Ring Theory - Quaternions and sets of inverses

[tomelwood]

Homework Statement

Two questions really, the first is about the ring of quaternions H and the second about a set of maps.
a) Find an element c in H such that the evaluation phi_c : C[x]-->H is not a ring homomorphism. In words that is: "the evaluation phi sub c from the ring of complex polynomials to the ring of quaternions"

b)Longer question, this one.
If R is a unitary ring and aE R (a is a member of R), let I_a (I sub a) = {bE R : ba = 1_R}, the family of left inverses of a. Prove that only one of the following three conditions holds:
i) a is not invertible on the left (ii) a is an identity element of R (ie there exists bE I_a such that ab=1_R) (iii) a has infinitely many left inverses

(NOTE This question comes after already proving that f:I_a -->I_a , b--> ab+b_0 - 1_R (b_0E I_a) is an injective map)
(NOTE This is a part b of a question, and to me it is unclear whether the ring R here is the same ring as used in part a. If so, then R is the ring of functions phi taking Q+ to Q+ such that phi(rs) = phi(r)phi(s) , r,sE Q+, and addition is defined between two maps as (phi + theta)(r) = phi(r)theta(r) and multiplication as (phi x theta)(r) = phi(theta(r)) Whether this is of any help I don't know. That's how much this question is confusing me!)

Homework Equations

The Attempt at a Solution

a) I think the answer is that if I have a polnomial in the ring of complex polynomials, eg f(x) = x^2 + 1 and I evaluate this at a point j in H, then I get f(j) = j^2 +1. However x^2 + 1 = (x+i)(x-i) which when x=j gives (j+i)(j-i) which is ij-ji = 2k =/= j^2 + 1. The problem I am having is how to formalise this, I think.

b) Really sorry, but I honestly have no idea where to start here as I don't entirely understand whether the ring is R as in the previous part of the question or not. I think even if I did know if it was, I still wouldn't know where to go from there! So I hate to ask for tips before attempting anything myself, although I have tried to verify (ii) by: We know ba = 1_R and want to show ab=1_R. So multiply ba on the right by b gives bab=1_R b. Now commute 1_R and b, so that bab = b 1_R and cancel the b to give ab=1_R as required? Though this seems veeery shaky to me, as I feel the b's should be different 'kinds' of b.

There is a third part to this question which says:
Let {pn}n>0 be the ordered sequence of all prime numbers. Prove that there exists a unique element phi in the ring R from part (a) [the one taking Q+ to Q+] such that phi(pn) = pn+1 for every n>0 and determine the family I_phi of left inverses of phi.
However this seems impossible and also too many questions to ask on this forum, so I don't expect any answers to this, I have just included it for completeness. However, obviously any pointers would be greatly appreciated, as I know of no way to map one prime number to the next!
Thanks very much in advance.

 

[ystael]

For (a), you have exactly the right idea. The formal part you are missing is: How do you prove that a map $$\varphi: R \to S$$ between two rings is not a ring homomorphism? You exhibit two elements $$r, r' \in R$$ such that $$\varphi(rr') \neq \varphi(r) \varphi(r')$$. So you need to choose $$c$$, and then fill in the blanks in the sentence: "$$\phi_c: \mathbb{C}[x] \to \mathbb{H}$$ is not a ring homomorphism because $$\phi_c(?) = {?} \neq {?} = \phi_c(?) \cdot \phi_c(?)$$."

 

[ystael]

For (b): The preamble "If $$R$$ is a ring with unity and $$a\in R$$" rebinds the variable $$R$$; you need not be discussing the same ring.

The condition (ii) is misstated; that there exists $$b\in I_a$$ (a left inverse $$b$$ of $$a$$) such that $$ab = 1_R$$ ($$b$$ is also a right inverse of $$a$$) means that $$a$$ is an invertible or unit element of $$R$$, not an identity element. (Perhaps confusingly, "unit" and "unity" mean totally different things in rings; the former means a multiplicatively invertible element, the latter means a multiplicative identity element.)

This problem is sometimes stated as "If an element of a ring has more than one left inverse, it has infinitely many." The fact that this isn't true in arbitrary monoids (a monoid being the multiplicative structure of a ring) should clue you in that somewhere you will need to use the distributive law.

The argument you gave to show that if $$ba = 1_R$$ then $$ab = 1_R$$ won't work; all it shows is that $$bab = b$$, which doesn't tell you anything about $$ab$$ if $$b$$ is not left cancellable. (Think about matrix rings and matrices that are not invertible to come up with examples where you encounter this kind of problem.)

 

[tomelwood]

Thanks for your reply.
OK. So I've got a) done now, thankyou (phi_j ((x+i)(x-i)) = j^2-1 =/=2k = phi_j(x+i)phi_j(x-i))

but unfortunately am still stumped on b) I'm afraid.
I can see that the question is in fact asking to show that the map either is not left invertible, an element has one left inverse, or that an element has infinitely many left inverses.
Is this the same as "If a has more than one left inverse then it has infinitely many"?
So if as you say this statement is not true, then that rules out having infinitely many left inverses. Though I don't know how to prove this at all...! And I don't see how an element can NOT have a left inverse, since the map is defined to be the family of left inverses of a?
Therefore a has only one left inverse...?
This all is very hand-wavey, I know. But once again it seems I need some pointers to formalise things, if indeed they are right...

 

[tomelwood]

I am still very confused on this question. Though I think I have to do somehting along the lines of. Let a have two right inverses a1 and a2. Show that a2 + a1a -1 is another right inverse. So a(1-a1a) = 0 I assume. Though now I don't know where to go. And if I prove this, does it answer the original question "show that only one of the following is true?" the answer being the third one.?
Thanks in advance.

 

[ystael]

I think you're confused about what the question is asking you to do. You are not supposed to pick one of the alternatives and prove it -- that is, you are not supposed to: prove $$I_a$$ is empty, or prove that $$a$$ is also right invertible, or prove that $$I_a$$ is infinite. Each of the three alternatives may hold for particular choices of $$R$$ and $$a$$. You are supposed to prove the disjunction: that whenever $$R$$ and $$a$$ are chosen as described in the hypotheses, one of the three alternatives holds.

One way to write this proof is to phrase it as "if alternative 1 does not hold, and alternative 2 does not hold, then alternative 3 must hold". That is, if $$a$$ has at least one left inverse, and does not have any right inverse, show that it must have infinitely many left inverses.