Ideal of the ring

**Greg98** · March 27th 2011, 11:04 AM

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!

**topspin1617** · March 27th 2011, 12:59 PM

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.

**Drexel28** · March 27th 2011, 09:15 PM

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 $\pi_2:\mathbb{R}^2\to\mathbb{R}<img src=$ a,b)\to b" alt="\pi_2:\mathbb{R}^2\to\mathbb{R}

a,b)\to b" /> since the kernel of this is clearly $\mathbb{R}\times\{0\}=X$ .

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