ring theory Archives - Epsilonify

Show that Q[sqrt(2)] is a field

The Mathematician — Wed, 09 Nov 2022 13:00:00 +0000

Show that $\mathbb{Q}[\sqrt{2}]$ is a field

Proof. We have that

\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2} \ | \ a,b \in \mathbb{Q} \}

. So we need to show that all elements in

\mathbb{Q}[\sqrt{2}]

do have an inverse. Indeed, if we take arbitrary

a,b \in \mathbb{Q}

such that

a + b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]

, then

\begin{align*} 1 &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a + b \sqrt{2})(a - b \sqrt{2})} \\ &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - ab\sqrt{2} + ab\sqrt{2} - 2b^2)} \\ &= (a + b \sqrt{2}) \frac{a - b \sqrt{2}}{(a^2 - 2b^2)} \end{align*}

and since

a^2 - 2b^2 \in \mathbb{Q}

and

a - b \sqrt{2} \in \mathbb{Q}[\sqrt{2}]

, this implies that

\frac{a - b \sqrt{2}}{(a^2 - 2b^2)}

is an inverse of

a + b \sqrt{2}

. So this implies that

\mathbb{Q}[\sqrt{2}]

is a field. This completes the proof.

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Boolean ring is commutative

The Mathematician — Wed, 28 Sep 2022 13:00:00 +0000

Show that a Boolean ring is commutative

By definition, a ring

R

is Boolean if

x^2 = x

\forall x \in R

.

Proof. We need to show that

xy = yx

for all

x,y \in R

. So first, we have:

\begin{align*} (x + y)^2 = (x + y) &\iff x^2 + xy + yx + y^2 = x + y \\ &\iff x + xy + yx + y = x + y \\ &\iff xy + yx = 0 \end{align*}

Now we have

xy = -yx

. We would like to prove that

-yx = yx

. We can check that by finding its inverse:

\begin{align*} (y + y) &= (y + y)^2 \\ &= y^2 + 2y + y^2 \\ &= y + y + y + y \\ &= 0 \end{align*}

which implies that

y + y = 0

. Now we get

-y = y

and therefore we have that

xy = -yx = yx

, which implies that Boolean ring

R

is indeed a commutative ring.

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Prove that (2,x) is not a principal ideal of Z[x]

The Mathematician — Mon, 19 Sep 2022 13:00:00 +0000

Prove that $(2,x)$ is not a principal ideal of $\mathbb{Z}[x]$

Proof. Assume by contradiction that $(2,x)$ is a principal ideal. Then $(2,x) = (f(x))$ , where $f(x) \in \mathbb{Z}[x]$ . Say we have $2 \in (f(x))$ . Then by the definition of an ideal, we have that $2 = f(x)g(x)$ for some $g(x) \in \mathbb{Z}[x]$ . By some basic calculus and that $\mathbb{Z}[x]$ is an integral domain, we have that $deg(f(x)g(x)) = deg(f(x)) + deg(g(x))$ . So we know $2 = f(x)g(x)$ possible if both $f(x)$ and $g(x)$ are constants. This implies that $f(x),g(x) \in \{\pm 1, \pm 2\}$ . We see the two different situations:

Assume that $f(x) = \pm 1$ . Then this implies that $(f(x)) = \mathbb{Z}[x]$ , which is not a proper ideal.
Assume that $f(x) = \pm 2$ . Then we have that if $x \in (f(x))$ , then $x = 2\cdot g(x)$ . This implies that $g(x) = \frac{1}{2}x$ , which is impossible since $g(x) \in \{\pm 1, \pm 2\}$ and secondly $\frac{1}{2} \not \in \mathbb{Z}$ .

So $(2,x)$ is not a principal ideal of $\mathbb{Z}[x]$ , which completes the proof.

Note that with the proof above, we also proved that $\mathbb{Z}[x]$ is not a PID.

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