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Abstract algebra: f(x) is reducible so is f(x+c)

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Let F be a field and f(x) in F[x]. If c in F and f(x+c) is irreducible, prove f(x) is irreducible in F[x]. (Hint: prove the contrapositive)

    2. Relevant equations

    So, im going to prove if f(x) is reducible then f(x+c) is reducible.

    3. The attempt at a solution

    f(x) = g(x)h(x) for g,h in F[x]. If [tex]f(x) = \sum{a_{i}x^i} = \sum{b_{i}x^i} \cdot \sum{c_{i}x^i} = g(x)h(x)[/tex] then [tex]f(x+c) = \sum{a_{i}(x+c)^i}[/tex]

    Now i just actually work through the algebra and after all is said and done, i should see that its equivalent to

    [tex]\sum{b_{i}(x+c)^i} \cdot \sum{c_{i}(x+c)^i}[/tex] ?

    Now, if this is a correct approach, i was thinking it involves alot of work.

    I was thinking about the evaluation homomorphism [tex]\phi_{x+c}:F[x] \rightarrow F
    [/tex] given by (including notational convenience) [tex]\phi_{x+c}(f(x)) = \phi(f,x+c) =
    f(x+c)[/tex] i.e. f evaluated at element x+c.

    I know [tex]\phi[/tex] is a surjective homomorphism; so if f(x) is reducible,
    [tex]\phi(f,x+c) = \phi(gh,x+c) = \phi(g,x+c)\phi(h,x+c)[/tex] but i cannot take that last part and equate it to [tex]g(x+c)h(x+c)[/tex] unless [tex]\phi[/tex] were injective.

    Now, I know that in infinite field F, F[x] is isomorphic to the ring of its induced functions. That is to say, if F is infinite, then any two polynomials that look the same, act the same.
    And vice versa, if two induced functions act the same, they are the same looking polynomial.

    I am desiring some kind of isomorphism, call it [tex]\gamma[/tex] so i could simply
    do this : [tex]\gamma(f, x+c) = \gamma(gh, x+c) = \gamma(g,x+c)\gamma(h,x+c) = g(x+c)h(x+c)[/tex] but i dont know how to get there.

    Someone had suggested adapting evaluation homomorphism instead of from F[x] to F, i have it go F[x] to F[x] by treating the x as a polynomial instead of an element of F
    such that [tex]\phi(f,x+c) = f(x+c)[/tex]; then showing if [tex]\phi(f(g),x)=\phi(f,\phi(g,x))[/tex] where that f(g) is formal composition and the right hand side is functional composition, it would be relevant.

    Please, any thoughts? Very curious to know more of whats going on. I know there are many things going on here. Please help. THanks.
  2. jcsd
  3. Oct 15, 2007 #2

    matt grime

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    Thus f(x+c)=.....

    There was no need to write more than a few symbols and, what, 4 words?
  4. Oct 15, 2007 #3
    isn't the factorization g(x)h(x) just for the polynomial described by f(x)?

    the polynomial f(x+c) is a fundamentally different polynomial? I realize that f evaluated at x+c can be thought of as g evaluated at x+c times h evaluted at x+c, but thats as far as i can get? just because the evaluation is the same doesn't mean i can go back and assume that i got their via the function induced by the polynomial g(x+c) and the function induced by h(x+c)?

    basically i feel that f(x) = g(x)h(x) expresses an identity namely, the polynomial [tex]\underbrace{\sum{a_{i}x^i}}_{f(x)} = \underbrace{\sum{b_{i}x^i}\cdot \sum{c_{i}x^i}}_{g(x)h(x)}[/tex], but i can't "plug in" x+c on the LHS because im not evaluating it. The equality is all that is given. this is why i was discussing a modified evaluation map [tex]\phi(f,x):F[x]\rightarrow F[x] = [/tex] the polynomial f "evaluated" not at x in F but at polynomial x in F[x].
    Last edited: Oct 15, 2007
  5. Oct 15, 2007 #4


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    Let's try this. x^2+x=x(x+1). Change x to x+c. (x+c)^2+(x+c)=(x+c)(x+c+1). Expand both sides. Magically both are x^2+x+2cx+c^2+c. f(x+c) factors the same way as f(x). It's not even really magic. x is just a symbol, and so is x+c. I can use them interchangably. x is just easier to write than x+c.
  6. Oct 15, 2007 #5
    Right, but thats just verifying one instance. It goes back to what i was saying earlier, that i have to write the formal product g(x+c)h(x+c) (meaning if g(x) = b_0 + b_1x^1 + .. b_nx^n and h(x) = c_0 + ... + c_mx^m then g(x+c)h(x+c) = b_0c_0 + ... + b_nc_mx^{n+m} which if i actually wrote it out more is much more work than that) and then showing that its in fact equal to f(x+c) = a_0 + a_1(x+c)^1 + ... + a_n(x+c)^n

    This is the basis of my question because i said that is only one method which involves alot of work. Thus i wanted to ask about alternative methods.

    really? i thought you can only use them interchangebly when you are talking about the polynomial function, not the polynomial.

    is it true that x in f(x) doesnt represent anyhting; the x in general represents an indeterminate in the ring of polynomials, but f(x) merely denotes a polynomial in F[x]. like, we can't "pull out" the x in f(x) and talk about that ? the x in the induced function f(x) from the ring of induced functions which is induced from the polynomial f(x) in F[x] represents a "pluggable" variable
    Last edited: Oct 15, 2007
  7. Oct 15, 2007 #6


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    It's not just verifying one instance - and I didn't HAVE to multiply it out, I just wanted to emphasize that it's concretely true. f(x+c) IS a different polynomial from f(x). You can see that in the example. But the factorization f(x)=g(x)h(x) is true for ALL x. So f(y)=g(y)h(y), f(z)=g(z)h(z), f(pi+1)=g(pi+1)h(pi+1) and even f(x+c)=g(x+c)h(x+c).
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