# Ideal of the ring

• March 27th 2011, 11:04 AM
Greg98
Ideal of the ring
Hello,
Let's assume that $\mathbb{R}^2$ is a ring, which operations are addition and multiplication using components (eg. $u + v = (u_1 + v_1, u_2 + v_2)$). 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 $f:R \rightarrow S$ kernel $Ker(f)$ is $R$'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!
• March 27th 2011, 12:59 PM
topspin1617
Well, the $x$-axis is, by definition, the set

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

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

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

$f:\mathbb{R}^2\rightarrow R$
$(a,b)\mapsto a$

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

You can do a completely analogous thing to show that the $Y$ axis is also an ideal.
• March 27th 2011, 09:15 PM
Drexel28
Quote:

Originally Posted by topspin1617
Well, the $x$-axis is, by definition, the set

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

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

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

$f:\mathbb{R}^2\rightarrow R$
$(a,b)\mapsto a$

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

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

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