Math 361, Spring 2014, Assignment 4

Carefully define the following terms, then give one example and one non-example of each:

1. Algebraic element (of an extension field).
2. Minimal polynomial (of an algebraic element).
3. Conjugate (of an algebraic element).
4. Subextension generated by a subset.
5. Finitely generated extension.
6. Simple extension.

Carefully state the following theorems (you need not prove them):

1. Classification of simple extensions.

Solve the following problems:

1. Section 29, problems 1, 5, 7, 11, and 13.

--------------------End of assignment--------------------

Questions:

1. Is the quiz still going to be on Monday now that you posted the homework so late?

No, there will be no quiz this Monday. --Steven.Jackson (talk) 21:22, 22 February 2014 (EST)

2. Is there a quiz tomorrow or this week?

No, the next quiz is Monday, March 10. --Steven.Jackson (talk) 22:00, 27 February 2014 (EST)

3. For the exam, will we be expected to know how to use the Eisenstein Criterion to prove a polynomial is irriducible over $\mathbb{Q}$ (which we could use as an argument to support forming a field extension)?--Robert.Moray (talk) 11:32, 2 March 2014 (EST)

It sounds like a good idea to know that. --Steven.Jackson (talk) 21:44, 2 March 2014 (EST)

Solutions:

Definitions:

1. Algebraic element (of an extension field).

   Suppose $F\rightarrow E$ is an extension and $\alpha\in E$. We say that $\alpha$ is algebraic over $F$, if there is a non-constant $p\in F[x]$ such that $p\left(\alpha\right)=0$.

   Example:

   In $\mathbb{Q}\rightarrow\mathbb{C}$, $i$ is algebraic /$\mathbb{Q}$ because it is a root of $x^2+1$</p><p>'''Non-example:'''</p><p>In $\mathbb{Q}\rightarrow\mathbb{C}$, $\pi$ is not algebraic, or transcendental / $\mathbb{Q}$ </p><br><p>''Note:'' In $\mathbb{R}\rightarrow\mathbb{C}$, $\pi$ is algebraic /$\mathbb{R}$ it is a root of x-$\pi$.</p><br><br> # Minimal polynomial (of an algebraic element).<p>For $\alpha\in E$, set the ''annhilator'', $ann_{F[x]}(\alpha)=\{f\in F[x]|f(\alpha)=0\}$. This is an ideal of $F[x]$.</p><p>$ann_{F[x]}(\alpha)=0\Leftrightarrow\alpha$is transcendental over $F$</p><p>Otherwise,(if $\alpha$ is algebraic over $F$) $ann(\alpha)$ has a unique monic generator, $irr(\alpha , F)$, which is called the minimal polynomial.</p><p>'''Example:'''</p><p>In $\mathbb{R}\rightarrow\mathbb{C}$, $irr(i,\mathbb{R}) = x^2+1$.</p><p>To determine the minimal polynomial, $irr(1+i,\mathbb{R})$,</p><p>Set $\alpha= 1+i$, then '"`UNIQ-MathJax1-QINU`"' '"`UNIQ-MathJax2-QINU`"' '"`UNIQ-MathJax3-QINU`"' '"`UNIQ-MathJax4-QINU`"' So $irr(1+i,\mathbb{R})=x^2-2x+2$.</p><p>'''Non-Example:'''</p><br><br> # Conjugate (of an algebraic element).<p>If $\alpha ,\beta\in E$ and are algebraic $/F$ we write $\alpha\sim_F\beta$ (read "$\alpha$ is conjugate to $\beta$") if and only if '"`UNIQ-MathJax5-QINU`"'.</p><p>'''Example:'''</p><p>In $\mathbb{Q}\rightarrow\mathbb{R}$, $\sqrt{2}\sim_\mathbb{Q}-\sqrt{2}$, since: '"`UNIQ-MathJax6-QINU`"' '"`UNIQ-MathJax7-QINU`"'</p><p>'''Non-example:'''</p><p>In $\mathbb{R}\rightarrow\mathbb{R}$, $\sqrt{2}\nsim_\mathbb{R}-\sqrt{2}$, since: '"`UNIQ-MathJax8-QINU`"' '"`UNIQ-MathJax9-QINU`"'</p><br><br> # Subextension generated by a subset.<p>Suppose $F\rightarrow R$, and $S\subseteq E$ is any subset. Then there is a unique "smallest" sub-extension containing S. It is called the subextension generated by $S$, written $F(S)$.</p><p>'''Example:'''</p><p>In $\mathbb{R}\rightarrow\mathbb{C}$:</p><p>$\mathbb{R}(i) =\mathbb{C}$</p><p>$\mathbb{R}(\sqrt{2}) =\mathbb{R}$</p><p>'''Non-example:'''</p><br><br> # Finitely generated extension.<p>$E$ is finitely generated over $F$ if $E=F(a_1,\ldots,a_n)$ for some $\alpha_1,\ldots,\alpha_n\in E$.</p><p>'''Example'''</p><p>'''Non-example'''</p><br><br> # Simple extension.<p>$E$ is a simpe extension if $E=F(\alpha)$ for some $\alpha\in E$</p><p>'''Example'''</p><p>'''Non-example'''</p> ==='"`UNIQ--h-6--QINU`"'Theorems:=== # Classification of simple extensions (*a theorem worth remembering)<p>Suppose $E=F(\alpha)$.</p><p>If $\alpha$ is algebraic, then $E\simeq F[x] /<irr(\alpha, F)>$</p><p>If $\alpha$ is transcendental, then $E\simeq Frac(F[x])$.