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I was reading the following solution here Prove that $E=F[\alpha^2]$ and I was not sure how is this statement in a solution is correct? "since $\alpha$ is a root of $x^2-\alpha^2$ in$F[\alpha^2][x],\dots$ " as far as I know that $\alpha \notin F[\alpha^2],$ could someone clarify this to me please? Why can not we say the $x^2 - \alpha^2$ is a minimal polynomial of $\alpha$ in $F[\alpha^2]$?

Also, is there any way that the degree of an extension be zero? And why is the extension of a field by itself has degree one?

Any clarification will be greatly appreciated!

Edit:

Here is the full answer written there:

Since $\alpha^2\in F[\alpha]$, $F[\alpha^2]\subseteq F[\alpha]$. Thus $[F[\alpha]:F]=[F[\alpha]:F[\alpha^2]]*[F[\alpha^2]:F]$ since $\alpha$ is a root of $x^2-\alpha^2$ in$F[\alpha^2][x]$, the extension $[F[\alpha]:F[\alpha^2]]\leq 2$. Thus it must be 1 since the total extension is of odd degree showing that $F[\alpha]=F[\alpha^2]$.

user26857
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Emptymind
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  • I think it would be wise to copy the statement (especially since it is not long) of the exercise. A hyperlink is not reliable because it can break. It also simplifies reading. After that, I’d be happy to answer the question. (PS: I am not the downvote.) – NaNoS Feb 07 '24 at 09:21
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    We have $[F:F]=1$ by definition, because $F$ is a $1$–dimensional $F$–vector space. This answers your last question. You should review the Definition from linear algebra. – Dietrich Burde Feb 07 '24 at 09:22
  • Thanks I will edit my post@NaNoS – Emptymind Feb 07 '24 at 09:22
  • Thank you, I was just scared from the problem that is why I forgot that . what about the rest of my questions @DietrichBurde? – Emptymind Feb 07 '24 at 09:26
  • @xxxxxxxxx I read them but nothing answered my first questions, I think $\alpha$ should be in $F[\alpha^2]$ for this to happen but this is not correct. What do you think? – Emptymind Feb 07 '24 at 09:28
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    I agree it's not clear but what is being asserted is not that $\alpha\in F[\alpha^2]$, but that $\alpha$ is a root of the polynomial $X^2-\alpha^2)$ which polynomial is a member of $F[\alpha^2][x]$. – ancient mathematician Feb 07 '24 at 09:33
  • why your variable in the polynomial is capital X? @ancientmathematician – Emptymind Feb 07 '24 at 09:46
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    @Emptymind that's a bit odd question, it's extremely common to use capital $X$ when talking about polynomial rings $F[X]$. Might be to distinguish polynomials from polynomial functions. – Ennar Feb 07 '24 at 09:51
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    @Emptymind I always call by polynomial placeholder $X$ and made a coupe of typos here, sorry. – ancient mathematician Feb 07 '24 at 14:13

1 Answers1

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To show that $\big[F[\alpha] : F[\alpha^2]\big] \leq 2$, it suffices to show that $\alpha$ is a root of a quadratic polynomial with coefficients in $F[\alpha^2]$. But now $T^2 - \alpha^2 \in F[\alpha^2][T]$ is appropriate.

For the second question, if you have a field extension $K \to L$ such that $[L : K] = 0$, it basically means that $L$ is the zero vector space over $K$, so $L = 0$ and $K = 0$ (because $L$ contains a copy of $K$ thanks to the extension). Now it depends on the definition of a field you choose, mine is that a field must have at least two elements (the neutral for addition and the neutral for multiplication) so it is not possible for me. If your definition is different and that $0$ is a field, then I have to think about it.

NaNoS
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  • what about my question about that $x^2 - \alpha^2$ is a minimal polynomial of $\alpha$? Can we say this? – Emptymind Feb 07 '24 at 09:50
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    We just proved that $F[\alpha] = F[\alpha^2]$ so actually $\alpha \in F[\alpha^2]$ and *the* minimal polynomial of $\alpha$ is just $T - \alpha$. There are situations where $T^2 - \alpha^2$ and $T - \alpha$ are the same polynomials (if char$(F) = 2$ and if $\alpha \in \mathbb F_2 \subset F$ but this is trivial). – NaNoS Feb 07 '24 at 10:04