I think I know what a root is. It's the value of elements in the domain of a function that map to 0 in the range of the function. At least that's how I see it. Do correct me if I'm wrong.

However I can't really tell what the purpose of finding roots is. I've tried to search for it, but can't find any conceptual overview or a practical example.

I mean what's so special about finding where the value of a function is 0?

EDIT: The answer/s below make some sense, but I was reading the "Princeton Companion to Mathematics", and in it is the following:

"What makes fields interesting, however, is not so much the existence of these basic examples as the fact that there is an important process of extension that allows one to build new fields out of old ones. The idea is to start with a field F, find a polynomial P that has no roots in F, and “adjoin” a new element to F with the stipulation that it is a root of P. This produces an extended field F , which consists of everything that one can produce from this root and from elements of F using addition and multiplication.'

The whole thing makes no sense. Specifically the part "one can produce from this root". What is production from roots? Very confusing.

I thought this must be because I don't fundamentally understand roots. This is the context in which I had this question.

  • $\begingroup$ Maybe to graph? $\endgroup$ – Jason Chen Oct 29 '15 at 15:58
  • $\begingroup$ We sometimes use roots given that broader definition, but the most common usage is for polynomials. $\endgroup$ – Thomas Andrews Oct 29 '15 at 15:59
  • $\begingroup$ The purpose of finding roots is to find the range of a function this tells us the maximum and minimum value of a function and where on coordinate axis the graph meets. $\endgroup$ – Archis Welankar Oct 29 '15 at 16:01
  • 1
    $\begingroup$ The clause you pointed out cannot be isolated from the rest of the sentence. The point is that you have a field and a polynomial, you introduce a new symbol for a root of that polynomial, and then you make an actual field from your old field and this new element by closing it over addition and multiplication. Perhaps the most well known example is making $C $ from $R $ by adjoining a root of $x^2+1$. $\endgroup$ – Ian Oct 30 '15 at 10:48
  • $\begingroup$ Note that the elements of the new field are all of the form $a+br$, where $r$ is this new root that we've made up and $a,b$ are elements of the old field $F$. One might wonder whether the new field is actually isomorphic to $\{ (a,b) : a,b \in F \}$ with the operations $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b) \cdot (c,d)=(a\cdot c,b \cdot d)$. (This is called the direct product of $F$ with itself.) $\endgroup$ – Ian Oct 30 '15 at 20:58

Finding solutions to an equation is equivalent to finding roots of a function.

Suppose you wanted to solve the equation $x^3 = 6 -7x^2$. That means you want to find all numbers $x$ such that $x^3 = 6 -7x^2$. Of course these are the exact same numbers which make $x^3 +7x^2 -6=0$ which are exactly the roots of the function $F(x) =x^3 +7x^2 -6$.

So finding roots are at least as important as solving equations. And the importance of finding solutions of an equation are determined by what the equation is being used to describe.

Edit: Your description of a field extension does not explain how those roots are "produced", nor does it try to. So you should not expect to understand how the roots are produced after reading it. What it means is that if $p(x)$ is a polynomial of degree $n$ with coefficients in some field $K$ then there is a larger field $F$, containing $K$, and containing $n$ elements $y_1, \ldots , y_n$ such that each $p(y_i)=0$. Indeed there is a smallest such field extension $F$. This means that if $K \subset G$ is any field extension which contains $n$ roots for $p$ then we must have $K \subset F \subset G$.

Edit Edit: Yeah, the production of the field containing the roots is different from the production itself -- and your description does not attempt to explain that construction either. It's easiest to understand if we already know some field where the roots live. For example every polynomial with coefficients in $\mathbb Q$ has roots in $\mathbb C$. Then just intersect all the subfields of $\mathbb C$ containing those roots to get the minimal field with that property. In the more general case where $q$ is irreducible with coeffieients in $K$ then it can be proved the quotient ring $K[x]/ (q)$ is a field and the polynomial $x$ is then a solution for $q$. Then we stack up these quotients to include enough roots.

  • $\begingroup$ Ok. I guess my question needs a bit more context. I've edited my question so it's a bit more clear in which direction I need some understanding. $\endgroup$ – ˆᵛˆ Oct 30 '15 at 7:59
  • $\begingroup$ It doesn't talk about production of roots. It's talking about production 'from' roots. Isn't that different? $\endgroup$ – ˆᵛˆ Oct 30 '15 at 11:57

The polynomial $x^2+x+1$ is never zero when $x$ is real. So invent a new number $\alpha$, where all we know about $\alpha$ is that $\alpha^2+\alpha+1=0$.
That gives rise to other numbers, including $(\alpha^2+3)(2\alpha-1)$ and many many more. These are the numbers you produce from the root $\alpha$.


Suppose you have a "number system" (i.e. field) $F$ and there is a polynomial $f(x)$ which has no root in this field. Then pretend $\alpha$ is a root of the polynomial in some bigger number system (field). By definition, fields are closed under addition, subtraction, multiplication and nonzero division, so we should be able to generate many more things in this bigger field than just $\alpha$ itself. In fact, we should be able to generate at least any polynomial in $\alpha$ (with coefficients in $F$).

(We actually want a stronger hypothesis on $f$ than merely not having a root in $F$: we want it to not be factorable any further in $F$.)

Denote by $F[\alpha]$ the collection of all such "polynomials in $\alpha$." Even if two polynomials are formally different, those two polynomials "in $\alpha$" can be represent the same thing. For instance, since we're pretending $\alpha$ is a root of $f(x)$, we know that $f(\alpha)$ and the zero polynomial $0$ represent the same thing, and further we can manipulate the equation $f(\alpha)=0$ however we want to get (all!) other instances of two polynomials in $\alpha$ representing the same element.

As another example, suppose $f(x)=c_nx^n+\cdots+c_1x+c_0$. Then go from $f(\alpha)=0$ to

$$\alpha^n=(-\frac{c_{n-1}}{c_n})\alpha^{n-1}+\cdots+(-\frac{c_1}{c_n})\alpha+(-\frac{c_0}{c_n}). $$

So $\alpha^n$ can be written as a linear combination of powers of $\alpha$ smaller than $n$. We can multiply the above equation by $\alpha$ over and over again, and every time we see $\alpha^n$ on the right side we can replace it with lower powers. As a result of this iterative process, we see that every power of $\alpha$ can be expressed as a linear combination of $\alpha^{n-1},\cdots,\alpha,1$. Conclusion: every polynomial can be expressed as a linear combination of these powers. Since $F[\alpha]$ has a finite spanning set, it is finite-dimensional as a vector space over $F$.

Our bigger field must contain all polynomials in $\alpha$, so it contains the ring $F[\alpha]$. But the bigger field is closed under nonzero division, so it also contains reciprocals $1/g(\alpha)$ of polynomials $g$ in $\alpha$ (as long as they're not $0$). It turns out that given any $g(\alpha)\in F[\alpha]$ which is not $0$, the "multiplication by $g(\alpha)$" function is a linear map $F[\alpha]\to F[\alpha]$ which is one-to-one, and hence a onto since $F[\alpha]$ is finite-dimensional, so that $1\in F[\alpha]$ has a preimage $h(\alpha)$ under this map, i.e. there is a polynomial $h(\alpha)\in F[\alpha]$ such that $g(\alpha)h(\alpha)=1$. Thus, the ring $F[\alpha]$ is actually a field!

In this way we've constructed the adjunction field $F(\alpha)$. One can make these ideas more formal by considering the quotient ring $F[x]/(f(x))$ where $F[x]$ is the polynomial ring and $(f(x))$ the ideal generated by the element $f(x)$.


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