0

I am reading field theory.

I dont understand the following fact:

What will be a basis of $\mathbb{R}(t)=\{\frac{f(t)}{g(t)}:f(t),g(t)\in \Bbb R[t]\}$.

I know that $\Bbb R(t)$ is a field and hence a vector space too over $\Bbb R$.

Also a basis of $\Bbb R[t]$ is $\{1,t,t^2,\ldots ,t^n,\ldots\}$

But how to find a basis of $\Bbb R(t)$?

permutation_matrix
  • 1,356
  • $\mathbb{R}[t]$ is not finite dimensional. You've only given a basis for the space of polynomials with degree bounded by $n$. – Randall Jan 11 '19 at 15:14
  • Do you know about the partial fraction decomposition of a polynomial? – quid Jan 11 '19 at 15:18
  • Because to find a basis of the space it's useful to have this presentation of polynomials at hand. I recommend you look it up. https://en.wikipedia.org/wiki/Partial_fraction_decomposition This might also help you clarify the question. – quid Jan 11 '19 at 15:21
  • It makes sense, since each extension field $L$ of a field $K$ can be viewed as a $K$-vector space. – Wuestenfux Jan 11 '19 at 15:24
  • There is not point to the "if"-clause. You say later on that $f,g$ are polynomials anyway. – quid Jan 11 '19 at 15:39
  • Yes. that's what we meant. – quid Jan 11 '19 at 15:43
  • See https://math.stackexchange.com/questions/126747/a-basis-for-kx-regarded-as-a-vector-space-over-k and note that over the reals the irreducible polynomials are those of degree one and those of degree two with negative discriminat. – quid Jan 11 '19 at 15:46

0 Answers0