Ideals of a two-dimensional algebra with a given basis

Question

My task is as follows:

Find all ideals in a two-dimensional algebra $A$ over $\mathbb{R}$ with basis 1, $e$ where 1 is the multiplicative identity and Case 1: $e^2=0$, Case 2: $e^2=1$.

My difficulty here is I am essentially unsure how to approach the problem initially. My only real thought was consider $A_1$ (corresponding to Case 1 where $e^2=0$) as elements of the form $\{a+eb : a,b \in \mathbb{R}\}$ then it seems I can prove, much in the same way as $a+b\sqrt{2}$, that this is a field, hence its only ideals are $(1), (0=e^2)$, but I feel queasy about this conclusion.

Further, this approach is less effective in Case 2 and I also would like a way to find the ideals more directly.

Thanks very much for any assistance or advice, and my apologies in advance for any ineptness on my part.

Answer 1

One way to look at this (I am not claiming that is the best ...) is to realize $A$ as either $$ \frac{{\Bbb R}[X]}{(X^2)}\qquad\text{or}\qquad \frac{{\Bbb R}[X]}{(X^2-1)} $$ and use the following fact:

If $f:R\rightarrow B$ is surjective homomorphism of rings there is a bijection between the ideals of $B$ and the ideals of $R$ containing $\ker(f)$.

Answer 2

Here I will present a different and direct approach to this intriguing question, one which does not depend on invoking the polynomial ring $\Bbb R[x]$ as was done by our colleague Andrea Mori in his excellent and accepted answer. Instead, we will work directly with the algebra $A$ and its basis $\{ 1_A, e \}$, where $1_A$ denotes the multiplicative unit of $A$.

Our first goal will be to characterize all 2-dimensional unital real algebras; in so doing, we will pick up the case

$e^2 = -1_A, \tag 1$

and find all the ideals of $A$ under this condition as well.

Preliminary remarks: since $\{1_A, e\}$ is a basis for $A$ over $\Bbb R$, $A$ is commutative, for every element of $A$ may be written in the form $a1_A + be$, $a, b \in \Bbb R$:

$(a1_A + be)(c1_A + de) = a1_A c1_A + a1_A de + be c1_A + bede = ac1_A^2 + ad1_Ae + bce1_A + bde^2$ $= ca1_A^2 + dae1_A + cb1_Ae + dbe^2 = c1_A a1_A + c1_Abe + dea1_A + debe = (c1_A + de)(a1_A + be); \tag 2$

also, the vector subspace

$\Bbb R 1_A \subsetneq A \tag 3$

is in fact a sub-algebra which is isomorphic to $\Bbb R$ under the mapping

$\Bbb R \ni r \mapsto r1_A \in \Bbb R1_A; \tag 4$

that is, we have

$r + s \mapsto (r + s)1_A = r1_A + s1_A, \tag 5$

and

$rs \mapsto (rs)1_A = rs1_A^2 = (r1_A)(s1_A). \tag 6$

Now consider the set $\{1_A, e, e^2 \}$; since

$\dim_{\Bbb R}A = 2, \tag 7$

$\{1_A, e, e^2 \}$ is linearly dependent over $\Bbb R$; thus there exist

$a, b, c \in \Bbb R, \tag 8$

not all $0$, with

$ae^2 + be + c1_A = 0; \tag 9$

I claim that

$a \ne 0; \tag{10}$

for in the contrary situation

$be + c1_A = 0; \tag{11}$

then

$b = 0 \Longrightarrow c1_A = 0 \Longrightarrow c = 0; \tag{12}$

but then $a = b = c = 0$, prohibited by assumption; and if $b \ne 0$, then

$be + c1_A = 0, \tag{13}$

which contradicts our hypothesis that $\{1_A, e \}$ forms a basis for $A$; therefore (10) binds and we may set

$\alpha = \dfrac{b}{a}, \; \beta = \dfrac{c}{a}, \tag{14}$

so that (9) may be written

$e^2 + \alpha e + \beta 1_A = 0. \tag{15}$

Denoting $e$ for the moment by $e_0$, we complete the square in (15) as follows: writing

$e_0^2 + \alpha e_0 = -\beta 1_A, \tag{16}$

we have

$e_0^2 + \alpha e_0 + \dfrac{\alpha^2}{4}1_A = \dfrac{\alpha^2}{4}1_A - \beta 1_A, \tag{17}$

yielding

$\left (e_0 + \dfrac{\alpha}{2}1_A \right )^2 = e_0^2 + \alpha e_0 + \dfrac{\alpha^2}{4}1_A = \dfrac{\alpha^2}{4}1_A - \beta 1_A = \left (\dfrac{\alpha^2}{4} - \beta \right )1_A. \tag{18}$

We distinguish three cases of this equation, according to whether

$\dfrac{\alpha^2}{4} - \beta >, =, < 0; \tag{19}$

we start by addressing the case

$\dfrac{\alpha^2}{4} - \beta > 0, \tag{20}$

in which (18) may be written

$\left ( \dfrac{e_0 + \dfrac{\alpha}{2}1_A}{\sqrt{\dfrac{\alpha^2}{4} - \beta}} \right )^2 = 1_A; \tag{21}$

now setting

$e_1 = \dfrac{e_0 + \dfrac{\alpha}{2}1_A}{\sqrt{\dfrac{\alpha^2}{4} - \beta}}, \tag{22}$

we readily see that

$A = \text{span} \{1_A, e_1 \} \tag{23}$

with

$e_1^2 = 1_A; \tag{24}$

likewise with

$\dfrac{\alpha^2}{4} - \beta = 0, \tag{25}$

we may take

$e_1 = e_0 + \dfrac{\alpha}{2}1_A, \tag{26}$

and we have

$A = \text{span} \{1_A, e_1 \}, \; e_1^2 = 0; \tag{27}$

finally, with

$\dfrac{\alpha^2}{4} - \beta < 0, \tag{28}$

we define

$e_1 = \dfrac{e_0 + \dfrac{\alpha}{2}1_A}{\sqrt{\beta- \dfrac{\alpha^2}{4}}}, \tag{29}$

and then

$A = \text{span} \{1_A, e_1 \}, \; e_1^2 = -1_A; \tag{30}$

'twixt (23)-(24), (27) and (30), every possible value of $\alpha^2 / 4 - \beta$ is covered, and we may proceed by analyzing each of these configurations of a two-dimensionl, real unital algebra $A$.

(30) is most readily dispensed with, since here $A$ is a field; indeed, for

$0 \ne a1_A + be_1 \in A, \tag{31}$

we have

$(a1_A + be_1) \left ( \dfrac{a1_A - be_1}{a^2 + b^2} \right ) = \dfrac{(a1_A + be_1)(a1_A - be_1)}{a^2 + b^2} = \dfrac{a^21_A + b^21_A}{a^2 + b^2} = 1_A; \tag{32}$

thus,

$(a1_A + be_1)^{-1} = \dfrac{a1_A - be_1}{a^2 + b^2}, \tag{33}$

and if follows that $A$ is a field; as such, it has no proper ideals; only $\{0\}$ and $A$ itself are ideals in this case; in fact the mapping

$A \ni a1_A + be_1 \mapsto a + bi \in \Bbb C \tag{34}$

is an isomorphism 'twixt $A$ and $\Bbb C$. We have thus dispensed with the case $e_1^2 = -1$.

We next turn to take up the case (27), in which $e_1^2 = 0$; by virtue of (7), we are assured that every proper ideal $I \subsetneq A$ is one-dimensional as an $\Bbb R$-subspace of $A$:

$\dim_{\Bbb R} I = 1; \tag{35}$

thus we may write

$I = (a1_A + be_1)\Bbb R = (a1_A + be_1)A = \langle a1_A + be_1 \rangle\tag{36}$

for some $a, b \in \Bbb R$; we observe that (27) implies

$a^21_A = a^2 1_A^2 = a^2 1_A^2 - b^2 e_1^2 = (a1_A - be_1)(a1_A + be_1) \in I; \tag{37}$

then if $a \ne 0$ we find

$1_A = a^{-2}a^21_A \in I \Longrightarrow I = A; \tag{38}$

thus any non-trivial proper ideal as in (36) must in fact be of the form

$I = \langle be_1 \rangle = e_1\Bbb R, \; b \ne 0; \tag{39}$

indeed we see that

$(c + de_1)be_1 = cbe_1 + dbe_1^2 = cbe_1 \in e_1\Bbb R, \tag{40}$

which shows that $\langle e_1 \rangle = \Bbb Re_1$ is closed under multiplication by arbitrary elements of $A$, as must be for every ideal $I \subset A$. We thus conclude that the only proper ideal of $A$ in the case (27) is $\langle e_1 \rangle = \Bbb Re_1$.

Finally, we turn to (23)-(24); again, we examine ideals one-dimensional $I$ of the form

$I = (a1_A + be_1) \Bbb R= \langle a1_A + be_1 \rangle; \tag{41}$

if

$a = 0, \tag{42}$

then

$I = \langle be_1\rangle = \langle e_1 \rangle, \; b \ne 0; \tag{43}$

for

$c + de_1 \in A, \tag{44}$

$d + ce_1 = (c + de_1)e_1 \in \langle e_1 \rangle, \tag{45}$

which shows that every $d + ce_1 \in A$ lies in $(e_1)$; thus

$I = \langle e_1 \rangle = A \tag{46}$

in this case; likewise, with

$b = 0 \tag{47}$

we obtain

$I = \langle a1_A \rangle, \tag{48}$

whence

$1_A = a^{-1}a1_A \in I \Longrightarrow I = A \tag{49}$

in this case as well; thus the generator of any proper

$I = \langle a1_A + be_1 \rangle \tag{50}$

must satisfy

$a \ne 0 \ne b; \tag{51}$

it follows that we may write

$I = \langle a1_A + be_1 \rangle = \langle 1 + \alpha e_1 \rangle, \; \alpha = a^{-1}b; \tag{52}$

for $c1_A + de_1 \in A$,

$(c1_A + de_1)(1_A + \alpha e_1) = c1_A + \alpha c e_1 + de_1 + \alpha d 1_A = (c + \alpha d)1_A + (\alpha c + d)e_1; \tag{53}$

thus an arbitrary $x1_A + ye_1 \in A$ is an element of $I$ provided that there are $c, d \in \Bbb R$ with

$(c + \alpha d)1_A + (\alpha c + d)e_1 = x1_A + ye_1, \tag{54}$

or

$c + \alpha d = x, \tag{55}$

$\alpha c + d = y; \tag{56}$

these two equations may be written as the $2 \times 2$ matrix system

$\begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} \begin{pmatrix} c \\d \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}, \tag{57}$

which has a solution for any given $x$, $y$ provided that

$1 - \alpha^2 = \det \left ( \begin{bmatrix} 1 & \alpha \\ \alpha & 1 \end{bmatrix} \right ) \ne 0; \tag{58}$

if this condition holds, every

$x1_A + ye_1 \in I \Longrightarrow I = A; \tag{59}$

therefore $I$ is a proper ideal when

$1 - \alpha^2 = 0 \Longleftrightarrow \alpha = \pm 1; \tag{60}$

returning to (52), we find that

$I_+ = \langle 1_A + e_1 \rangle, \; I_- = \langle 1_A - e_1 \rangle, \tag{61}$

are the proper ideals of $A$. In more concrete terms,

$(a1_A + be_1)(1_A + e_1) = a1_A + ae_1 + be_1 + be_1^2 = (a + b)1_A + (a + b)e_1, \tag{62}$

$(a1_A + be_1)(1_A - e_1) = a1_A - ae_1 + be_1 - be_1^2 = (a - b)1_A + (b - a)e_1; \tag{63}$

thus,

$I_+ = \{c(1_A + e_1), \; c \in \Bbb R \},\tag{64}$

$I_- = \{c(1_A - e_1), \; c \in \Bbb R \},\tag{65}$

are the two proper ideals of $A$; we further note that, though

$I_+ \ne \{0\} \ne I_- \tag{66}$

(as may be see by taking $c = 1$ in (64), (65)),

$I_+ \cap I_- = \{0\}; \tag {67}$

for if

$r(1_A + e_1) = s(1_A - e_1), \tag{68}$

then

$r(1_A + e_1)^2 = s(1_A - e_1)(1_A + e_1) = s(1_A^2 - e_1^2) = s(1_A - 1_A) = 0, \tag{69}$

and

$(1_A + e_1)^2 = 1_A^2 + 21_Ae_1 + e_1^2 = 1_A + 2e_1 + 1_A = 2(1_A + e_1) \ne 0; \tag{70}$

(69) and (70) together yield

$2r(1_A + e_1) = 0 \Longrightarrow r = 0 \Longrightarrow s(1_A - e_1) = 0 \Longrightarrow s = 0, \tag{71}$

whence

$r(1_A + e_1) = s(1_A - e_1) = 0, \tag{72}$

and thus (67) binds.

So concludes our discussion of the structure of the two-dimensional algebras over $\Bbb R$, including the classification of their ideals.