Solving advanced algebra related problems

aa285312415 注册会员
2023-02-28 02:52

This answer partially references GPT, GPT_Pro to solve the problem better
Proof: If f(x)g(x) is not all zero, then at least one of f(x) or g(x) is not zero.

Let f(x)=a_nx^n+a_(n-1)x^(n-1)+... +a_0,g(x)=b_mx^m+b_(m-1)x^(m-1)+... Plus b_0.

According to the formula:

(f(x)g(x)) ^n=(a_nx^n+a_(n-1)x^(n-1)+... +a_0)(b_mx^m+b_(m-1)x^(m-1)+... +b_0)^n
=a_n^nb_m^nx^{mn}+(a_nb_m^{n-1}+a_{n-1}b_m^n)x^{mn-1}+(a_nb_{m-1}^{n-1}+a_{n-1}b_{m-1}^{n-2}+... +a_0b_m^n)x^{mn-2}+... +(a_{n-1}b_0^{n-2}+a_0b_0^{n-1})x^{m+n-2}+a_0b_0^{n}x^{m+n-1}

and(f(x), g(x)) ^ n = a_nb_mx ^ {}(m + n) - 1 +(a_(n - 1} b_} {m - 1) x ^ {2} m + n - +(a_ b_ {n - 2} {2} m -) x ^ {3} m + n - +... +(a_0b_0)x^{m+n-(m+1)}

It follows that

When f(x) or g(x) is not zero, the coefficients of the above two formulas are the same, it can be proved.
If the answer is helpful, please accept it.

dingyan124 注册会员
2023-02-28 02:52

Lemma: Let $A$and $B$be elements of F[x], so ( A , B ) = ( g c d ( A , B ) ) , < / span > < / span > < / span > where $(A, B) $said A and B generated ideal.

Proof: Since $F[x]$is an Euclidean field, it is also a PID field, that is, every ideal can be generated by an element. Set $d = GCD(A, B) $, have the $d | A | B $$and $d, therefore $(A, B) \ subseteq(d) $. On the other hand, since $d$is the greatest common factor, it can be expressed as A linear combination of $A$and $B$, i.e. $d = aA + bB$, thus $d \in(a,B)$. Thus, $(d) \subseteq(A,B)$, i.e. $(A,B) =(d)$.

Now, let's go back to our equation. inherit; ( f ( x < /span> n , g ( x < /span> n < /span> = ( f ( x , g ( x ) < /span> n ) .

For any polynomial $a(x)$and $b(x)$, let $d = gcd(a(x),b(x))$, then according to the lemma, ( a (

cwg620 注册会员
2023-02-28 02:52

reference GPT and our thinking, we need to prove that two polynomial equals the ideal:(f(x) ⁿ, g(x) = ⁿ)(f(x)(x, g) ⁿ).

in the first place, we can prove a lemma: for any polynomial f(x), g(x), h(x) ∈ f(x), with(f(x), g h(x))(x) =(f(x), g(x))(f(x), h(x)). The proof is as follows:

f(x), g(x), h(x) the greatest common factor of d(x), is f(x) = d(x) p(x), g(x) = d(x) q(x), h(x)(x) = d r(x), with p(x), q(x), r(x) is a relatively prime polynomials. Then,(f(x), g(x), h(x)) =(d(x) p(x), d(x) q(x) r(x))(x) = d(p(x), q(x) r(x)). Since p(x) and q(x) r(x) relatively prime, so(p(x), q(x) r(x)) =(p(x), q(x))(p(x), r(x)), so(f(x), g(x) h(x)) =(f(x), g(x))(f(x), h(x)). The certificate is completed.

Now back to the subject, we have:

(f(x) ⁿ, g(x) ⁿ) =((f(x), g(x)) ⁿ)(according to lemma)

(f(x)(x, g) ⁿ, g(x) ⁿ) =((f(x)(x, g), g(x)) ⁿ)(according to the same lemma)

notice(f(x) ⁿ, g(x) ⁿ) and(f(x)(x, g) ⁿ, g(x) ⁿ) in f [x] generated on the same ideal, because they contain f(x) ⁿ and g(x) ⁿ, and according to lemma are:

((f(x), g(x)) ⁿ) ⊆((f(x)(x, g), g(x)) ⁿ)

((f(x)(x, g), g(x)) ⁿ) ⊆((f(x), g(x)) ⁿ)

Therefore, the two ideals are equal. The certificate is completed.

duanchigui 注册会员
2023-02-28 02:52

Answer completely, don't stand up, OK?

dht8064 注册会员
2023-02-28 02:52

  • First, we define an ideal I on the polynomial ring F[x] as the set generated by the polynomial f(x) in F[x], denoted I =(f(x)). That means that any member of I can be represented as a linear combination of f(x).
  • Next, let's prove the equation in question.
  • First, since F is a field of characteristic zero, F[x] is an entire ring. That means that if f(x) and g(x) are not all zero, then f(x) and g(x) are not all zero.
    < li > then, we come to prove(f(x) ⁿ, g(x) ⁿ) ⊆(f(x)(x, g) ⁿ).
  • For any polynomial a(x) and b(x), we have:

a ⁿ f(x)(x) ⁿ + b(x)(x) ⁿ ⁿ g =(a(x) f(x) + b(x) g(x)) ⁿ

  • This is a direct application of the binomial theorem. Therefore, a ⁿ f(x)(x) ⁿ and b(x)(x) ⁿ ⁿ g in(f(x)(x, g) ⁿ). Therefore, arbitrary(f(x) ⁿ, g(x) ⁿ) are the elements of(f(x)(x, g) ⁿ).
    < li > then, we have to prove(f(x)(x, g) ⁿ ⊆(f(x) ⁿ, g(x) ⁿ).
    < li > hypothesis h(x)(f(x)(x, g) ⁿ. Then, there exist polynomials p(x) and q(x) such that:

h(x) = p(x) f(x))(x, g + q(x) g(x)(x, g)

  • According to the binomial theorem, we have:
ⁿ = ∑

h(x) = 0 to n(k), C(n, k) [f(x) ⁿ - k g(x) ⁿ] [g f(x)(x) ⁿ - k ⁿ] ^ k

    < li > this expansion will each item in the respectively with p(x) f(x)(x, g) and q(x) g(x, g(x)), we can get: < / li >
    < li > for each k, C(n, k) f(x) ⁿ - k g(x) ⁿ in(f(x) ⁿ, g(x) ⁿ).
    < li > because(f(x) ⁿ, g(x) ⁿ) is an ideal, so arbitrary linear combination are(f(x) ⁿ, g(x) ⁿ). Therefore, p(x)(x, f, g(x)) in(f(x) ⁿ, g(x) ⁿ).
    < li > equally, for each k, C(n, k) g f(x)(x) ⁿ - k ⁿ in(f(x) ⁿ, g(x) ⁿ). Therefore, q(x) g(x, g(x))(f(x) ⁿ, g(x) ⁿ).
    < li > therefore, h(x) ⁿ in(f(x) ⁿ, g(x) ⁿ). This means(f(x)(x, g) ⁿ) ⊆(f(x) ⁿ, g(x) ⁿ).
    < li > what has been discussed above, we proved that(f(x) ⁿ, g(x) ⁿ) =(f(x)(x, g) ⁿ).

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Publish Time
2023-02-28 02:52
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2023-02-28 02:52