Fdtxjr

My task is as follows:

Find all ideals in a two-dimensional algebra $A$ over $mathbbR$ 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 $a+eb : a,b in mathbbR$ then it seems I can prove, much in the same way as $a+bsqrt2$, 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.

One way to look at this (I am not claiming that is the best ...) is to realize $A$ as either
$$
fracBbb R[X](X^2)qquadtextorqquad
fracBbb R[X](X^2-1)
$$
and use the following fact:

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

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

$dim_Bbb RA = 2, tag 7$

$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

$a ne 0; tag10$

for in the contrary situation

$be + c1_A = 0; tag11$

then

$b = 0 Longrightarrow c1_A = 0 Longrightarrow c = 0; tag12$

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

$be + c1_A = 0, tag13$

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

$alpha = dfracba, ; beta = dfracca, tag14$

so that (9) may be written

$e^2 + alpha e + beta 1_A = 0. tag15$

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

$e_0^2 + alpha e_0 = -beta 1_A, tag16$

we have

$e_0^2 + alpha e_0 + dfracalpha^241_A = dfracalpha^241_A - beta 1_A, tag17$

yielding

$left (e_0 + dfracalpha21_A right )^2 = e_0^2 + alpha e_0 + dfracalpha^241_A = dfracalpha^241_A - beta 1_A = left (dfracalpha^24 - beta right )1_A. tag18$

We distinguish three cases of this equation, according to whether

$dfracalpha^24 - beta >, =, < 0; tag19$

we start by addressing the case

$dfracalpha^24 - beta > 0, tag20$

in which (18) may be written

$left ( dfrace_0 + dfracalpha21_Asqrtdfracalpha^24 - beta right )^2 = 1_A; tag21$

now setting

$e_1 = dfrace_0 + dfracalpha21_Asqrtdfracalpha^24 - beta, tag22$

we readily see that

$A = textspan 1_A, e_1 tag23$

with

$e_1^2 = 1_A; tag24$

likewise with

$dfracalpha^24 - beta = 0, tag25$

we may take

$e_1 = e_0 + dfracalpha21_A, tag26$

and we have

$A = textspan 1_A, e_1 , ; e_1^2 = 0; tag27$

finally, with

$dfracalpha^24 - beta < 0, tag28$

we define

$e_1 = dfrace_0 + dfracalpha21_Asqrtbeta- dfracalpha^24, tag29$

and then

$A = textspan 1_A, e_1 , ; e_1^2 = -1_A; tag30$

'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

$0 ne a1_A + be_1 in A, tag31$

we have

$(a1_A + be_1) left ( dfraca1_A - be_1a^2 + b^2 right ) = dfrac(a1_A + be_1)(a1_A - be_1)a^2 + b^2 = dfraca^21_A + b^21_Aa^2 + b^2 = 1_A; tag32$

thus,

$(a1_A + be_1)^-1 = dfraca1_A - be_1a^2 + b^2, tag33$

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

$A ni a1_A + be_1 mapsto a + bi in Bbb C tag34$

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 $A$:

$dim_Bbb R I = 1; tag35$

thus we may write

$I = (a1_A + be_1)Bbb R = (a1_A + be_1)A = langle a1_A + be_1 rangletag36$

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

$a^21_A = a^2 1_A^2 = a^2 1_A^2 - b^2 e_1^2 = (a1_A - be_1)(a1_A + be_1) in I; tag37$

then if $a ne 0$ we find

$1_A = a^-2a^21_A in I Longrightarrow I = A; tag38$

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

$I = langle be_1 rangle = e_1Bbb R, ; b ne 0; tag39$

indeed we see that

$(c + de_1)be_1 = cbe_1 + dbe_1^2 = cbe_1 in e_1Bbb R, tag40$

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

$I = (a1_A + be_1) Bbb R= langle a1_A + be_1 rangle; tag41$

if

$a = 0, tag42$

then

$I = langle be_1rangle = langle e_1 rangle, ; b ne 0; tag43$

for

$c + de_1 in A, tag44$

$d + ce_1 = (c + de_1)e_1 in langle e_1 rangle, tag45$

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

$I = langle e_1 rangle = A tag46$

in this case; likewise, with

$b = 0 tag47$

we obtain

$I = langle a1_A rangle, tag48$

whence

$1_A = a^-1a1_A in I Longrightarrow I = A tag49$

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

$I = langle a1_A + be_1 rangle tag50$

must satisfy

$a ne 0 ne b; tag51$

it follows that we may write

$I = langle a1_A + be_1 rangle = langle 1 + alpha e_1 rangle, ; alpha = a^-1b; tag52$

for $c1_A + de_1 in A$,

$(c1_A + de_1)(1_A + alpha e_1) = c1_A + alpha c e_1 + de_1 + alpha d 1_A = (c + alpha d)1_A + (alpha c + d)e_1; tag53$

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

$(c + alpha d)1_A + (alpha c + d)e_1 = x1_A + ye_1, tag54$

or

$c + alpha d = x, tag55$

$alpha c + d = y; tag56$

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

$beginbmatrix 1 & alpha \ alpha & 1 endbmatrix beginpmatrix c \d endpmatrix = beginpmatrix x \ y endpmatrix, tag57$

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

$1 - alpha^2 = det left ( beginbmatrix 1 & alpha \ alpha & 1 endbmatrix right ) ne 0; tag58$

if this condition holds, every

$x1_A + ye_1 in I Longrightarrow I = A; tag59$

therefore $I$ is a proper ideal when

$1 - alpha^2 = 0 Longleftrightarrow alpha = pm 1; tag60$

returning to (52), we find that

$I_+ = langle 1_A + e_1 rangle, ; I_- = langle 1_A - e_1 rangle, tag61$

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

$(a1_A + be_1)(1_A + e_1) = a1_A + ae_1 + be_1 + be_1^2 = (a + b)1_A + (a + b)e_1, tag62$

$(a1_A + be_1)(1_A - e_1) = a1_A - ae_1 + be_1 - be_1^2 = (a - b)1_A + (b - a)e_1; tag63$

thus,

$I_+ = c(1_A + e_1), ; c in Bbb R ,tag64$

$I_- = c(1_A - e_1), ; c in Bbb R ,tag65$

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

$I_+ ne 0 ne I_- tag66$

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

$I_+ cap I_- = 0; tag 67$

for if

$r(1_A + e_1) = s(1_A - e_1), tag68$

then

$r(1_A + e_1)^2 = s(1_A - e_1)(1_A + e_1) = s(1_A^2 - e_1^2) = s(1_A - 1_A) = 0, tag69$

and

$(1_A + e_1)^2 = 1_A^2 + 21_Ae_1 + e_1^2 = 1_A + 2e_1 + 1_A = 2(1_A + e_1) ne 0; tag70$

(69) and (70) together yield

$2r(1_A + e_1) = 0 Longrightarrow r = 0 Longrightarrow s(1_A - e_1) = 0 Longrightarrow s = 0, tag71$

whence

$r(1_A + e_1) = s(1_A - e_1) = 0, tag72$

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.