# I Field Extensions and "Free Parameters"

1. May 30, 2017

### Math Amateur

I am reading "Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with some remarks of Lovett pertaining to Theorem 7.1.10 ...

The statement of Theorem 7.1.10 reads as follows (page 325) :

The remarks pertaining to Theorem 7.1.10 read as follows (page 326) :

I do not understand the use of 't' in the above text ... nor do I fully understand the analysis involving it ...

My specific question is as follows:

In the above remarks, Lovett writes the following:

" ... ... For example keeping $t$ as a free parameter, $F[t]$ is a subring of $F(t)$. ... ... "

What is $'t'$ and why exactly are we introducing it?

Why not stick with $x$ and $F[x]$ and $F(x)$ ... ... ?

I note that Lovett does not usually introduce a "free parameter" (whatever that is?) and happily deals with the indeterminate $x$... ... ?? ... ... so ... indeed, one may ask when is a "free parameter" necessary and when is it not needed ...

Hope someone can help ...

Peter

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2. May 31, 2017

### andrewkirk

$F(\alpha)$ is the field generated by $F\cup\{\alpha\}$, and $F[x]$ is the ring of polynomials with coefficients in $F$.

But what does $F[\alpha]$ denote? A quick internet search for use of that notation draws a blank.

Does the author explain what he means by $F[\alpha]$? I feel we need to make sense of that before we try to make sense of $F[t]$.

Failing that, does the other book you were reading about this recently (Dummitt & Foote?) use that notation and if so do they define it? I have a vague memory of seeing something like $F[\alpha]$ in one of your earlier posts about the other book, but I may be misremembering.

3. May 31, 2017

### Staff: Mentor

In general $\mathbb{F}[.]$ denotes a ring over $\mathbb{F}$ and $\mathbb{F}(.)$ denotes a field, the quotient field of $\mathbb{F}[.]$.

In case it is $\mathbb{F}[t]$ or $\mathbb{F}[x]$ we usually mean indeterminates, i.e. a variable $t$ or $x$ and polynomial rings where $t^{-1} \notin \mathbb{F}[t]$ and $x^{-1} \notin \mathbb{F}[x]$, and their quotient fields where $t^{-1} \in \mathbb{F}(t)$, resp. $x^{-1} \in \mathbb{F}(x)$.

The indeterminates behave like transcendental numbers, as they don't obey any algebraic equation. The only difference is an intuitive one: we do not replace numbers - algebraic or transcendental - by other numbers, but we are used to replace indeterminates by numbers, such as $\mathbb{F}[x] \longrightarrow \mathbb{F}[\sqrt{2}]$.

Now what the author means is, that $\mathbb{Q}[\pi]$ and $\mathbb{Q}(\pi)$ are not the same algebraic object. The first is only a ring, an infinte ring extension of the rationals, because $\frac{1}{\pi} \notin \mathbb{Q}[\pi]$, whereas the second is a field: $\frac{1}{\pi} \in \mathbb{Q}(\pi)$.

On the other hand, the situation changes, if we talk about algebraic numbers. In this case is $\mathbb{F}[\alpha] = \mathbb{F}(\alpha)$.
Can you show, why $\frac{1}{\alpha} \in \mathbb{F}[\alpha]$ if $\alpha$ is algebraic over $\mathbb{F}$?

4. May 31, 2017

### andrewkirk

I'm afraid I still don't follow this. Are you able to write a formal definition of the symbol string $F[\alpha]$?

A formal definition of the symbol string $F[x]$, where $F$ denotes a field, is as the set $F^*\triangleq \bigcup_{k=1}^\infty F^k$ (ie the set of all finite $k$-tuples of elements of $F$, where $k$ is any positive integer) with operations of addition and multiplication defined so as to correspond with those operations for polynomials that have coefficients equal to the components of the $k$-tuples.

With this definition the symbol string $[x]$ is a postfix, unary operator that maps a field to its corresponding polynomial ring. Writing any letter other than $x$ between square brackets is an undefined symbol string. So we need a new definition to make sense of symbol strings like $F[\alpha]$ or $F[t]$.

Thank you

5. May 31, 2017

### Staff: Mentor

Another remark: Sometimes the quotient fields $\mathbb{F}(x)$ or $\mathbb{Q}(\pi)$ are written $\mathbb{F}[x,\frac{1}{x}]$ and $\mathbb{Q}[\pi,\frac{1}{\pi}]$ which explicitly makes sure that the inverse elements are adjoint as well.

6. May 31, 2017

### Staff: Mentor

Yes, it's the image of the projection $F[x] \twoheadrightarrow F[x]/(m_\alpha(x))$ where $m_\alpha(x)$ is a unitary irreducible polynomial over $F$. The elements in this image can be written as polynomials in $\alpha$ where a certain polynomial equation equals zero, namely $m_\alpha(x)=0$ in which case we call the zeros of it, e.g. $\alpha$ algebraic. All "numbers" $\tau$ which do not have such an equation satisfy $F[x] \cong F[\tau]$ and are called transcendental numbers. Numbers in this context refer to elements in some field extension of $F$ which has to be given otherwise. One could also start with such a field extension, i.e. a given algebraic number $\alpha$ and define $F[\alpha]$ as the image of $F[x] \longrightarrow F[\alpha]$ by substitution with the arithmetic in the extension field.
The most formal definition of a polynomial ring, yes.
$x$ is not always the indeterminant. Some authors use $t$. I don't think that it is an undefined symbol string, as for writing it $x, t$ or $\alpha$ is only a convention. So formally $F[\alpha]$ would be the polynomial ring in $\alpha$ over $F$. But I agree, that $\alpha$ usually denotes a number in some extension field of $F$ that satisfies a polynomial equation, i.e. the quotient ring. So formally we introduce an algebraic dependency among the coefficients of elements in $F[x]$.

I'm not sure if this answers the question. I only wanted to say that (usually) $F[.]$ brackets denote rings whereas $F(.)$ parantheses denote their quotient fields. In case of algebraic numbers they are the same, in case of transcendental numbers (or indeterminants) they are not.

7. May 31, 2017

### Math Amateur

Thanks Andrew and fresh_42 ... the picture is becoming clearer ...

I just thought I would present Lovett's explanation of $F[ \alpha ]$ and $F( \alpha )$ ... and also $F[x]$ ...

Lovett's first significant mention of such symbols is in Section 5.2: Rings Generated by Elements (pages 216 - 225) ...

In Section 5.2.1 Generated Subrings, Lovett defines $R$ where $R$ is a subring of a commutative ring $A$, and $S$ is a subset of $A$ ... ... as follows:

Lovett then goes on to define polynomial rings $R[x]$ ... in section 5.2.2 ...

The start of Section 5.2.2 reads as follows:

Lovett then goes on to define/discuss addition and multiplication of polynomials ...

I note that in defining polynomial rings Lovett uses the term "the variable $x$" and not the mysterious term "indeterminate" for $x$ ... ...

In the context of fields, Lovett mentions $F[ \alpha ]$ and $F( \alpha )$ in Section 7.1 Introduction to Field Extensions ... . The reference is in the context of ways to construct some field extensions ... and reads as follows:

Hope the above helps ...

Thanks again for your clarifications and help ...

Peter

EDIT ...I have no idea why my text is crossed out ???

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8. May 31, 2017

### Math Amateur

Hi Andrew, fresh_42 ... ...

In order to further clarify my puzzlement at the differences between $F[x]$ and $F[t]$ I thought i would present the theorem which first convinced me I need a better understanding of t as a "free parameter" ... ...

The Theorem concerned is Theorem 7.6.3 in Lovett, Section Splitting Fields and Algebraic Closure ... ... the theorem and its proof read as follows:

In the text above from Lovett, in the third paragraph of the proof, we read the following:

" ... ... In $F'[t], (t - \overline{x} )$ is a linear factor of $f(t)$ ... ... "

I am not sure why we are introducing $t$ ... and, further, I am not sure of the nature of $t$ ...

Is it just a variable in $F'$ where we want to distinguish it from $x$ which is a variable in $F$ ... is that all there is to it ..

Mind you, later we read ...

" ... ... If

$f(x) = \sum_{ i = 0 }^{ n + 1 } a_i x^i$

then $f(t) = 0$ is equivalent to $f(t) - f( \overline{x} ) = 0$ ... ... "

I cannot see why $f(t) = 0$ is equivalent to $f(t) - f( \overline{x} ) = 0$ ... ... ?

Can you help?

Peter

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9. May 31, 2017

### andrewkirk

OK it's starting to take shape now. This is how it looks to me, based on yours and fresh's posts:

• Let $F$ be a field that is a sub-field of field $E$.
• Let $\alpha$ denote an element of $E-F$. In what follows $we can replace$\alpha$by any letter other than$x$•$F[x]$denotes the ring of polynomials with coefficients in$F$. In this notation, the use of$x$is essential. It cannot be replaced by any other letter, unless the author provides their own definition about what they mean by that. •$F(\alpha)$denotes the field generated by$F\cup\{\alpha\}$. It will be a subfield of$E$. •$F[\alpha]$denotes the ring generated by$F\cup\alpha$. This is a subring of$F(\alpha)$because the latter is the field generated by the generators and the former is the ring generated by them. Field generation generates a (not necessarily strictly) bigger set than ring generation because it includes the creation of inverses, which ring-generation doesn't. • When Lovett writes$F[\alpha]=F(\alpha)$he is saying that, given the premise that$[F(\alpha):F]$is finite, ring generation gives the whole field. When Lovett writes$F[t]$or$F(t)$, he is trying to indicate that$t$is a general element of$E-F$. But I don't think he needs to change letters to do that, as no particular properties have been ascribed to the symbol$\alpha$. So when he says for instance that '$t$is not a unit of$F[t]$' he is just saying that, for any$t\in E$, the ring$F[t]$does not necessarily contain a multiplicative inverse for$t$. Nor would we expect it to, since the creation of mult inverses is not part of the ring generation process (unlike the field generation process). If$t\in F$it will have a mult inverse in$F[t]$, because$F\subseteq F[t]$and$F$is a field. But if$t\in E-F$it will not necessarily have a mult inverse in$F[t]$, We can replace the$t$s in the previous para by$\alpha$s if we like. 10. May 31, 2017 ### Math Amateur Thanks Andrew ... I think that cleared things up a lot .. Just a small point ... you write ... " ... ...$F[x]$denotes the ring of polynomials with coefficients in$F$. In this notation, the use of$x$is essential. It cannot be replaced by any other letter, unless the author provides their own definition about what they mean by that. ... " Yes, OK ... but ... what is the special nature of x ... what are the defining properties of this special symbol x ... Just one other point ... You write... ... " ... ... When Lovett writes$F[t]$or$F(t)$, he is trying to indicate that$t$is a general element of$E-F$. ... ... " So t is just what we normally refer to as a "variable" ... in contrast to the special nature of x ... is that right? Hope you can help ... Peter . 11. Jun 1, 2017 ### andrewkirk In symbolic logic we talk in terms of strings of symbols. The symbols are all either • variable symbols, which means what you think it would mean: things like$x$,$t$,$F$,$\alpha$• constant symbols. These are things like$\pi$,$0$,$\emptyset$,$\mathbb R$,$\aleph_0$and$e$and refer to fixed elements of the domain of discourse • function symbols, which are things like the$f$in$f(x)$• predicate symbols, which are like function symbols except that they return values of True or False • logical symbols, which are things like$\to$for implication,$\neg$for negation and$\forall$and$\exists$for quantification When we write$F[x]$to denote the ring of polynomials with coefficients in$F$, the squiggle$x$is not any of those sorts of symbols. Rather, it is a part of the indivisible function symbol$[x]$, because$F[x]$denotes, in postfix notation, the application of the function$[x]$to the argument$F$. So$x$in that context has no meaning, just as the horizontal bar in the middle of a capital A has no meaning. It is a 'cross mark', not a letter of the roman alphabet. Replacing that cross mark by the curly squiggle that looks like an$\alpha$would be analogous to replacing the horizontal bar in the capital A by an$\alpha$squiggle. We would get a letter/symbol that is simply not in our alphabet. In order to be able to interpret the symbol string$F[\alpha]$we need an additional definition, and that's what we worked towards above. With that preamble, consider the two sets of markings$F[x]$and$F[t]$.$F[x]$denotes the result of applying the function$[x]$to the argument$F$. The function$[x]$takes as input a field and returns as output/image the ring of polynomials over that field. This is a unary function, meaning it has only one argument. The symbolism is postfix, which means the function symbol is written after the argument, which contrasts with our usual approach of writing the function symbol first (as in$f(x)$). In contrast, we could regard$[\ \ ]$as a binary function that takes as arguments a field and an element of a ring containing that field and returns the subring generated by those two. It is written in infix notation, which means the function symbol$[\ \ ]$is between the first and second arguments, as when we write$F[t]$. It's not exactly infix because the second bracket is after the$t$, but you get the idea. A proper infix function is the addition symbol '+', which is placed between its first and second arguments, as in$a+b$. So$F[x]$and$F[t]$are different things. One is the application of a unary function to the argument$F$, and the cross squiggle has no stand-alone meaning in it. The other is the application of a binary function to the arguments$F$and$t$. In this format we are unable to label a field element as$x$and then compactly denote the ring generated by$F\cup \{x\}$because if we try to do so we will write$F[x]$and the reader will interpret that as the application of$[x]$to$F$. If we were using proper postfix notation, we could make the distinction. We'll keep$[x]$with its current meaning, and change the symbol for the second function from$[\ \ ]$in infix, to$[]$in postfix. Then we write$F[x]$as$F[x]$, we write$F[t]$as$Ft[]$(two arguments and then the function symbol) and we can also write$Fx[]$to denote the ring generated by$F\cup\{x\}$. This may seem confusing. That's because it's bad strategy notationally to use$F[t]$to indicate what they do, which makes it look like it's the same sort of thing as$F[x]$when it isn't. They could have written it as$F^{[t]}$or$F_t$or$F\langle t\rangle$or any of a wide variety of other things. EDIT: On reflection, I can see why they want to write$F[\alpha]$though. If my hurried thinking is correct,$F[\alpha]$is the set \left\{p(\alpha)\ :\ p\in F[x]\right\} where in this case we are interpreting each polynomial$p$as a function from$F$to$F$, rather than as just a formal$k$-tuple of elements of$F$If we think of what happens when we generate a ring, we can see that that is the result we will get. There is no division in a ring. We can raise$\alpha$to any power we like, and we can add and multiply by elements of$F$, which gives us a polynomial expression in$\alpha$. Last edited: Jun 1, 2017 12. Jun 1, 2017 ### Math Amateur Thanks Andrew ... that was extremely helpful and quite a learning experience ... I was reflecting on what you said about the nature of$x$... and was puzzled about how we should view how many texts deal with$\overline{x}$as a root of the quotient field$F[x]/ ( p(x) )$... ... given what you have said about$x$, it does not seem to me to make sense to say that$\overline{x}$is a root of the quotient field$F[x]/ ( p(x) )$... ... how should we make sense of such statements ... ... Can you help ... ... ? As an example of$\overline{x}$being portrayed as a root of$F[x]/ ( p(x) )$I am providing Dummit and Foote's Theorem 3, Section 13.1 : Basic Theory of Field Extensions ... as follows: Hope you can help ... Peter #### Attached Files: • ###### D&F - 1 - Theorem 3 - Section 13.1 - PART 1 ... ....png File size: 70.6 KB Views: 37 • ###### D&F - 2 - Theorem 3 - Section 13.1 - PART 2 ... ....png File size: 32.7 KB Views: 32 13. Jun 1, 2017 ### andrewkirk This touches on another notational trap that can cause confusion, which is the symbols used to denote a polynomial. Above they have used$p(x)$to denote a polynomial, which is fairly standard, but creates ambiguity because, when$p(x)$is considered as an element of a polynomial ring, the$x$is again just a squiggle that is part of the indivisible symbol$p(x)$. That's why I prefer to use just a$p$to refer to a polynomial instead, which you might notice that I did at the end of my previous post. That is against common usage though. I'm a bit of an iconoclast I'm afraid. They say you can't fight City Hall but I see no reason why not. One way that the notation$p(x)$causes confusion is that when that is used to denote a polynomial, the author then at some stage will likely write something like$p(\alpha)$which is actually a number, that is the result of applying the polynomial (qua function) to the field element$\alpha$. So again we have on the one hand$p(x)$being an indivisible symbol, and on the other,$p(\alpha)$being the application of a function$p$to an argument$\alpha$- so it's two separate symbols. In the above,$p(x)$is an indivisible symbol denoting a polynomial, while$p(\bar x)$is something else entirely, which we'll get to in a couple of paras. What about$\pi(x)$? To work out what that means, we look at the domain of the function$\pi$. The domain is$F[x]$, which is a polynomial ring, which means that the$x$inside the parentheses must denote a polynomial. It is the polynomial that may be written as$x+0$or, in$k$-tuple notation, as$(1,0)$or in function notation (considering the polynomial as a function) as$x\mapsto x$. So$\bar x=\pi(x)$is the element of the range$K=F[x]/\langle p(x)\rangle$that is the image under map$\pi$of the polynomial$(1,0)$(I prefer tuple notation where possible because it is less vulnerable to ambiguity). What, then, is$p(\bar x)$? Strictly speaking, it's a meaningless symbol string, because the domain of the polynomial$p$, qua function, is$F$and$\bar x$is in$K$, which is definitely not$F$or any subset thereof. What it actually means is$P(\bar x)$(note the use of upper case here) where$P:K\to K$is the analog of the function$p:F\to F$, given by P(q)=\sum_{k=0}^n p_k q^k where$n$is the degree of$p$, and$p_k$is the coefficient in the term of order$k$in$p$, and all arithmetic operations on the RHS are in the ring$K$. So I would write the author's line p(\bar x)=\overline{(p(x))} as P(\pi((1,0)))=\pi(p)which is the same as P((1,0)+\langle p\rangle)=p+\langle p\rangle and the RHS is$\langle p\rangle$, which is$0_K$. and there is not an$x$anywhere in sight! I would add that I find it confusing for the author to write$\phi=\pi|_F$. The domain of$\pi$does not contain$F$so we cannot restrict it to$F$as the notation suggests. What he means is that$\phi=\pi\circ\psi$where$\psi:F\to F[x]$such that$\psi(\alpha)$is the constant (degree zero) polynomial$\alpha$14. Jun 1, 2017 ### Stephen Tashi That's a good question. I conjecture that in previous passages the text has used "$x$" ambiguously to stand for either a "pure" indeterminate (meaning one that satisfies no equations except for identities like$x^3 = (x)(x^2)$) or an indeterminate "with side conditions" ( amounting to one of a specific set of numbers defined implicitly by requiring that$x$satisfy some polynomial equation such as$x^2 = 2$). When I see the notation$F[x]$, I think of "$x$" as being a pure indeterminate, but that might be my own bad habit. An author who makes a statement about "$F[x]$" may intend that the statement applies even if "$x$" satisfies a particular polynomial equation. In the passage in question, "$t$" is being used to emphasize that it represents an "pure" indeterminate. The author uses the terminology "free parameter" in place of my "pure indeterminate". I don't know whether either terminology is standard. This is an unsophisticated way of re-stating what the others have said. 15. Jun 6, 2017 ### fresh_42 ### Staff: Mentor I have seen an example in which$x$has been an indeterminate and$t$a transcendental number, in order to have$F(t)[x]$. But a transcendental number over$F$behaves exactly the same way the indeterminate does:$F(t)[x] \cong F(y)[x] \cong F(x)[y] \cong F(x)[t]$. There is absolutely no benefit in applying language theory and mix it into algebra and ring theory at this stage. The rules are easy:$F(\textrm{ whatever })$denotes a field,$F[\textrm{ whatever }]$denotes a ring, the former being its quotient field. To distinguish between transcendental numbers and indeterminates is not really necessary but makes life easier, especially if polynomials are involved. Whether one writes polynomials as$p(x) \, , \, p(t)\, , \,p(\alpha)\, , \,p(a)$doesn't make any difference. Of course we are simply used to use$x$or$y$or$x_1, x_2,\ldots $and often$t$as well, which would make the usage of$\alpha$or$a$as variables confusing, but in the end, it means whatever which meaning we attach to it. So assuming a field$F$, a transcendental number$t$over$F$, an indeterminate$x$and an algebraic number$a$over$F$with (irreducible monic) minimal polynomial$m_a(x)$, we get: •$F[x] \cong F[t]$are isomorphic rings, e.g.$\mathbb{Q}[x] \cong \mathbb{Q}[\pi]$. •$F(x) \cong F(t)$are isomorphic fields, e.g.$\mathbb{Q}(x) \cong \mathbb{Q}(\pi)$. •$F(x)$is the quotient field of$F[x]$and$F(t)$the quotient field of$F[t]$•$F[x] \neq F(x)$and$F[t] \neq F(t)$•$F(a) = F[a]$because the ring$F[a]$contains already all multiplicative inverses and is thus a field. •$F[a] \cong F[x]/\langle m_a(x) \rangle \cong F(a)$as the ideal$\langle m_a(x) \rangle$is prime as well as maximal. • We could even write$\mathbb{Q}[\pi]/\langle m_\sqrt{2}(\pi) \rangle \cong \mathbb{Q}[\sqrt{2}]=\mathbb{Q}(\sqrt{2})$to illustrate how transcendental numbers as$\pi$have to be treated. However, we are simply used to write rational polynomials as elements of$\mathbb{Q}[x]$or$\mathbb{Q}[t]$, depending on whether the author prefers$x$as indeterminate or$t$, e.g. as it's often the case if$t$can play the role of time. There is no rule. Even$\mathbb{Q}[a]$would be allowed as polynomial ring but probably had to be considered as a real bad habit. But if$a$is transcendental over$\mathbb{Q}$then$\mathbb{Q}(x)$and$\mathbb{Q}(a)$would be the same. 16. Jun 7, 2017 ### Math Amateur Thanks fresh_42 ... really appreciate your help on this set of issues ... most helpful!!! Reflecting on what you have said ... Peter 17. Jun 7, 2017 ### Math Amateur Hi fresh_42 ... thanks again for all your help ... but ... could you help further ... You write: "I have seen an example in which$x$has been an indeterminate and$t$a transcendental number, in order to have$F(t)[x]$. But a transcendental number over$F$behaves exactly the same way the indeterminate does:$F(t)[x] \cong F(y)[x] \cong F(x)[y] \cong F(x)[t]$... ... " I do not understand why a transcendental number over$F$behaves exactly the same way the indeterminate does ... in fact I am not sure of the analysis in the case of a transcendental number ... can you explain in simple terms ... Hope you can help ... Peter 18. Jun 7, 2017 ### fresh_42 ### Staff: Mentor The point is that neither an indeterminate nor a transcendental number fulfills any algebraic equation. So from the view point of a ring, they multiply and add in the same way - only formally, i.e.$x + x^2 = x + x^2$and$\pi + \pi^2 = \pi +\pi^2$. No way to simplify those expressions. And other operations aren't available. That's what I mean by "the same", the same ring. You can substitute them with one another and still have the same object. Of course we are used to replace indeterminates by numbers$x \mapsto 2$which we don't do with transcendental (= non algebraic) numbers. Nobody substitutes$\pi \mapsto 2$. Therefore we distinguish them, but as algebraic objects$F[x]$and$F[\pi]$are the same thing (as long as$\pi \notin F$). And we reserve indeterminates for polynomials to avoid confusion, because there are fields which contain a number like$\pi$, in which case the analog breaks down. The entire example should only illustrate why$F[x] \neq F(x)$whereas$F[a]=F(a)$if$a$is algebraic over$F$. And to demonstrate that some notations are only "what we are used to" and not necessarily a given. Obviously I don't want you to write polynomials with$\pi$as their indeterminate, but whether it is$x,X,x_1,X_1,t$or$T$isn't really important. And in a case like$F(x)[t]$we have$F(x)$as the field, which makes$x$an element of the field and$t$the indeterminate. Here you are free to think of$x$as a transcendental number which makes$F(x)$a number field, or$x$as a variable which makes$F(x)$the quotient field of the polynomial ring$F[x]$and$t$a second variable. 19. Jun 7, 2017 ### andrewkirk If a number$t$is transcendental over$F$then •$\{t^k\ :\ k\in\mathbb Z,\ k\geq 0\}$is a basis for the ring$F[t]=\langle F\cup\{t\}\rangle_{Ring}$as a vector space over$F$. Although it is not obvious, we need to here use the observation that$t$solves no algebraic equation to conclude that that set is a basis - ie linearly independent. If there were a nontrivial linear combination of those elements that was equal to zero, we'd have a nontrivial polynomial over$F$with$t$as root. We also know that •$\{x^k\ :\ k\in\mathbb Z,\ k\geq 0\}$is a basis for the ring$F[x]$as a vector space over$F$. We can make a vector-space isomorphism$f:F[t]\to F[x]$as the linear extension of the map that takes$t^k$to the polynomial$x^k$. We then prove that this is also a ring isomorphism, by showing that it preserves multiplicative structure, ie that$f(p(t)\times q(t))=pq = f(p(t))f(q(t))$. We can do that by showing it to be the case for monomials$t^k$and then applying the distributive law, together with the vector space homorphism properties we already have, to show it applies to polynomial functions of$t$. Hence$F[t]\cong_{Ring} F[x]$. Given isomorphic rings, we can construct isomorphic fields$F(t)$and$F(x)$in the same way for both, and we note that the isomorphism maps$t$to$x##.

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