# Abstract Nonsense

## Algebraic Extensions (Pt. II)

Point of Post: This is a continuation of this post.

The fundamental fact about minimal polynomials is the following:

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Theorem: Let $k/F$ be an extension and $\alpha\in k$ algebraic. Then, $F(\alpha)=F[\alpha]\cong F[t]/(m_\alpha(t))$ as $F$-algebras.

Proof: Clearly $\text{ev}_\alpha:F[t]\to F[\alpha]$ is a surjective $F$-map and since $(m_\alpha(t))=\ker\text{ev}_\alpha$ the rest follows from the first isomorphism theorem for $F$-algebras (which, if you aren’t aware of is just a combination of the first ring and module isomorphism theorem).

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It remains to prove that $F[\alpha]=F(\alpha)$, but this is simple enough. Namely, we have the inclusion $F[\alpha]\subseteq F(\alpha)$. Conversely, note that $m_\alpha$ is necessarily irreducible since $F[\alpha]$ (being a subring if and integral domain) is an integral domain so that $F[t]/(m_\alpha)$ is an integral domain, and so $(m_\alpha)$ is a prime ideal but since $F[t]$ is a PID we know that all prime ideals are maximal and so $(m_\alpha)$ is maximal and thus $F[t]/(m_\alpha)\cong F[\alpha]$ is a field. Thus, $F[\alpha]$ being a field inside of $F(\alpha)$ implies that $F[\alpha]=F(\alpha)$ since $F(\alpha)$ is the minimal subfield of $k$ containing $\alpha$ and $F$. $\blacksquare$

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Remark: There’s an important element in this last paragraph that is important. Indeed, for any algebraic element in a field extension we have that the minimal polynomial is irreducible and monic, and moreover it’s clear that the reverse is true–any monic irreducible polynomial annihilating $\alpha$ must be the minimal polynomial.

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One of the most important things about the above theorem is that it allows us to conveniently count the dimension of an extension of the form $F(\alpha)$ with $\alpha$ algebraic. Indeed, we have the following theorem:

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Theorem: Let $F$ be a field and $f(t)\in F[t]$ be non-constant. Then, $F[t]/(f(t))$ is a $F$-vector space of dimension $\deg f(t)$.

Proof: Let $\deg f(t)=n\geqslant 1$, we claim that $\{\overline{1},\cdots,\overline{t^{n-1}}\}$ (where $\overline{g(t)}$ is the coset $g(t)+(f(t))$ in the quotient ring) is a basis for $F[t]/(f(t))$.

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Indeed, it’s clear that this set is linearly dependent for if $a_0\overline{1}+\cdots+a_{n-1}\overline{t^{n-1}}$ is zero in $F[t]/(f(t))$ then this tells us that $a_0\overline{1}+\cdots+a_{n-1}\overline{t^{n-1}}\in (f(t))$ and so $f(t)$ divides $a_0+\cdots+\cdots+a_{n-1}t^{n-1}$ which is impossible if $a_0=\cdots=a_{n-1}=0$ by degree considerations.

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To see that $\{\overline{1},\cdots,\overline{t^{n-1}}\}$ spans $F[t]/(f(t))$ we merely note that for any $g(t)\in F[t]$ the division algorithm in $F[t]$ tells us that we can write $g(t)=q(t)f(t)+r(t)$ where $\deg r(t). Thus, we see that

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$\overline{g(t)}=\overline{q(t)f(t)+r(t)}=\overline{r(t)}\in\text{span}_F\{\overline{1},\cdots,\overline{t^{n-1}}\}$

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and so the conclusion follows. $\blacksquare$

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Let’s put this to use. In our last post we pointed a cool consequence of the multiplicativity of the degrees in a tower. We said that $\sqrt{2}$ can’t be a polynomial in $\sqrt[5]{2}$ since this would imply that $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt[5]{2})$ and this can’t happen since we would need $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ to divide $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$. Of course, we (as of then) had no reason to believe that these indices were correct. Of course, using the above we can do this easily. Indeed, the above tells us that we just need to find the minimal polynomials of $\sqrt{2}$ and $\sqrt[5]{2}$ in $\mathbb{Q}$. So, what are they? Well we have some pretty obvious candidiates in the form $x^2-2$ and $x^5-2$ respectively. Since these are monic we only have to prove they are irreducible to conclude that they are, in fact, the minimal polynomials. To do this we appeal to Eisenstein’s criterion which says that since $2,5$ divide all but the leading term of $x^2-2$ and $x^2-5$ but $4,25$ don’t divide the constant term of $x^2-2$ and $x^2-5$ then the polynomials must be irreducible. Since $\deg(x^2-2)=2$ and $\deg(x^5-2)=5$ I have not lied to you about the degrees of $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt[5]{2})$.

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Now what we’d like to discuss are the field extension that every element is algebraic. Unimaginatively we call such an extension algebraic.

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For example, the extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is algebraic. To see this we merely note that if you give me any nonzero $\alpha\in\mathbb{Q}(\sqrt{2})$ then the fact that $1,\alpha,\alpha^2$ are three distinct elements in a two dimensional $\mathbb{Q}$-space tells me that there exists $a_0,a_1,a_2$ such that $a_0+a_1\alpha+a_2\alpha^2=0$  so that $\alpha$ must satisfy the polynomial $f(t)=a_0+a_1t+a_2t^2$.

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References:

[1] Morandi, Patrick. Field and Galois Theory. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. Algebra. New York: Springer, 2002. Print.

[4] Conrad, Keith. Collected Notes on Field and Galois Theory. Web. <http://www.math.uconn.edu/~kconrad/blurbs/&gt;.

[5] Clark, Pete. Field Theory. Web. <http://math.uga.edu/~pete/FieldTheory.pdf&gt;.

March 25, 2012 -