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I'm reading a proof of the theorem that $F(a)$ is isomorphic to $ F[x]/\langle p(x) \rangle$ in the book called Contemporary Abstract Algebra by Gallian, and I don't understand some key elements of the proof [I think this book is too confusing in numerous places in the sections on rings and fields, but it was clear and straightforward on groups]. Here are the theorem and the proof:

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Here's what I do not understand:

(1) If $\langle p(x) \rangle$ is a maximal ideal in $F[x]$ and $\ker \phi\ne F[x]$, then how does it follow from the First Isomorphism Theorem for Rings and the corollary (below) that $\phi(F[x])$is a subfield of $F(a)$?

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(2) When the author writes that

$\phi(F[x])$ contains both $F$ and $a$ and recalling that $F(a)$ is the smallest such field ...

what does he refer to by "the smallest such field"? The smallest what field?

(3) Can't one immediately see the image of $\phi$ is indeed $F(a)$, just by definition? $\phi$ takes all polynomials in $F[x]$ and evaluates them at $a$. I don't understand the author's argument at all.

(4) The author also writes that

enter image description here

But isn't the basis for $F(a)$ one-dimensional, after all?

Sorry for the long message. Would really appreciate your help in understanding this proof. In my opinion the author is skipping too many details. In my university, I'd get reduced marks for unproved statements like "clearly, [...]".

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  • $\begingroup$ $F(a)$ is a $F$-vector space, this is what we are going to prove, by showing that $F[x]/(p(x))$ is a field (and obviously a $F$-vector space) and that $\phi : F[x]/(p(x)) \to F[a]$ is a surjective ring homomorphism with $\phi(c) = 0 \implies c= 0$ so that it is an isomorphism. Hence $F[a]$ is a field, i.e. $F[a] = F(a)$. $\endgroup$ – reuns Mar 12 '17 at 7:53
  • $\begingroup$ @user1952009 What do you mean by "$F$-vector space"? $\endgroup$ – sequence Mar 12 '17 at 9:05
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    $\begingroup$ If $F$ is a field then $F^d$ is a $d$-dimensional $F$-vector space. Here $F[a] = \{ \sum_{n=0}^{d-1} c_n a^n, c_n \in F\}$ is a $d$-dimensional $F$-vector space, where $d = deg(p)$. And since we are proving $F[a]$ is a field, we get $F[a] = F(a)$. $\endgroup$ – reuns Mar 12 '17 at 10:24
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(1) $\langle p(x)\rangle$ is a maximal ideal in $F[x]$. According to Theorem 17.5, $p(x)$ is irreducible over $F$. According to Corollary 1, $F[x]/\langle p(x)\rangle$ is a field. But $\ker{\phi}=\langle p(x)\rangle$. According to the First Isomorphism Theorem of Rings, $\phi(F[x])\cong F[x]/\langle p(x)\rangle$, showing that the image of $\phi$ is a subfield of $F(a)$.

(2) $F(a)$ is the smallest field containing both $F$ and $a$ by definition (see the discussion above the Definition in page 340).

(3) Yes, and the author proved it easly as you say (end of first paragraph). But since the image of $\phi$ is $F(a)$, it follows from the result in (1) that $F[x]/\langle p(x)\rangle\cong F(a)$

(4) Why do you think that $F(a)$ is one-dimensional? Is it because you see it as a vector space over itself? $\mathbb{C}$ is one dimensional as a $\mathbb{C}$-vector space, but $2$-dimensional as an $\mathbb{R}$-vector space. Even $\mathbb{R}$ can be infinite dimensional if you consider it as a vector space over $\mathbb Q$. $F(a)$ is surely one dimensional as a vector space over itself, like any field. But over $F$, that's not the case because $a\notin F$ (otherwise $p(x)$ wouldn't be irreducible), but given the way any element can be expressed in $F(a)$ as shown in the theorem, the dimension of $F(a)$ as a vector space over $F$ is exactly the degree of $p(x)$.

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    $\begingroup$ (1) You should know what things are before moving on to their properties. Always refer to definitions. According to the paragraph right above the definition in page 340, $F(a)$ is by definition the smallest field containing $F$ and $a$. In particular, it is a field. (2) Concerning $F[a]$, that's just an error. I meant $F(a)$, not $F[a]$. Now, concerning what you said "Isn't $F(a)$ an evaluation of all polynomials in $F[x]$ at $a$?" The answer is yes; in fact that's the meaning of $\phi(F[x])=F(a)$. But by definition in your book, $F(a)$ is THE smallest field containing $F$ and $a$... $\endgroup$ – Scientifica Mar 12 '17 at 9:39
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    $\begingroup$ ... It turns out that that's the same thing as shown in the theorem. (3) One doesn't prove what's a definition, that's right? But as I said, by definition, $F(a)$ is the smallest field containing $F$ and $a$. So, the obvious fact that $F(a)=\phi(F[x])$ is not a definition, so it should be proven. $\phi$ is a map from $F[x]$ to $F(a)$ given by $\phi(f(x))=f(a)$. It makes sense, i.e, for any $f(x)\in F[x]$, $f(a)$ is indeed in $F(a)$ because it is a sum and products of elements in $F\subset F(a)$ and $a\in F(a)$. Hence, $\text{Im}\phi\subset F(a)$. Why equality holds?... $\endgroup$ – Scientifica Mar 12 '17 at 9:43
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    $\begingroup$ ... To show equality, we can show that $F(a)\subset\text{Im}\phi$, right? What does it mean that $F(a)$ is the *smallest field containing $F$ and $a$? It means the following: "For any field $K$ such that $F\subset K$ and $a\in K$, we have $F(a)\subset K$". As shown in (1), $\text{Im}\phi$ is a subfield of $F(a)$. In particular, $\text{Im}\phi$ is a field. Furthermore, it contains $F$ and $a$. One has $F\subset\text{Im}\phi$ because for any $g\in F$, let $f(x)=g\in F[x]$. Then $f(a)=g\in\text{Im}\phi$. One also has $a\in\text{Im}\phi$ because for $f(x)=x\in F[x]$, $f(a)=a\in\text{Im}\phi$... $\endgroup$ – Scientifica Mar 12 '17 at 9:47
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    $\begingroup$ ...Thus $\text{Im}\phi$ is a field containing $F$ and $a$. By definition of $F(a)$, we have $F(a)\subset\text{Im}\phi$. Combining this with $\text{Im}\phi\subset F(a)$, we conclude that $F(a)=\text{Im}\phi$. If you have another question, feel free to ask it, @sequence . But make sure you understand the definitions of things ;) $\endgroup$ – Scientifica Mar 12 '17 at 9:50
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    $\begingroup$ You're welcome @sequence :) Well you should know that in mathematics, some things can have many definitions, but all of which are equivalent, i.e, if you have two statements that represents definitions of something, the two are said to be equivalent definitions if when you consider the definition as one of them, you can prove that what's stated in the other holds. In your case, you gave two different definitions of $F(a)$: One is "the smallest field containing $F$ and $a$". The other is $F(a)=\left\{\frac{g(a)}{h(a)}\mid g(x),h(x)\in F[x]\right\}...$ $\endgroup$ – Scientifica Mar 12 '17 at 23:43
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(1) The First isomorphism theorem tells you that $\;F[x]/\ker\phi\cong\phi(F[x])\;$ , and corollary 1 tells you that $\;F[x]/\langle p(x)\rangle\;$ is a field whenever $\;p(x)\;$ is irreducible. Both things, together with $\;\ker\phi=\langle p(x)\rangle\;$ gives you the result you ask about.

(2) Smallest such field" means here " the smallest field containing both $\;F\;$ and $\;a\;$" .

(3) Well, by its mere definition, $\;\phi(g(x)):=g(a)\;$ , so $\;\phi(F[x])=F[a]\;$ . Yet, as $\;a\;$ is algebraic over $\;F\;$ , we have that $\;F[a]=F(a)\;$ and we're done.

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  • $\begingroup$ For (1), the problem is that it is not obvious that $\phi(F[x])$ is a subfield of $F(a)$ from the corollary and the First Isomorphism Theorem for rings. Can you please explain how exactly it may follow so? $$ $$ For (2), how do we know that $F(a)$ is the smallest field containing $F$ and $a$? I don't think this is obvious. I.e., how do we know that $F(a)$ actually contains $F$? $$$$ For (3), Sorry, didn't understand. What does it mean to say "algebraic over"? This is upcoming in the book, but it's not been covered up to the page I'm reading. $\endgroup$ – sequence Mar 12 '17 at 7:43
  • $\begingroup$ @sequence What isn't clear? I think you may be missing some details: The FIT tells you the first isomorphism. That much must be clear. And now the corollary gives you that the left hand (and thus the right one, too) in that isomorphism is a field. (2) That is the definition of that, but you can also almost trivially prove it: any field containing both $\;F,a\;$ must contain any rational function over $\;F\;$ in $\;a\;$ , i.e.: $\;F(a)\;$ ...If you don't know what is "algebraic element" then I think you're attempting something you're still not ready for. $\endgroup$ – DonAntonio Mar 12 '17 at 7:51
  • $\begingroup$ Cont. : "An element $\;a\;$ is algebraic over a field $\;F\;$ " means there existis $\;0\neq p(x)\in F[x]\;$ s.t. $\;p(a)=0\;$ . $\endgroup$ – DonAntonio Mar 12 '17 at 7:51
  • $\begingroup$ (1) Im $\phi$ is isomorphic to $F[x]/\langle p(x)\rangle$, that is clear. That Im $\phi$ is a field is also clear. But it's not clear why $\phi(F[x])$ must be a subfield of $F(a)$. I don't see how this follows from the preceding. (2) How do you prove that any field containing $F$ and $a$ must contain any rational function over $F$ in $a$? For algebraic elements, this has not yet been covered in the book (up to and including the page I'm reading on). So the author must be assuming something that has been covered so far, I believe. @DonAntonio $\endgroup$ – sequence Mar 12 '17 at 9:17
  • $\begingroup$ @sequence Definition: $\;F(a):=\left\{\,\frac{p(a)}{q(a)}\;/\;p(x),\,q(x)\in F[x]\;,\;\;q(a)\neq0\,\right\}\;$ , and since $\;\phi(h(x)):=h(a)\;$ , I think the inclusion is now clear... $\endgroup$ – DonAntonio Mar 12 '17 at 9:25

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