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In most of books on elementary algebra, intermediate algebra and college algebra, the degree of the non-zero polynomial $$f(x)=a_nx^n+\cdots a_1x+a_0$$ with $a_n\neq 0$ is defined to be $n$.

But I am not sure this is well defined. Namely, shall we prove to the students that if the same $f(x)$ has another expression $b_mx^m+\cdots+b_1x+b_0$ with $b_m\neq 0$, then $m=n$ before we can actually define the degree of the polynomial? Otherwise we say we are defining the degree but the definition is illogical.

It seems to be that at the level of elementary algebra, intermediate algebra and college algebra, this proof may be a bit too complicated to the students. On the other hand, without the proof the definition of the degree of polynomials is not even logically established.

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    $\begingroup$ I feel like unless your teaching a course technical enough that the proof would be withing range most students are going to take $f(x)$ as uniquely defined by that particular polynomial. Can you give an example of a context where you need to use the uniqueness of degree? $\endgroup$ – Nate Bade Nov 5 at 3:07
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    $\begingroup$ Think about the definition of $f(x)-g(x)$, and how this difference can be 0. $\endgroup$ – user52817 Nov 5 at 3:23
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    $\begingroup$ This comes up in linear algebra. The issue is that we say all the $x^j$ are linearly independent/give a basis. In some sense polynomials are defined in such a way as for this to be true, as there can be no (linear) relation among different $x^j$. Of course you are right that this property must hold. But if we just treat it as a special type of vector then it's not so bad. (Not that this would help in college algebra, I guess.) $\endgroup$ – kcrisman Nov 5 at 3:49
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    $\begingroup$ That a polynomial has a unique expression in the way indicated in the question follows from any formal definition of "polynomial". See Lang's Algebra, chapter II, section 3 for one such (the polynomials with coefficients in a commutative ring $R$ are defined to be the set $R[t]$ of functions from the monoid $N = \{0, 1, 2, \dots, \}$ to $R$ that are zero except on a finite subset of $N$; for $r \in R$, $rt^{n}$ denotes the function assigning the value $r$ to $n$ and $0$ to all other elements of $N$; it follows that $P(t) \in R[t]$ has a unique expression $a_{0} + a_{1}t + \dots + a_{n}t^{n}$). $\endgroup$ – Dan Fox Nov 5 at 9:24
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    $\begingroup$ @Zuriel It's been a long time since I've taught algebra so I'm a little out of my depth here, but are we saying the function $f$ has a degree? Or that the polynomial $2x^3 + 1$ does? Because not all functions have a finite degree (take $\sin(x)$), so what good is degree serving here? The polynomial $2x^3 + 1$ certainly has a degree, and we can use that to find solutions, asymptotic, etc. But is seems like we can make these statements without uniqueness. Do we need the function to have a degree? $\endgroup$ – Nate Bade Nov 7 at 2:42
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I'm going to rewrite this answer to clarify what I think the issue is. I think the OP is imagining a different definition of the ring $k[x]$ than most answerers are. Here are two reasonable definitions:

  1. $k[x]$ is the ring of formal expressions of the form $\sum_{j=0}^{\infty} p_j x^j$ with $p_j \in k$ and we require that $p_j$ is $0$ for $j$ sufficiently large. See Wikipedia for the details.

  2. Let's define $\mathrm{Poly}(k)$ to be the ring of functions $k \to k$ which are of the form $x \mapsto \sum_{j=0}^d p_j x^j$ for some $p_j \in k$.

In any book on modern algebra which is sophisticated enough to have careful definitions (Herstein, Artin, Dummit and Foote...), the definition of $k[x]$ will be something like the first one. With this definition, it is very clear that degree is well defined because equality of polynomials is defined to mean equality of coefficients.

But the OP talking about a polynomial having two different representations makes me think that he or she is imagining a definition where that might be possible, and Definition 2 strikes me as the most likely. Indeed, if $k$ is the field with $q$ elements, then $x^q$ and $x$ represent the same function in $\mathrm{Poly}(k)$, even though one is a polynomial of degree $q$ and the other has degree $1$. This is one of the reasons that Definition 1 is the standard definition.

In a course which is sophisticated enough to have careful definitions, it would be appropriate to prove that the obvious map $k[x] \to \mathrm{Poly}(k)$ is an isomorphism for $k$ infinite. I could also imagine a course where the ground field is always $\mathbb{R}$ and one might prove that $\mathbb{R}[x] \cong \mathrm{Poly}(\mathbb{R})$ without bringing up that finite fields exist.

But we don't have to raise the issue when we define degree, because Definition 1 is the standard definition of $k[x]$ and this avoids the issue. Also, as Henry Townser says, many courses are at too low a level to reach this level of precision.

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  • $\begingroup$ I upvote this! We should be aware of the different definitions. You point out that for a finite field $F$ of order $q$, the two polynomial functions $x$ and $x^q$ are equal. However, viewed as functions in an extension of $F$, these polynomial functions need not be equal. $\endgroup$ – user52817 Nov 14 at 17:56
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This: "On the other hand, without the proof the definition of the degree of polynomials is not even logically established." is not quite right. What is needed to establish the definition is the fact that the degree is well-defined, but this fact is stated (implicitly) by stating the definition.

The expectation that all facts be proven is out of place in a course at the level of elementary/intermediate/college algebra - it is neither useful nor appropriate to expect that all facts will be rigorously proven in courses at this level.

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In general, I agree with @Henry Towsner on the fact that the proofs should not always be presented in an elementary course.

However, I have to disagree on the implicit "well-definition property" of any definition. Such a definition would require that some sort of uniqueness property has been proved, which cannot - or should not - be done prior to the definition of the degree of a polynomial.

However, if you want to find a "balance" between formal and school mathematics, why not use the following rephrasing of the definition you proposed:

A polynomial $p(x)$ is said to be of degree $n$, where $n\in\mathbb{N}$ if it is non-trivial if it can be represented in the form: $$p(x)=a_n x^n+\ldots+a_1x+a_0,\ a_n\neq0.$$

Thus, you may avoid uniqueness which could be either discussed with the class or be taken for granted.

P.S.: In case you have talked about equality between polynomials, proving that a polynomial's degree is unique is almost obvious for a student.

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I do not believe that this is a concern that would surface, or be worth surfacing, in the courses named by this question's title. In my reading, the question is analogous to worrying about whether you can ask about e.g. the thousands place of $7521$: To do so assumes that the base ten representation of $7521$ is unique, or, in particular, that "thousands place" is well-defined.

[In fact, this cooked-up analog might be viewed as an example of the main question: Write coefficients as $a_k$ and evaluate the polynomial at $x=10$ to get a number represented in base ten, i.e., consider the polynomial $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $x \mapsto 7x^3 + 5x^2 + 2x + 1$, and observe that $f(10) = 7521$, where the "biggest place" is the thousands place, i.e., $\deg f = 3$.]

Anyway: I don't imagine any of the named courses would need to specify that "thousands place" is well-defined; so, aiming for a type of rigor that ensures elementary/intermediate/college algebra students see "degree" is well-defined appears to me at least as unnecessary.

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Soon after introducing polynomials, students learn to add and subtract polynomials. Notice that if $f(x)$ and $g(x)$ are polynomials with degrees $n$ and $m$ respectively, and if $f(x)-g(x)=0$, then $n=m$. It follows that the degree of a polynomial is well-defined.

So the proof that degree is well-defined is not difficult at all. On the other hand, students at this level are typically not able to see the need to establish that it is well-defined.

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