is a two-sided ideal of $A$.
We check that $\im(ϕ)$ is a subalgebra of $B$ (see Definition 1.14). Since $ϕ$ is a $K$-linear map, the image $\im(ϕ)$ is a subspace. It is also closed under multiplication and contains the identity element; in fact, we have \begin{aligned}ϕ(a)ϕ(b)=ϕ(ab)∈ \im(ϕ)\mbox{ ~and ~} 1_B=ϕ(1_A)∈ \im(ϕ) \end{aligned} since $ϕ$ is an algebra homomorphism.
(a) If $a+I=a'+I$ then $a−a'∈I$ and since $I⊆\ker(ϕ)$ we have \begin{aligned}0 = ϕ(a-a') = ϕ(a) - ϕ(a'). \end{aligned} Hence $\bar {ϕ }$ is well defined, and its image is obviously equal to $\im(ϕ)$. It remains to check that $\bar {ϕ }$ is an algebra homomorphism. It is known from linear algebra (and easy to check) that the map is $K$-linear. It takes the identity $1_A+I$ to the identity element of $B$ since $ϕ$ is an algebra homomorphism. To see that it preserves products, let $a,b∈A$; then \begin{aligned} \bar{ϕ}((a+ I)(b+ I)) = \bar{ϕ}(ab+ I) = ϕ(ab) = ϕ(a)ϕ(b) = \bar{ϕ}(a+ I)\bar{ϕ}(b+I). \end{aligned}

(b) We see directly that $\ker(\bar {ϕ }) = \ker(ϕ )/I$. The homomorphism $\bar {ϕ }$ is injective if and only if its kernel is zero, that is, $\ker(ϕ)=I$. The last part follows.□

Example 1.27.

  1. (1)
    Consider the following evaluation homomorphism of $ℝ$-algebras (see Example 1.24), \begin{aligned}Φ:ℝ[X] → ℂ~,~~ Φ(f)=f(i) \end{aligned} where $i^2=-1$. In order to apply the isomorphism theorem we have to determine the kernel of $Φ$. Clearly, the ideal $(X^2+1)=ℝ[X](X^2+1)$ is contained in the kernel. On the other hand, if $g∈ \ker (Φ )$, then division with remainder in $ℝ[X]$ yields polynomials $h$ and $r$ such that $g=h⋅(X^2+1)+r$, where $r$ is of degree ≤ 1. Evaluating at $i$ gives $r(i)=0$, conjugation gives $r(-i)=0$, but $r$ has degree ≤ 1, so $r$ is the zero polynomial, that is, $g∈(X^2+1)$. Since $Φ$ is surjective by definition, the isomorphism theorem gives \begin{aligned} ℝ[X]/(X^2+1) ≅ ℂ \end{aligned} as $ℝ$-algebras.
     
  2. (2)
    Let $G$ be a cyclic group of order $n$, generated by the element $a∈G$. For a field $K$ consider the group algebra $KG$; this is a $K$-algebra of dimension $n$. Similar to the previous example we consider the surjective evaluation homomorphism \begin{aligned}Φ:K[X] → KG~,~~ Φ(f)=f(a). \end{aligned} From the isomorphism theorem we know that \begin{aligned}K[X]/\ker(Φ) ≅ \mathrm{im}(Φ) =KG. \end{aligned} The ideal generated by the polynomial $X^n-1$ is contained in the kernel of $Φ$, since $a^n$ is the identity element in $G$ (and then also in $KG$). Thus, the algebra on the left-hand side has dimension at most $n$, see Example 1.21. On the other hand, $KG$ has dimension $n$. From this we can conclude that $\ker (Φ )=(X^n-1)$ and that as $K$-algebras we have \begin{aligned}K[X]/(X^n-1) ≅ KG. \end{aligned}
     
  3. (3)
    Let $A$ be a finite-dimensional $K$-algebra and $a∈A$, and let $A_a$ be the subalgebra of $A$ as in Example 1.16. Let $t∈ ℕ_0$ such that {1, $a,a^2$, …$a^t$} is linearly independent and $a^{t+1} = \sum _{i=0}^t λ _ia^i$ for $λ_i∈K$. The polynomial \begin{aligned}m_a:= X^{t+1}-\sum_{i=0}^tλ_iX^i ∈ K[X] \end{aligned} is called the minimal polynomial of $a$. It is the (unique) monic polynomial of smallest degree such that the evaluation map (see Example 1.24) at $a$ is zero. By the same arguments as in the first two examples we have \begin{aligned}K[X]/(m_a) ≅ A_a. \end{aligned}
     
  4. (4)
    Suppose $A=A$1 ×… × $A_r$, the direct product of $K$-algebras $A$1…, $A_r$. Then for every $i$ ∈{1, …, $r$} the projection $π_i:A→A_i$, $(a$1, …, $a_r)↦a_i$, is an algebra homomorphism, and it is surjective. By definition the kernel has the form \begin{aligned}\ker(π_i) = A_1× … × A_{i-1} × 0 × A_{i+1} × … × A_r. \end{aligned} By the isomorphism theorem we have $A/\ker(π_i)≅A_i$. We also see from this that $A$1 ×… × $A$$i$−1 × 0 × $A$$i$+1 ×… × $A_r$ is a two-sided ideal of $A$.
     
  5. (5)
    We consider the following map on the upper triangular matrices \begin{aligned}Φ:T_n(K)→ T_n(K)\mbox{ , ~} \begin{pmatrix} a_{11} & & ∗ \\ & ⋱ & \\ 0 & & a_{nn} \end{pmatrix} ↦ \begin{pmatrix} a_{11} & & 0 \\ & ⋱ & \\ 0 & & a_{nn} \end{pmatrix}, \end{aligned} which sets all non-diagonal entries to 0. It is easily checked that $Φ$ is a homomorphism of $K$-algebras. By definition, the kernel is the two-sided ideal \begin{aligned}U_n(K):= \{ a=(a_{ij})∈ M_n(K)∣ a_{ij}=0\mbox{ for }i≥j\} \end{aligned} of strict upper triangular matrices and the image is the $K$-algebra $D_n(K)$ of diagonal matrices. Hence the isomorphism theorem yields that \begin{aligned} T_n(K)/U_n(K) ≅ D_n(K) \end{aligned} as $K$-algebras. Note that, moreover, we have that $D_n(K)≅K×…×K$, the $n$-fold direct product of copies of $K$.
     
  6. (6)
    We want to give an alternative description of the $ℝ$-algebra $ℍ$ of quaternions from Sect. 1.1.1. To this end, consider the map \begin{aligned}Φ:ℍ→ M_4(ℝ)\mbox{ , ~} a+bi+cj+dk ↦ \begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{pmatrix}. \end{aligned} Using the formula from Example 1.8 for the product of two elements from $ℍ$ one can check that $Φ$ is an $ℝ$-algebra homomorphism, see Exercise 1.11.

    Looking at the first row of the matrices in the image, it is immediate that $Φ$ is injective. Therefore, the algebra $ℍ$ is isomorphic to the subalgebra $\im(Φ)$ of $M_4(ℝ)$. Since we know (from linear algebra) that matrix multiplication is associative and that the distributivity law holds in $M_4(ℝ)$, we can now conclude with no effort that the multiplication in $ℍ$ is associative and distributive.

     

Exercise 1.7.

Explain briefly how examples (1) and (2) in Example 1.27 are special cases of (3).