Let $L/k$ be a field extension of degree $n$. Then it is not difficult to see that $L$, as a $k$-algebra, can be embedded in $M_n(k)$. Now is it ever possible to embed $L$ in $M_{n-1}(k)$?
33999
Suppose $K \subset L$ is a finite field extension of degree $m$. Is it true that there exists some natural $n$ such that $L$ is isomorphic to a subfield of $M^{n\times n}(K)$, the ring of $n\times n$ matrices with entries in $K$? What is the smallest dimension (as a function of $m$) for which an isomorphism exists?  The example I had in mind was the 2-dimensional extension $\mathbb{R} \subset \mathbb{C}$ with $\mathbb{C}$ being isomorphic to $\{ \begin{pmatrix}a&amp;-b\\b&amp;a\end{pmatrix} : a, b \in \mathbb{R}\}$.
40688
Is there a (proper) subfield $K$ of $\mathbb{R}$ such that $\mathbb{R}$ is an algebraic extension of $K$?  From this question, , it is clear that for such an $K$, $[\mathbb{R}:K]=\infty$.  This question seems to be non-trivial, and I suspect any existence result will be non-constructive. Any references will be appreciated.
72517
I have problems to prove the following statement.     Let $K$ be a field and let $K(u,v)$ be an algebraic extension of $K$. If $u$ is separable over $K$ then $K(u,v)$ is a simple extension.   (My incomplete) proof: We can suppose that $char K=p$, because in characteristic $0$ all finite  extensions are simple. Let's divide the problem in $3$ cases (the third is problematic):  1) $v$ is separable. In this case $K(u,v)$ is a separable extension of $K$ and so it is simple.    2) $v$ is purely inseparable. We prove that $K(u,v)=K(u+v)$; by definition exists $m\in\mathbb N$ such that $v^{p^m}\in K$, so $(u+v)^{p^m}=u^{p^m}+v^{p^m}\in K(u^{p^m})$. So $K(u^{p^m})\subseteq K(u+v)$, but $u$ is separable on $K$ and follows that $K(u^{p^m})=K(u)$. Moreover $v=(u+v)-u\in K(u+v)$ and we can conclude that $K(u+v)=K(u,v)$.  3) $v$ is neither separable nor purely inseparable. ???    I would appreciate any hint about the point 3), but also a more organic proof without any subdivision.  Thanks in advance.
63271
Here is another problem of the book Fields and Galois Theory by Patrick Morandi, page 37.     Let $f(x)$ be an irreducible polynomial over $F$ of degree $n$, and let $K$ be a field extension of $F$ with $[K:F]=m$. If $\gcd(n,m)=1$, show that $f$ is irreducible over $K$.   I'm preparing for my midterm exam so I'm trying to solve as many as this book problems. Thanks for your helps.
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
86430
Problem: Consider the subfield $$E = \mathbb{Q} \big(\left\{\sqrt[n]{2} \mid n \in \mathbb{N} \setminus \left\{0\right\} \right\} \big)$$ of $\mathbb{R}$. Prove that $[E: \mathbb{Q}]$ is not finite.  Attempt: I proved that $E$ is algebraic over $\mathbb{Q}$. Because let $u \in E$. Then there is a $n \in \mathbb{N} \setminus \left\{0\right\}$ such that $u = \sqrt[n]{2}$. Then $f(x) = x^n - 2 \in \mathbb{Q}[x]$ is the minimal polynomial of $u$ over $\mathbb{Q}$ such that $f(u) = 0$. Hence $E$ is algebraic over $\mathbb{Q}$.  I now want to prove the degree of extension is not finite. I assume it is finite. Suppose then $[E: \mathbb{Q}] = k \in \mathbb{N}$. I don't know how to derive a contradiction from this.   I know that if $u$ is algebraic over $\mathbb{Q}$ and $f$ is the minimal polynomial of $u$ over $\mathbb{Q}$, then $[\mathbb{Q}(u) : \mathbb{Q}] = \deg(f)$.   Help/suggestions are appreciated.
40814
I recently saw in class that the degree $2$ extensions of a field of characteristic $\neq 2$ are given by square roots of non-squares in the base field.   I wonder what happens in the case of characteristic $2$ fields. For finite fields of characteristic $2$, it's clear what's going on, but what about others like $(\mathbb{Z}/2\mathbb{Z})(X)$?
25784
Let $\mathbb F$ be a field and $\mathbb  K $ be an extension field of $\mathbb F$  such that $\mathbb K$ is algebraically closed.   Let $\mathbb L$ be the field of all elements of $\mathbb K$ which are algebraic over $\mathbb F$. Then $\mathbb L_{|\mathbb F}$ is an algebraic extension.   My question is : Is $\mathbb L$ algebraically closed ?  I am trying to prove the existence of algebraic closure, so please don't assume that every field has an algebraic closure.
65879
Suppose that $L/K$ is a finite separable extension of fields and let $M$ denote the Galois closure of $L$. Let $\textrm{Hom}_K(L,M)$ denote the set of all $K$ - algebra homomorphisms from $L$ to $M$. Since $L/K$ is separable we know that the number of elements in $\textrm{Hom}_K(L,M)$ is equal to $[L:K]$. Now I want to prove that we have an isomorphism of $M$ - algebras   $$\varphi : M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$$  Proof that they are isomorphic as $K$ - modules : Write $L = K(a)$ for some $ a\in L$ (we can do this via the primitive element theorem). Then  $$\begin{eqnarray*} M \otimes_K L &amp;=&amp; M \otimes_K K[a] \\ &amp;\cong&amp; M\otimes_K K[x]/(f) \hspace{3mm} \text{where $f$ is the minimal polynomial of $a$ over $K$} \\ &amp;\cong&amp; M[x]/(f)\\ &amp;\cong&amp; M[x]/(f_1\ldots f_{[L:K]}) \hspace{3mm} \text{where the $f_i$ are the distinct}\\ &amp;&amp; \hspace{1.5in} \text{irreducible factors of $f$ since $M/L$ is Galois} \\ &amp;\cong&amp; M^{[L:K]}\end{eqnarray*}$$  where  the last step was using the Chinese remainder theorem. For the third last step, we consider the ses   $$ 0 \to (f) \to K[x] \to K[x]/(f) \to 0$$  and tensor with the exact functor $-\otimes_K M$ to get $$0 \to (f) \otimes_K M \to K[x] \otimes_K M \to K[x]/(f) \otimes_K M \to 0$$  and so $$\begin{eqnarray*} M \otimes_K K[x]/(f) &amp;\cong&amp; K[x]/(f) \otimes_{K} M \\ &amp;\cong&amp; \frac{K[x] \otimes_{K} M}{f \otimes_K M}\\ &amp; \cong&amp; M[x]/(f) \end{eqnarray*}$$ where $(f)$ is now viewed as an ideal of $M[x]$.          My question is: The tensor product $M \otimes_K L$ is a left $M$ - module, but why is it also an $M$ - algebra? Also why are the isomorphisms above isomorphisms of $M$ - algebras and not just $K$ - modules?
92660
Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.   Here $[F:K]$ denotes the dimension of $F$ over $K$ as a vector space.  I already know it is true for $K=F$, so the problem here is when $f$ is irreducible in $K[X]$.
123274
Suppose $K$ is a field, and $K\subset E$ is an algebraic extension, and $K\subset F$ is any extension.  Suppose that $L$ is some field which contains both $F$ and $E$ as subfields.  I want to show that $FE$ (the smallest subfield of $L$ containing both $F$ and $E$) is algebraic over $F$.  Lang mentions that the result is true because an element remains algebraic under lifting, but I can't prove the claim with this.  I interpret his comment to mean that if $x\in E$ is algebraic over $K$, then $x\in FE$ is algebraic over $F$.  This is certainly true.    But this doesn't seem enough if I start with any element of $FE$.
91629
Suppose $K/F$ is an extension field, and $E_1$ and $E_2$ are two intermediate subfields of finite degree. Let $E$ be the subfields of $K$ generated by $E_1$ and $E_2$. I'm trying to prove that $$ [E:F]\leq[E_1:F][E_2:F].$$  Since $E_1$ and $E_2$ are finite extensions, I suppose they have bases $\{a_1,\dots,a_n\}$ and $\{b_1,\dots,b_m\}$, respectively. If $E_1=F$ or $E_2=F$, then the inequality is actually equality, so I suppose both are proper extension fields. I think $E=F(a_1,\dots,a_n,b_1,\dots,b_m)$. Since $\{a_1,\dots,a_n,b_1,\dots,b_m\}$ is a spanning set for $E$ over $F$,  $$[E:F]\leq n+m\leq nm=[E_1:F][E_2:F]$$ since $m,n&gt;1$.  Is this sufficient? I'm weirded out since the problem did not ask to show $[E:F]\leq [E_1:F]+[E_2:F]$ which I feel will generally be a better upper bound.
105973
Let $X$ be an algebraic variety over field $k$ . $X$ can be embedded in a complete variety by Nagata's compactification theorem.    Moreover, can we embed $X$ in a projective space $\mathbb{P}_k^n$  $???$ Please give me references which contain either proof or counter examples.
55141
An old qual problem asks us to     Show that for every positive integer $n$, there exists a cyclic extension of $\mathbb{Q}$ of degree $n$ which is contained in $\mathbb{R}$.   A first thought might be towards Kummer theory: we could adjoin an $n^\text{th}$ root of, say, a prime number.  But when $n&gt;2$, $\mathbb{Q}$ lacks the full cohort of roots of unity that would make this work.  If $n$ is a power of $2$ we can get what we want by adjoining ($\mathbb{Q}$-linearly independent) square roots to $\mathbb{Q}$, and I think some  things can be done in other degrees ( at least $n=3$ and $n=5$)  but a more general $n$ has me stumped.  Could I get a nudge in the right direction on this problem?
64815
Let $F$ be a field and let $(K,*)$ be an associative $F$-algebra which, as a vector space, is finitely generated over $F$. Given an element $a\in K$, do there necessarily exist elements $a_1,a_2 \in K$ satisfying $a_1*a_2=a$?  I'm not sure how to go about this. Since we have an associative $F$-algebra we know that $v*(w*y)=(v*w)*y$ for all $v,w,y \in K$.
51426
If $E/F$ is a field extension, and $H$ is a subgroup of $\mathrm{Aut}(E/F)$, it is quite trivial to see that $H\subset \mathrm{Aut}(E/E^H)$.   Since the theorem only shows the inclusion relationship, I think there must be a lot of examples where $\subset$ is actually $\subsetneq$. But due to lack of knowledge I can't come up with one.   Could you help me with this? Best regards.
87694
I am doing exercise 3.24 of Marcus which is the following.           Let $L,K$ be number fields with $L/K$ a finite extension (of degree $[L:K] = n$) with $R = \mathcal{O}_K$ and $S = \mathcal{O}_L$. A  prime $P$ of $R$ is said to be totally ramified in $L$ iff $PS = Q^n$ for some prime $Q$ of $S$.           (a) Show that if $P$ is totally ramified in $L$ then it is totally ramified in $M$ for any intermediate field $M$, $K \subseteq M \subseteq L$.          (b) If $P$ is totally ramified in $L$ and unramified in another extension $L'$ of $K$ then $L \cap L'= K$.           (c) Give a new proof that $\Bbb{Q}(w)$ where $w = e^{2\pi i/m}$ has degree $\varphi(m)$ over $\Bbb{Q}$ (First do it for $m = p^r$ by using the fact that $(p) = (1- w)^{\varphi(m)}$ and then build up to any $m$ by using (b) above.      Now I have done (a) and (b) but when I come to (c) I fail to see how the problem is essentially circular. Even in the case $m = p^r$, I would have to use the fact that the ring of integers of $\Bbb{Q}(w)$ is $\Bbb{Z}[w]$. The proof of this fact already uses knowledge of $[\Bbb{Q}(w) : \Bbb{Q}]$ being equal to $\varphi(m)$. Am I missing something here?  Thanks.