cubic polynomial – Problems in Mathematics https://yutsumura.com Thu, 10 Aug 2017 15:39:21 +0000 en-US hourly 1 https://wordpress.org/?v=5.3.6 https://i2.wp.com/yutsumura.com/wp-content/uploads/2016/12/cropped-question-logo.jpg?fit=32%2C32&ssl=1 cubic polynomial – Problems in Mathematics https://yutsumura.com 32 32 114989322 Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$ https://yutsumura.com/cubic-polynomial-x3-2-is-irreducible-over-the-field-qi/ https://yutsumura.com/cubic-polynomial-x3-2-is-irreducible-over-the-field-qi/#respond Fri, 05 May 2017 02:29:45 +0000 https://yutsumura.com/?p=2838 Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.   Proof. Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion (with prime $p=2$). This implies that if $\alpha$...

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]]> Problem 399

Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.

 

Proof.

Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the degree of the field extension $\Q(\alpha)$ over $\Q$ is $3$:
\[[\Q(\alpha) : \Q]=3. \tag{*}\]

Seeking a contradiction, assume that $x^3-2$ is reducible over $\Q(i)$.
Then $x^3-2$ has a root in $\Q(i)$ as it is a reducible degree $3$ polynomial. So let us call the root $\alpha \in \Q(i)$.

Then $\Q(\alpha)$ is a subfield of $\Q(i)$ and thus we have
\[2=[\Q(i) :\Q]=[\Q(i): \Q(\alpha)][\Q(\alpha):\Q]\geq 3\] by (*). Hence we have reached a contradiction.
As a result, $x^3-2$ is irreducible over $\Q(i)$.

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]]> https://yutsumura.com/cubic-polynomial-x3-2-is-irreducible-over-the-field-qi/feed/ 0 2838 Application of Field Extension to Linear Combination https://yutsumura.com/application-of-field-extension-to-linear-combination/ https://yutsumura.com/application-of-field-extension-to-linear-combination/#respond Wed, 15 Mar 2017 02:33:20 +0000 https://yutsumura.com/?p=2433 Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$. Let $\alpha$ be any real root of $f(x)$. Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in...

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]]> Problem 335

Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.
Let $\alpha$ be any real root of $f(x)$.
Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.

 

Proof.

We first prove that the polynomial $f(x)=x^3-x+1$ is irreducible over $\Q$.
Since $f(x)$ is a monic cubic polynomial, the only possible roots are the divisors of the constant term $1$. As we have $f(1)=f(-1)=1\neq 0$, the polynomial has no rational roots. Hence $f(x)$ is irreducible over $\Q$.

Then $f(x)$ is the minimal polynomial of $\alpha$ over $\Q$, and hence the field extension $\Q(\alpha)$ over $\Q$ has degree $3$.
If $\sqrt{2}$ is a linear combination of $1, \alpha, \alpha^2$, then it follows that $\sqrt{2}\in \Q(\alpha)$. Then $\Q(\sqrt{2})$ is a subfield of $\Q(\alpha)$.

Then the degree of the field extension is
\begin{align*}
3=[\Q(\alpha): \Q]=[\Q(\alpha): \Q(\sqrt{2})] [\Q(\sqrt{2}): \Q].
\end{align*}
Since $[\Q(\sqrt{2}): \Q]=2$, this is impossible.
Thus, $\sqrt{2}$ is not a linear combination of $1, \alpha, \alpha^2$.

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