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# 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.

dingyan124 注册会员
2023-02-28 02:52

Lemma: Let $A$and $B$be elements of F[x], so

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;

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

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).

ctc1027 注册会员

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