Supposed $f(x), g(x) \in F[x]$ for some field $F$ are polynomials of degrees $m, n $ respectively. Moreover assume that they are relatively prime. By Euclidean algorithm I can find $a'(x), b'(x)$ such that $$a'(x)f(x) + b'(x)g(x) = 1$$ and thus for any $p(x)$, I can find $a(x), b(x)$ such that $$a(x)f(x) + b(x)g(x) = p(x)$$ My question is what if you restrict the degree of $p(x)$ to be less than $d$. Then can I find $a(x), b(x)$ of degrees $d-m, d-n$ respectively that satisfy the equation above?
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Yes this is a consequence of the Euclid's extended algorithm * – Elaqqad Apr 25 '15 at 00:21
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Actually this is the The statement of the theorem I don't have time to give a proof now but maybe someone else could provide it – Elaqqad Apr 25 '15 at 00:25
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I think I'm missing something. I see that this applies for the gcd of $f$ and $g$ (in this case 1), but not for a general polynomial. – Alex Apr 25 '15 at 00:46
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Let $f(x) = x^2$ and $g(x) = x^2 + 1$, so that $-f(x) + g(x) = 1$. Note that $m=n=2$.
Let $p(x) = x$, so the degree of $p(x)$ is less than $d=2$. However there do not exist coefficients $a(x),b(x)$ of degree zero (constants) such that:
$$ a(x)f(x) + b(x)g(x) = p(x) $$
since any such linear combination would be a polynomial in $x^2$ (and $p(x) = x$ is not).
hardmath
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Thanks for the counterexample. However, I was more interested in the case where $d \geq n+m$, in which case Luis' answer below proves what I want. – Alex Apr 25 '15 at 21:11
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In a principal ring A, like F[x], all pair of elements f, g have m.c.d. and l.c.m. French call “Identité de Bezout” the property: if d = (f, g) then there exist a, b in A such that af + bg = d. When d = 1 we have for all p, apf + bpg = p. On the other hand we look at the equation fX + gY = p and since we have a solution ap, bp we can make X = ap + hg and Y = bp – hf with the parameter h in F[x] because (X – ap)f + (Y – bp)g = 0. What you have now is to see the question of degrees to be considered.
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