I'm working on the following question from Shafarevich's Basic Algebraic Geometry I and wanted to check that I understand what's going on.
The set $X\subset{\mathbb A}^2$ is defined by the equation $f:x^2+y^2=1$ and $g:x=1$. Find the ideal $\mathfrak{U}_X$. Is it true that $\mathfrak{U}_X = (f,g)$?
First, is $X$ defined as the common zeros of $f$ and $g$? That is what I would assume but in this case, $X$ would only be a single point $(1,0)$ and that doesn't sound right for this problem.
If $X$ just is a single point, I'm not exactly sure what $\mathfrak{U}_X$ is. I would be surprised if it was $(f,g)$ but I am not sure how to prove that there are more polynomials in $\mathfrak{U}_X$ than there are in $(f,g)$. Could I say for example that $y$ is not in $(f,g)$ because elements of $(f,g)$ have even degrees as polynomials with coefficients in $k[x]$?
You have the right idea. As $X$ consists of a single point, it should not be hard to give explicit generators for $\mathfrak{U}_X$. If you want to get hands-on with the details, you can start by noting that $$\mathfrak{U}_X\supset(f,g)=(x^2+y^2-1,x-1)=(y^2,x-1).$$ Can you find explicit generators for $\mathfrak{U}_X$ from here? Can you show that the inclusion above is not an equality?