Originally Posted by

**topspin1617** Well, the $\displaystyle x$-axis is, by definition, the set

$\displaystyle X=\{(a,0)\mid a\in \mathbb{R} \}$.

If you want to prove that $\displaystyle X$ is an ideal using the statement you have written, we need a ring $\displaystyle R$ and a ring homomorphism $\displaystyle f:\mathbb{R}^2\rightarrow R$ such that $\displaystyle X=\ker f$.

How about taking $\displaystyle R=\mathbb{R}$, and defining

$\displaystyle f:\mathbb{R}^2\rightarrow R$

$\displaystyle (a,b)\mapsto a$

(projection onto the first coordinate)? It should be fairly simple to prove that $\displaystyle f$ is indeed a ring homomorphism, and that the kernel of $\displaystyle f$ is exactly the set of points located on the $\displaystyle x$-axis.

You can do a completely analogous thing to show that the $\displaystyle Y$ axis is also an ideal.