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Let $(F, +, \circ)$ be a field whose zero is $0_F$ and whose unity is $1_F$.

Let $X$ be transcendental over $F$.

Let $F[X]$ be the ring of polynomials in $X$ over $F$.

Let $d$ be an element of $F[X]$ of degree $n \geq 1$.

Then $\forall f \in F[X]: \exists q,r \in F[X]: f = q \circ d + r$ such that either:

(1) $r=0_F$ or:

(2) $r \neq 0_F$ and $r$ has degree that is less than $n$.

"Let $X$ be transcendental over $F$." -Why does the the theorem require $X$ not be the zero of any non-zero polynomial over $F$? I am not ever sure what $X$ means here. Does it just mean polynomials in $1$ variable where that variable is called $X$?

Please help me understand these ingredients of the theorem.

source: https://proofwiki.org/wiki/Division_Theorem_for_Polynomial_Forms_over_Field

Bill Dubuque
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jenny9
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  • What does "$0F$" mean? If you mean the zero element of $F$, just write $0$. Whether that is $0$ in $F$ or in $F[X]$, it's basically the same, just as nobody gets bothered when we think of integers as fractions via $n = n/1$. You need $X$ trancendental over $F$ since otherwise you can write a polynomial in more than one way, so the notion of the degree of a polynomial or its coefficients are ambiguous. Try looking at polynomials in $\mathbf Q[\sqrt{2}]$. – KCd Nov 22 '22 at 02:34
  • @KCd what does $X$ represent here? – jenny9 Nov 22 '22 at 02:39
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    $X$ is the variable. – Will Jagy Nov 22 '22 at 02:43
  • how can a variable be trancenental over a field? – jenny9 Nov 22 '22 at 02:45
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    I think the statement “Let $X$ be transcendental over $F$” could be omitted and the next line could be replaced with “Let $F[X]$ be the ring of polynomials with coefficients in $F$.” – littleO Nov 22 '22 at 02:54
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    I just don't understand how a variable could be the root of a polynomial over a field. – jenny9 Nov 22 '22 at 03:00
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    @jenny9 the distinctions here are subtle. Suggest you stick with a text at a consistent level while not worrying about wikipedia. – Will Jagy Nov 22 '22 at 03:16
  • In $\Bbb Q[X]/(X^2-2)$ the "variable" $X$ is a root of the polynomial $,Y^2-2,,$ that's how. Thus $X$ is algebraic over $\Bbb Q$, i.e. not transendental (= indeterminate). For more on the motivation behind the definition of rings of formal polynomials see here and here – Bill Dubuque Nov 22 '22 at 13:33

1 Answers1

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Does it just mean polynomials in 1 variable where that variable is called X?

Yes.

The best way to understand this theorem is to think that that is exactly what is meant and that Proofwiki is making things needlessly complicated.

(If we want to give Proofwiki some credit I think they formulate in this way because they don't want to casually use the term 'variable' and then get sucked into questions like of 'but what is a variable?' which is indeed a slippery concept. By saying that $X$ is not algebraic over $F$ they don't answer the question what $X$ is, but they do give the information about the relation between $F$ and $X$ that is needed to make sense of the notions in the theorem like degree.

But if you are already comfortable with the notion of 'polynomial in one variable' then all these things made already sense and the best path forward is to conclude that they are indeed talking about polynomials in one variable and focus on the rest of the proof.)

The wonderful theorem here is that the notion of division with remainder and all its surprising consequences such as unique factorization that we know from the integers also have an analogue in the ring of polynomials in one variable. It is a miraculous and beautiful example of how, by looking a bit more abstractly, seemingly unrelated things are suddenly the same in mathematics. It is best to enjoy (and understand) this theorem without being distracted by things like the transcendentality of $X$.

(If $X$ is a variable in the ordinary sense of the word, then it is automatically transcendental. Conversely, in situations like the one where $F = \mathbb{Q}$ and $X = \pi$ where we don't think about $X$ as a variable, but know it is transendental, we still have that, as a ring, $\mathbb{Q}[\pi]$ is isomorphic to the ring of polynomials $\mathbb{Q}[X]$ and results like the division theorem still apply. So a second way of giving credit to Proofwiki is saying that they treat this case and the 'standard' $F[X]$ case in one swoop.)

Vincent
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