/sci/ - Science & Math

I'm trying to compute the [math]\check{\text{C}}[/math]ech cohomology of [math]\mathbb P^1[/math] and of the affine line with the double origin over a field [math]K[/math] 'by hand', by using the cover [math]\mathfrak U[/math] by two affine lines, where [math]\text{Spec}(K[t])[/math] is glued to another copy of itself on the open subset [math]\text{Spec}(K[t,t^{-1}])[/math] via the morphisms [math]t\mapsto t^{-1}[/math], respectively [math]t\mapsto t[/math].

The projective line has [math]\check{\text{C}}[/math]ech complex
[eqn]0\to K[t]\times K[t]\xrightarrow{d^0} K[t,t^{-1}]\xrightarrow{d^1} 0[/eqn] with [math]d^0(p(t),q(t))=q(t^{-1})-p(t)[/math]. We have that [math]\ker d^0 = K[/math], since only the subset [math](k,k)[/math] maps to [math]0[/math] for [math]k\in K[/math]. Also, we have that [math]\text{im} \;d^0=K[t,t^{-1}][/math] clearly, ie [math]d^0[/math] is surjective. In particular, [math]H^0(\mathfrak U, \mathbb P^1)=K[/math] and [math]H^1(\mathfrak U, \mathbb P^1)=0[/math], and the rest are zero.

The affine line [math]\tilde{\mathbb A}^1[/math] has the same complex but instead [math]d^0(p(t),q(t))=q(t)-p(t)[/math]. This gives us that [math]\ker d^0 = K[t][/math] and [math]\text{im}\; d^0=K[t][/math]. Therefore [math]H^0(\mathfrak U, \tilde{\mathbb A}^1)=K[t][/math] and [math]H^1(\mathfrak U,\tilde{\mathbb A}^1)=K[t,t^{-1}]/K[t][/math].

My problem lies in interpreting the last quotient. I want to think that this quotient is [math]K[t^{-1}]_+[/math], ie, the ideal consisting of all degree [math]\geq 1[/math] polynomials in [math]t^{-1}[/math]. I think this because the set of constant polynomials is within the image of [math]d^0[/math], but I could be wrong.

Is the previous correct?