MHB Can every subring of an algebraic extension be a field?

[mathmari]

Hey!

$K \leq E$ an algebraic extension. I am asked to show that each subring of $E$ that contains $K$ is a field.

I have done the following:

$K \leq E$ algebraic $\Rightarrow \forall a \in E, \exists f(x)\in K[x]:f(a)=0$

A subfield of $E$ that contains $K$ is $K[a], \forall a \in E$.

Since $a$ is algebraic over $K$,it stands that $K(a)=K[a]$.

Therefore, a subring of $E$ that contains $K$ is $K(a),\forall a\in E$, where $K(a)$ is a field since it is a ring and $1\in K(a)$ so $1=f(a)g^{-1}(a)\Rightarrow f^{-1}(a)=g^{-1}(a)\in K(a)$.

Is this correct??

Or is there something I could improve??

 

[mathbalarka]

You have only proved that there is a subring of $E$ containing $K$ which is a field, but that is not what the questions asks for. It wants you to prove that *every* subring of $E$ containing $K$ is a field.

First go through the fundamental definitions. What are the properties which makes a ring a field? Being commutative, having the identity element *and* having inverses. The first two properties follow naturally, as the ring is contained in a field.

Assume $\alpha \in R \subset E$. You have to prove that $\alpha^{-1} \in R$ too. You need to look at some of the rings to get a good idea of these : think, for example, about $\Bbb Q[\omega]$ where $\omega$ is one of the cube roots of unity. This, of course, is a field$(+)$ but ignore that harsh reality for a second. Can we find an expression for $\omega^{-1}$? Well note that the minimal polynomial is $\omega^2 + \omega + 1 = 0$ so, $\omega(\omega + 1) = -1$ which implies $\omega(-1-\omega) = 1$. Thus, by definition of an inverse, $-1-\omega$ is the inverse of $\omega$.

Can you do similar with your ring $R$? Maybe consider the minimal polynomial of $\alpha$ in $K$? What can you do with it? Remember that $K$ is contained in $R$ and is "spanned" by $K$, i.e., rings (fields) $K[\gamma_i]$ are all sitting inside $R$ for $\gamma_i \in R$. Use these.

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$(+)$[CRANK]PS : I have been trying to find a ring which is also an algebraic extension of a field for years, and it's pretty sad that the search result is inconclusive as of now. Anyone who can help me with a good algorithm on GAP for doing this will get a share of 1000 dollars from the prize money[ENDCRANK]

 

[mathmari]

Can you do similar with your ring $R$? Maybe consider the minimal polynomial of $\alpha$ in $K$? What can you do with it? Remember that $K$ is contained in $R$ and is "spanned" by $K$, i.e., rings (fields) $K[\gamma_i]$ are all sitting inside $R$ for $\gamma_i \in R$. Use these.

Since $a$ is algebraic over $K$, it solves the minimal polynomial $Irr(a,K)=\gamma_0+\gamma_1x+\dots +\gamma_{n-1}x^{n-1}+\gamma_nx^n$.

$$\gamma_0+\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n=0$$

Since $Irr(a,K)$ is irreducible, it has to be $\gamma_0 \neq 0$, otherwise it could be written as $x \cdot q(x)$.

$$-(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)=\gamma_0 \\ \Rightarrow -\gamma_0^{-1}(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)=1 \\ \Rightarrow a \left ( -\gamma_0^{-1}(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)\right )=1$$

Therefore, $a^{-1}=-\gamma_0^{-1}(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n) \in R$.

Is this correct??

 

[mathbalarka]

Nope, you have made an algebra mistake on the 3rd line :

$$-(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)=\gamma_0 \\ \Rightarrow -\gamma_0^{-1}(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)=1 \\ \Rightarrow a \left ( -\gamma_0^{-1}\color{red}{(\gamma_1+ \dots +\gamma_{n-1}a^{n-2}+\gamma_na^{n-1})}\right )=1$$

Thus it follows that $a^{-1} = -\gamma_0^{-1} (\gamma_1 + \cdots + \gamma_n a^{n-1})$ but the right hand side is a polynomial in $a \in R$ with constant coefficients from $K \subset R$, thus in $R$. Hence, $a^{-1} \in R$.

 

[Deveno]

You have only proved that there is a subring of $E$ containing $K$ which is a field, but that is not what the questions asks for. It wants you to prove that *every* subring of $E$ containing $K$ is a field.

First go through the fundamental definitions. What are the properties which makes a ring a field? Being commutative, having the identity element *and* having inverses. The first two properties follow naturally, as the ring is contained in a field.

Assume $\alpha \in R \subset E$. You have to prove that $\alpha^{-1} \in R$ too. You need to look at some of the rings to get a good idea of these : think, for example, about $\Bbb Q[\omega]$ where $\omega$ is one of the cube roots of unity. This, of course, is a field$(+)$ but ignore that harsh reality for a second. Can we find an expression for $\omega^{-1}$? Well note that the minimal polynomial is $\omega^2 + \omega + 1 = 0$ so, $\omega(\omega + 1) = -1$ which implies $\omega(-1-\omega) = 1$. Thus, by definition of an inverse, $-1-\omega$ is the inverse of $\omega$.

Can you do similar with your ring $R$? Maybe consider the minimal polynomial of $\alpha$ in $K$? What can you do with it? Remember that $K$ is contained in $R$ and is "spanned" by $K$, i.e., rings (fields) $K[\gamma_i]$ are all sitting inside $R$ for $\gamma_i \in R$. Use these.

---

$(+)$[CRANK]PS : I have been trying to find a ring which is also an algebraic extension of a field for years, and it's pretty sad that the search result is inconclusive as of now. Anyone who can help me with a good algorithm on GAP for doing this will get a share of 1000 dollars from the prize money[ENDCRANK]

Isn't $\text{End}_{\Bbb R}(\Bbb R^n)$ such a ring, by the Cayley-Hamiltion theorem? Or did you mean something else?

 

[mathbalarka]

Ahhh this is a sneaky catch. I meant a ring which is a *finite* algebraic extension of a field ($K[x_1, x_2, \cdots, x_n] = K(x_1, x_2, \cdots, x_n)$ so there is none, which was the supposed to be the punch-line of the crank). Never studied Cayley-Hamilton, so that must be fun.

Thanks for the information!

 

[mathmari]

Nope, you have made an algebra mistake on the 3rd line :

$$-(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)=\gamma_0 \\ \Rightarrow -\gamma_0^{-1}(\gamma_1a+ \dots +\gamma_{n-1}a^{n-1}+\gamma_na^n)=1 \\ \Rightarrow a \left ( -\gamma_0^{-1}\color{red}{(\gamma_1+ \dots +\gamma_{n-1}a^{n-2}+\gamma_na^{n-1})}\right )=1$$

Thus it follows that $a^{-1} = -\gamma_0^{-1} (\gamma_1 + \cdots + \gamma_n a^{n-1})$ but the right hand side is a polynomial in $a \in R$ with constant coefficients from $K \subset R$, thus in $R$. Hence, $a^{-1} \in R$.

You`re right! Thank you for your answer! So, since R is a subring the only thing that we have to show is that each element of R has an inverse in R, right?? To do that we used the fact that each a is an algebraic element over K, since the extension is algebraic. What if the extension were not algebraic?? Would it still stand that each subring of E that contains K is a field??

 

[mathbalarka]

What if the extension were not algebraic?? Would it still stand that each subring of E that contains K is a field??

That's not nessesary. Pick up $E = \Bbb{Q}(X)$ and $K = \Bbb Q$ for some transcendental $X$ over $K$. $E/K$ thus is not algebraic. Let $R$ be the ring $\Bbb {Q}[X]$. This contains $K$, but is not a field : $R$ is the ring of rational polynomials, and $X \in R$. The inverse of $X$ in $E$ is $1/X$, which is not a polynomial, and thus not in $R$.

 

[mathmari]

That's not nessesary. Pick up $E = \Bbb{Q}(X)$ and $K = \Bbb Q$ for some transcendental $X$ over $K$. $E/K$ thus is not algebraic. Let $R$ be the ring $\Bbb {Q}[X]$. This contains $K$, but is not a field : $R$ is the ring of rational polynomials, and $X \in R$. The inverse of $X$ in $E$ is $1/X$, which is not a polynomial, and thus not in $R$.

I see... Thank you very much! (Sun)