Simple Field Extensions Misleading Question?

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Simple Field Extensions... Misleading Question?

This is from Pinter's A Book of Abstract Algebra. The section of exercises is called "Simple Extensions."

[page 278]

3. a+c is a root of p(x) iff a is a root of p(x+c) ...

I was thinking about the question in terms of field extensions. I attempted to use the fact that F(a)=F(a+c), which I proved in a preceding exercise. That is, the field generated by F and a is equal to the field generated by F and (a+c). But, I simply noticed that ...

If a+c is a root of p(x), then p(a+c)=0. Now, considering the polynomial p(x+c), if we plug in a we get p(a+c), which we know equals 0. [The converse is similar.]

Seemed way too easy, and thus I am very nervous that I missed something, especially since the section focused on properties of simple field extension.

Am I correct? Thanks for the help! =)
 
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Yeah, your proof seems correct. It's a pretty weird question, let me check pinter.
 


I guess that Pinter includes that exercise as some kind of "hint" towards exercise 4.
 


micromass, thank you so much! I am glad you feel its a little weird too. It seems misplaced. The previous chapter was entitled "Substitution in Polynomials." This question, in my opinion, would fit nicely there. One of the great things about Pinter is that the heading of each exercise set gets the reader thinking in a certain direction. But, when an odd question shows up, the novice starts to feel a little doubtful when the answer required nothing from the section. =)
 


Oh, there we go! Just took a look at exercise 4. Good ol' Pinter, never disappoints! :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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