1. Ideal of the ring

Hello,
Let's assume that $\displaystyle \mathbb{R}^2$ is a ring, which operations are addition and multiplication using components (eg. $\displaystyle u + v = (u_1 + v_1, u_2 + v_2)$). The task is to prove that x-axis and y-axis are ideals of this ring. Following theorem must be used in the proof:
"Ring homomorphism's $\displaystyle f:R \rightarrow S$ kernel $\displaystyle Ker(f)$ is $\displaystyle R$'s ideal".

I think the problem isn't very hard, when the correct homomorphism is found. I tried some, but I couldn't discover anything that works. So, any help is welcome. Thank you very much!

2. 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.

3. 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.
I think topspin made a slight clerical error. To show that $\displaystyle X\unlhd \mathbb{R}$ one should consider the second canonical projection $\displaystyle \pi_2:\mathbb{R}^2\to\mathbb{R}a,b)\to b$ since the kernel of this is clearly $\displaystyle \mathbb{R}\times\{0\}=X$.