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Decide whether the $\{a+b\sqrt{2} \mid a,b \in \Bbb Z\}$ is a ring with usual addition and multiplication. If a ring is formed, state whether the ring is commutative, whether it has unity, and whether it is a field.

I am confused about how this is a ring closed under the binary operation of "+". If a = b = 1, then don't we have some number that is not in the integers? Thanks for your help!

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  • $\begingroup$ In the set $\{a+bi\mid a,b\in \Bbb R, i^2=-1\}=\Bbb C$, don't we have numbers that are not in the real numbers? :) $\endgroup$
    – rschwieb
    Mar 21, 2014 at 15:46

2 Answers 2

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The condition is that the coefficients are integers, not that the elements of the ring are integers. The element $1+\sqrt 2$, in your example, is perfectly valid because the coefficients, $a=1$ and $b=1$, are integers. On the other hand, something like $$ \frac{1}{2}-\frac{1}{2}\sqrt2 $$ is not in the ring because it has coefficients that are not integers.

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  • $\begingroup$ So to prove this is closed would I say phi((a+b)+(c+d)) = a + bsqrt(2) + c + d sqrt(2) = phi(a+b) + phi(c+d) under phi? $\endgroup$ Mar 21, 2014 at 3:23
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    $\begingroup$ What is $\varphi$? $\endgroup$
    – user134824
    Mar 21, 2014 at 3:34
  • $\begingroup$ Nevermind, J. W. Perry answered my question about that, but in what direction should I step in to try and prove the left and right distributive laws of the above ring? $\endgroup$ Mar 21, 2014 at 3:40
  • $\begingroup$ You have to show that if $s$, $r_1$, and $r_2$ are ring elements then $sr_1+sr_2=s(r_1+r_2)$. You can do this by writing these elements as a sum of an integer and a multiple of $\sqrt2$ and multiplying everything out, or you can use the fact that $\mathbb R$ satisfies the distributivity law. $\endgroup$
    – user134824
    Mar 21, 2014 at 3:55
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Consider arbitrary elements $a_1+b_1\sqrt{2}$ and $a_2+b_2\sqrt{2}$ with $a_1,a_2,b_1,b_2 \in \mathbb{Z}$.

\begin{equation} a_1+b_1\sqrt{2} + a_2+b_2\sqrt{2} = (a_1+a_2)+(b_1+b_2)\sqrt{2}. \end{equation}

Since $a_1,a_2,b_1,b_2 \in \mathbb{Z}$, $a_1+a_2 \in \mathbb{Z}$ and $b_1+b_2\in \mathbb{Z}$ (closure of integers under $+$).

Therefore $(a_1+a_2)+(b_1+b_2)\sqrt{2} \in \{a+b\sqrt{2} | a,b \in \mathbb{Z}\}$, and the set is closed under $+$.

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