One of the coolest functional equations I have seen! 

Yeah, but not only shows that this is homomorphism but actually shows that there is only one such homomorphism from Z[x] into Z

@@M4DA. I think you might be a little bit confused. There are infinitely many different homomorphisms from Z[x] to Z, all of them are fully characterized by the image of x (which can be any integer, so that all evaluation maps are homomorphisms). You need to use the fact that a map from Z[x] to Z is a homomorphism if and only if it is evaluation at some integer in order to conclude that a map Z[x] to Z is a homomorphism if and only if it satisfies the two conditions given in the video.

In general, the set of ring homomorphism from Z[x] to R corresponds to the elements of R via the evaluation map. The ring Z[x] plays the same role among rings that the one element set plays among sets (maps from Z[x] correspond to elements of the target). A direct way to see this connection is via the free-forgetful adjunction between rings and sets. More generally. Fix a (commutative?) ring k, and let [R, S] denote the k-linear ring homomorphisms R -> S, where R and S are k-algebras (note that being a ring is equivalent to being a Z-algebra). Then there is an isomorphism [k[x], R] = R given by the evaluation map.

As for the recursive and divisibility aspect, this unfortunately doesn't generalize as well. Z is special here because it can be generated in a recursive way. To get a similar statement with another ring, you'll need a similar recursive definition of that ring.

Yes. This. I see it all the time. Nobody remembers the n=0 base case because it's too easy.

Take two perfect squares to be a and b , and then multiply both by the same product of primes (where each prime appears at most once), then check that they are less than n. As to the algorithm, I think you can precompute all such products of primes

Related to USAJMO 2010 Problem 1

I will add to Angel's great explanation that when division is also well-defined (division by _non-zero_ elements, of course) and multiplication is commutative, we call this kind of structure a "field" (if you have division but multiplication is not commutative then this is called a skew-field) - so for example, the rational numbers, the real numbers, and the complex numbers are fields, and an example of a skew-field is the quaternions (however the next step up the "complex ladder", which are called the octonions, are not a skew-field). And I will also note that there is another property of general rings besides possibly-non-commutative multiplication which is unintuitive: the notion of "zero-divisors". A zero-divisor is a non-zero element _c_ which "divides zero" in the sense that there exists another non-zero element _d_ such that _cd_ = 0. Simply: in a general ring the product of non-zero elements can be zero - for example this happens with matrices, or a slightly easier example if you know what modular arithmetic is is that 2=/=0 mod 6 and 3=/=0 mod 6 but 2*3=0 mod 6. A commutative ring in which this kind of thing doesn't happen (in which there are no zero-divisors) is called an "integral domain"* - for example the integers are an integral domain, as are polynomials with real coefficients. *or without the commutativity requirement, a "domain" (but then one can to make a distinction between left zero-divisors and right-zero-divisors) - for example polynomials whose coefficients are quaternions form a domain.

oooh I see the answer in previous comments. Question is gone. Thanks

Bruh.

nice name

@@user-fi6if8gx3g thank you!

When doing algebra, for a R a ring, you should view R[x] as being a RING, not as a SET of functions. Here's why: Suppose R = Z/(3Z) = ring of equivalence classes of integers modulo 3. For any a in R, a^3 = a (Check: [0]^3 = [0], [1]^3 = [1], [2]^3 = [8] = [2+3(2)] = [2].) Let p in R[x] be p(x) = x^3 - x. Let q in R[x] be q(x) = 0 (which is of course 0 in R, so probably should be written [0]). Then if you think of p as a FUNCTION from R to R, then p is the zero function... p maps every element of R = { [0], [1], [2] } to the value [0] in R. Thus thought of AS FUNCTIONS, p and q are the same... they both map every element of R to [0]. But they aren't the same AS ELEMENTS of R[x]... p is a degree 3 polynomial, while q is a constant polynomial, p has 2 terms while q has only 1, and so on. When it becomes useful to consider them as functions, what you're actually doing is applying an evaluation homomorphism to them. If f in R[x], a in R, then E(a) : R[x] -> R is a ring homomorphism defined by E(a)(f) = f(a) in R, meaning what you'd expect: replace x everywhere by a in f, and then compute the result.

What part are you talking about?

@@jussari7960 it is the theorem that says that if we have a P € Z[X] polynomials with integer roots r there exist Q € Z[X] s.t P=(X-r)Q

@@matthieumoussiegt Take any p ∈ Z[X] (of degree ≥1) with a root at r ∈ Z and let q(x) be the polynomial that satisfies p(x) = (x-r)q(x). Write p and q as p(x) = a_n*x^n + ... + a_1*x + a_0 and q(x) = b_(n-1)*x^(n-1) + ... + b_1*x + b_0 and note that a_i are all integers. Then from p(x) = (x-r)q(x) we get a_i = b_(i-1) + r*b_i for all 0≤i

@@jussari7960 the problem is not that it is not in Q[X] the problem that it is specified that q is in Z[X]

@@matthieumoussiegt Oh, I see what you mean. i think that is true too, (though not immediately obvious). Take the notation from my previous comment. Then we have b_(n-1) = a_n is an integer, and inductively b_(i-1) = a_i - r*b_i is an integer for i≥0. (I initially messed up the formulas for a_i but they should hopefully be correct now)

In this context, the polynomials in Z[x] are being treated as formal expressions as opposed to functions. However, in the solution it gets established that theta must be equal to the map sending each polynomial to its evaluation at some integer. Theta cannot be evaluation for any domain other than the integers. If it were, then it would map the polynomial p(x) = x to some object that is not an integer, which would contradict the assumption that the codomain of theta is the integers.

from 8:11 we assume f is an arbitrary polynomial, so the proof works for all polynomials

We never made the assumption, we just defined a to be the value of θ(x), (which is some integer) and then from that proved that for any function it must be evaluation at a.

i believe this problem is more algebraic than analytic what about you ?

Definitely algebra

I think it's in the realm of analytic number theory?

@@andersen9044 I agree, but he did mention open intervals...

There's nothing analytic about this other than the examples of derivative and integral evaluation in the beginning

I suspect Gauss's Lemma is relevant here.

yes they are. *Z* [X] is the smallest ring which contains *Z* and the indeterminate X, while *Q* (root 2) is the smallest field that contains *Q* and root 2. We use square brackets for "smallest ring containing" and round brackets for "smallest field containing". In the case of root 2, the smallest ring containing *Q* and root 2 turns out to be a field, so we have *Q* [root 2] = *Q* (root 2). (note: there shouldn't be gaps between *Z* and [X] and between *Q* and (root 2) but RU-vid won't let me format it without a space)

@@schweinmachtbree1013 Very informative. Thanks.

The reply from schweinmachtbree is about the difference in meaning between the square bracket [] and parentheses () notation. If that was the extent of your question, then what follows might be a waste, ut I'm not sure exactly what you were asking. If you were asking if there's some relationship between polynomials (adjoining the indeterminate x) and these special rings/fields (adjoining a specific value like sqrt(2), then the answer is yes... there's a BIG TIME connection between the two ideas: Let s = square root of 2 (just to make writing it easier). First notice that Q(s) = Q[s]: If a,b in Q, a,b not both 0, then it's impossible that a^2 = 2b^2. That's because, if assume a^2 = 2b^2, then if one is 0, then both are, so neither is 0... and since neither is 0, a^2 = 2b^2 => (a/b)^2 = 2, and a/n in Q, and so 2 has a rational square root. Contradiction. If a,b in Q, a,b not both 0, then (a + b s) ( [a/(a^2-2b^2)] - [b/(a^2-2b^2)] s ) = 1. Thus every non-zero element of Q[s] has an inverse, making Q[s] a field. Since Q and s are in the field Q[s], have that Q(s) is a subset of Q[s]. Since Q and s are in the ring Q(s) (it's a field, hence it's a ring), have that Q[s] subset of Q(s). Therefore Q(s) = Q[s]. Now let J = < x^2 - 2 > = the ideal generated by the polynomial x^2 - 2 in Q[x]. If f in Q[x], then f + J in Q[x]/J (which is a field, since J is a maximal, since Q[x] a PID... but that's not important yet). By the Remainder Theorem, there exists unique q, r in Q[x] s.t. f(x) = q(x) (x^2 - 2) + r(x) with deg(r)

@@mathboy8188 Well i don't know enough to know the full extent of what i was asking. I just wanted to know if adjoint being used in both cases are the same mathematical concept. Any information helps thanks.

@@mathboy8188 1) phi is a ring homomorphism It evaluates the polynomial at s which clearly respects addition and multiplication. It also evaluates 1 as 1. 2) The kernel is all polynomials with s as a root. Polynomials with x^2 -2 as a factor. The kernel is the ideal generated x^2 - 2. 3) phi is onto. For any rational a,b. a+bx evaluates to a+bs. By the ring isomorphism theorem we have Q[s] isomorphic to Q[x]/J

(Disclaimer: I am not knowledgeable in math and am not sure of what I'm talking about) The polynomial in ℚ[x] has coefficients in ℚ, and integrating over it from a to b just means integrating it over [a ; b], because x is totally allowed to be any number, there's not restriction on that. For example p₀(x) ≔ 3x² - x/8 ∈ ℚ[x], and since p₀(x) is defined for all x ∈ [0 ; 1], you could define the function: Θ : ℚ[x] ⟶ ℚ p(x) ⟼ ∫ p(x)dx with bounds 0 and 1 and then get that Θ(p₀(x)) = ∫ p₀(x)dx, bounds 0 & 1 = ∫ (3x² - x/8)dx, bounds 0 & 1 = 15/16 (∈ ℚ) Hope this answers your question!

@@endersteph if you extend the polynomial function on R that is where the integration normally defined and you define the function that it is p(x) on rational points and 0 on irrational points and you integrate it then you receive always the value 0 for the integral, because Q is a Lebesgue zero set in R.

@chaparral82 I know but this doesn't apply here. The function you are describing, 𝟙_ℚ × p, is not a polynomial. If the coefficients of p are in ℚ, then p is in ℚ[X], and integrating it over real numbers yields a rational number. Therefore, the function which takes a polynomial with rational coefficients and outputs its integration over say [a ; b] for reals a and b - which is what Michael was talking about - is a function from polynomials with rational coefficients (ℚ[X]) to ℚ. Again what I want to get at is that polynomials are integrable over real numbers, so there is no problem in doing this. For reminder: the coefficients of x ↦ a_n · x^n + a_n-1 · x^n-1 + ... + a_2 · x^2 + a_1 · x + a_0 are the constants a_0 through a_n. If A is a set that contains all of those coefficients, then this polynomial is in A[X]. This does not change the fact that the polynomial is defined over the reals, and that you can integrate it over [a ; b].