Question about vector space isomporhism

Let p1(t), p2(t) be two polynomials in the F[t] polynomial ring, where F is a field. Let ß1 be a congruence relation for: r(t) ß1 s(t) <==> p1(t)|r(t) - s(t), and ß2 the same for r(t) ß2 s(t) <==> p2(t)|r(t) - s(t).

When will be F[t]/ß1 and F[t]/ß2 isomorph vector spaces. Give necessary and sufficient condition.

Thank you!

Re: Question about vector space isomporhism

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**gotmejerry** Let p1(t), p2(t) be two polynomials in the F[t] polynomial ring, where F is a field. Let ß1 be a congruence relation for: r(t) ß1 s(t) <==> p1(t)|r(t) - s(t), and ß2 the same for r(t) ß2 s(t) <==> p2(t)|r(t) - s(t).

When will be F[t]/ß1 and F[t]/ß2 isomorph vector spaces. Give necessary and sufficient condition.

Thank you!

So, here's what you're saying. You have the polynomial ring for some field and you're thinking of it as a vector space over in the natural way. You're asking then when given how can you tell if as vector spaces (where denotes the ideal generated--not that this really is a subspace since sits nicely inside ). Well, we know that vector spaces are exceedingly simple objects, in that their only invariant, the only thing that distinguishes them is their dimension. So, rephrasing your question we need only figure out when . So, let's figure out in general, given any polynomial what is. Note that since generates that generates . What I claim though is that if then is a linear combination of the . We can do this by induction. For example, and so

Now, if we assume that is a linear combination of for all then noting that

Allows us to conclude that is a linear combination of the and since each of these are a linear combination of the the induction follows.

From the fact that every element of a generating set of is a linear combination of allows us to conclude that is a generating set for . What we now claim though is that is a linearly indepdent set. Indeed, the basic idea is that if we could write

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then this tells us that but this is ridiculous since .

From all of this we may conclude that is a basis for and so we may conclude that .

Thus, if and only if which by the above is true if and only if .

*Remark:* Why you should have thought this to be true was to look at and look at polynomials with and then use the CRT to conclude that .

Re: Question about vector space isomporhism