Characteristic proof

Let A be an integral domain. Prove that if (x+1)^2 = x^2 + 1 in A[x], then A must have characteristic 2.

Any thoughts? Thanks for your time!

Well $\displaystyle (x+1)^2 = x^2 + 2x + 1 $, which means $\displaystyle x^2 + 2x + 1 = x^2 + 1 $ which means $\displaystyle 2x = 0 $. since (1+1)x = 0x = 0, which means (1+1) in A is 0, since Char(Integral Domain) is the smallest n such that 1 + 1 + ... + 1 = 0 (where 1+1.. was done n times), 2 is the characteristic of A,

Quote:

---

Originally Posted by jakncoke

Well $\displaystyle (x+1)^2 = x^2 + 2x + 1 $, which means $\displaystyle x^2 + 2x + 1 = x^2 + 1 $ which means $\displaystyle 2x = 0 $. since (1+1)x = 0x = 0, which means (1+1) in A is 0, since Char(Integral Domain) is the smallest n such that 1 + 1 + ... + 1 = 0 (where 1+1.. was done n times), 2 is the characteristic of A,

---

i think one (small) thing is missing from this: which can be supplied in some different ways:

you say (1+1)x = 0...it's important to realize that that identifying this with 2x (in the square) isn't always justified (unless you are considering (A[x],+) as a Z-module). now, A is an integral domain, but to derive 1+1 = 0 from (1+1)x = 0, we need that A[x] is an integral domain. while this is true, it ought to be shown, like so:

suppose that f(x)g(x) = 0, but g(x) ≠ 0 (the 0 in both cases is the 0-polynomial, not the 0 of A).

since g(x) ≠ 0, we have the leading coefficient non-zero, so g(x) = ax^m+k(x) (where deg(k) < deg(g), or k = 0).

now if f(x) ≠ 0, we have the leading coefficient of f non-zero, so f(x) = a'xⁿ + h(x) (where deg(h) < deg(f), of h = 0).

thus f(x)g(x) = (aa')x^m+n + k(x)f(x) + h(x)g(x) + h(x)k(x)

if h = 0, k ≠ 0

deg(kf + hg + hk) = deg(kf) = deg(k) + deg(f) < deg(g) + deg(f)

if h ≠ 0, k = 0

deg(kf + hg + hk) = deg(hg) = deg(h) + deg(k) < deg(f) + deg(g)

if h ≠ 0, k ≠ 0

deg(kf + hg + hk) = max(deg(kf),deg(hk)) < deg(f) + deg(g) (see above)

if h = 0, k = 0

kf + hg + hk = 0.

so if we write s(x) = k(x)f(x) + h(x)g(x) + h(x)k(x), then f(x)g(x) = (aa')x^m+n + s(x), where either deg(s) < m+n, or s = 0.

since this is the 0-polynomial, we have aa' = 0, which is impossible since A is an integral domain. thus f must not have any non-zero leading coefficient, so f is the 0-polynomial. thus A[x] is an integral domain.

**********
alternate proof:

use the homomorphism (evaluation map) A[x]-->A given by f(x)--->f(1). under this map we have (1+1)x = x + x --> 1 + 1.

since x + x = 0,and homomorphisms preserve the additive identity, 1 + 1 = 0 in A, so char(A) = 2.

**********
the point being we either need to know A[x] is an integral domain, or work inside A (which we know is an integral domain), to rule out the possibility x might be a zero-divisor in A[x].