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

[rookandpawn]

Homework Statement

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)

Homework Equations

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

The Attempt at a Solution

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


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

\sum{b_{i}(x+c)^i} \cdot \sum{c_{i}(x+c)^i} ?

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

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

I know \phi is a surjective homomorphism; so if f(x) is reducible,
\phi(f,x+c) = \phi(gh,x+c) = \phi(g,x+c)\phi(h,x+c) but i cannot take that last part and equate it to g(x+c)h(x+c) unless \phi 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 \gamma so i could simply
do this : \gamma(f, x+c) = \gamma(gh, x+c) = \gamma(g,x+c)\gamma(h,x+c) = g(x+c)h(x+c) but i don't 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 \phi(f,x+c) = f(x+c); then showing if \phi(f(g),x)=\phi(f,\phi(g,x)) 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 what's going on. I know there are many things going on here. Please help. THanks.

 

[matt grime]

f(x) = g(x)h(x) for g,h in F[x].

Thus f(x+c)=...

There was no need to write more than a few symbols and, what, 4 words?

 

[rookandpawn]

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 that's 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 \underbrace{\sum{a_{i}x^i}}_{f(x)} = \underbrace{\sum{b_{i}x^i}\cdot \sum{c_{i}x^i}}_{g(x)h(x)}, but i can't "plug in" x+c on the LHS because I am not evaluating it. The equality is all that is given. this is why i was discussing a modified evaluation map \phi(f,x):F[x]\rightarrow F[x] = the polynomial f "evaluated" not at x in F but at polynomial x in F[x].

 

[Dick]

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.

 

[rookandpawn]

Right, but that's 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 a lot of work. Thus i wanted to ask about alternative methods.

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.

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) doesn't 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

 

[Dick]

Right, but that's 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 a lot 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) doesn't 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

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).