Question

Prove that $\mathit{T}\mathbf{=}\mathbf{\left\{}\mathbf{a}\mathbf{+}\mathbf{b}\sqrt{\mathbf{2}}\mathbf{|}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\in }\mathbf{\mathbb{Z}}\mathbf{\right\}}$ is a subring of$\mathbf{ℝ}$ and$\mathit{M}\mathbf{=}\mathbf{\left\{}\mathbf{a}\mathbf{+}\mathbf{b}\sqrt{\mathbf{2}}\mathbf{\left|}\mathbf{5}\mathbf{\right|}\mathbf{a}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{5}\mathbf{|}\mathbf{b}\mathbf{\right\}}$ is a maximal ideal in$\mathit{T}$

Short answer

Itis proved that$M$is a maximal ideal in$T$.

Step-by-step solution

Subring of a ring

Let $\mathit{R}$be a ring and $\mathit{S}$ be a subset of $\mathit{R}$. Then, S is said to be a subring of the ring $\mathit{R}$ if it is closed under subtraction and multiplication. That is, if $\mathit{a}\mathbf{,}\mathit{b}\mathbf{\in }\mathit{S}$then,

• $\mathit{a}\mathbf{-}\mathit{b}\mathbf{\in }\mathit{S}$
• $\mathit{a}\mathit{b}\mathbf{\in }\mathit{S}$
  

Maximal Ideal:

An ideal $\mathit{M}$ in a ring$\mathit{R}$ is said to be maximal if$\mathit{M}\mathbf{\ne }\mathit{R}$ and whenever$\mathit{J}$ is an ideal such that$\mathit{M}\mathbf{\subseteq }\mathit{J}\mathbf{\subseteq }\mathit{R}$ then,$\mathit{M}\mathbf{=}\mathit{J}$ , or $\mathit{J}\mathbf{=}\mathit{R}$.

Prove subring

Since,$0\in T$, so $T$is non-empty.

Let,$a_1+b_1\sqrt{2},a_2+b_2\sqrt{2}\in T$, then

$(a_1+b_1\sqrt{2})-(a_2+b_2\sqrt{2})=(a_1-a_2)+(b_1-b_2)\sqrt{2}\in T$and

$(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})=(a_1{a}_2+2b_1{b}_2)+(a_1{b}_2+a_2{b}_1)\sqrt{2}\in T$,

which proves that $T$is a subring of $\mathrm{ℝ}$.

Conclusion

By definition of an ideal, it is clear that$M$is an ideal of $T$.

Now, we have to prove that$M$is maximal ideal of $T$.

Let $p+q\sqrt{2}\notin M\Rightarrow 5\nmid pand5\nmid q$.

And therefore,${\left[p\right]}_5\ne {\left[0\right]}_5,{\left[q\right]}_5\ne {\left[0\right]}_5$in ${\mathrm{\mathbb{Z}}}_5$.

We know that, in ${\mathbb{Z}}_5$${\left[0\right]}_5^2={\left[0\right]}_5,{\left[1\right]}_5^2={\left[1\right]}_5,{\left[2\right]}_5^2={\left[4\right]}_5,{\left[3\right]}_5^2={\left[4\right]}_5,{\left[4\right]}_5^2={\left[1\right]}_5$and so on.

So,${[p^2-2q^2]}_5={\left[p\right]}_5^2-{\left[2q\right]}_5^2={\left[0\right]}_5$if and only if ${\left[p\right]}_5={\left[0\right]}_5={\left[q\right]}_5$.

Hence, we arrived at a contradiction. Therefore,

$p^2-2q^2=(p+q\sqrt{2})(p-q\sqrt{2})$is not divisible by 5.

So, any ideal $I$that contains $(p+q\sqrt{2})$and $M$, must contain $(p^2-2q^2)$and $5$.

By the division algorithm, we must have some $r$and $i$, such that,

$(p^2-2q^2)=5r+i$where, $i=1,2,3,4$.

Whatever the case may be,$1\in I$ so we have, $I=T$, which shows that$M$ is a maximal ideal of $T$.