Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?
15731
This is an exercise in some textbooks.     Let $E$ be an algebraic extension of $F$. Suppose $R$ is ring that contains $F$ and is contained in $E$. Prove that $R$ is a field.   The trouble is really with the inverse of $r$, where $r\in R$. How to prove that $r^{-1}\in R$, in apparent lack of a characterization of $R$.  It occurred to me to use the smallest field containing $R$ ($R$ is easily shown to be an integral domain), that's the field of quotients, and proving that it's $R$ itself, but I don't really know how to proceed.  A not-too-weak, not-too-strong hint will be much appreciated.   Beware $ $ Readers seeking only hints should  beware that there is now a complete answer.
13280
Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra.  (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite?
32348
If $R$ is a commutative integral domain and it also satisfies descending chain condition on its ideals then how will we show that such ring $R$ will be a field?
42912
It's well known that if $K$ is a finitely generated extension of some field $E$, then any intermediate field $F$, $E\subseteq F\subseteq K$, is also finitely generated over $E$.  I'm curious, does the same hold for rings?      Say $S$ is a ring, and $R\supset S$ is a finitely generated extension of $S$. If $T$ is any intermediate ring, is it necessarily true that $T$ is finitely generated over $S$ as a ring?   Is it as simple as saying that for any $t\in T$, $T$ can be generated by the generators of $R$ over $S$? I feel unsure about this statement, since it's not clear to me that the generators of $R$ over $S$ need be in $T$.  If not, what is an example what shows otherwise? Thanks.
22758
Let $F=\mathbb{F}_p(X,Y)$ be the field of rational functions in variables $X,Y$ over the finite field of $p$ elements. Let $K=\mathbb{F}_p(X^p,Y^p)$ be a subfield. Note that for any $f\in F$, $f^p\in K$. Deduce from this that $F/K$ is not a simple extension.
54222
Let $R$ be a commutative, unital ring and let $I$ be an ideal of $R.$ Prove that if $I$ is maximal then $R/I$ is a field.   This problem is probably trivial for most of you abstract algebra fans out there, but remember that I'm new to the field, so please don't use some next-level math that I can't understand at all. I think this problem could be solved using next-level math very quickly though.   For this problem, I know that fields are commutative, unital, and every nonzero element is a unit/invertible. Also, $I$ is maximal wrt subset, so if there is an ideal that contains $I,$ then it must be $R.$
27428
Let $K$ be a field and $f \in K[X]$ of degree $n$ with Galois group $S_n$. Let $a$ be a root of $f$, $L = K(a)$, and let $E$ be a intermediate field of the extension $K \subset L$. Prove that $E = K$ or $E = L$.   Any suggestions on where to start? Thanks.
98625
Let $K$ be an algebraic extension field (not necessarily finite) of $\mathbb{Q}$. Let $\mathscr{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$.  Then, is $\mathscr{O}_K$ Noetherian?  If $K$ is a finite extension of $\mathbb{Q}$, then $\mathscr{O}_K$ can be shown to be Noetherian. However, if we do not assume that finiteness condition, is it still true?  (It can be shown that $\mathscr{O}_K$ is an integrally closed domain of Krull-dimension $1$. Hence, if it is Noetherian, it is a Dedekind domain.)
115087
Let $K/F$ be a field extension, I know that if $K/F$ is a finite extension then a simple argument from linear algebra shows that since every homomorphism of fields from $K$ to $K$ that fixes $F$ is 1-1 it is also onto, i.e. an automorphism of $K$.  Can someone please give an example that this is not the case if $K/F$ is not a finite extension ?
1484
Let $R$ be a commutative ring with identity. Show that $R$ is a field if and only if the only ideals of $R$ are $R$ itself and the zero ideal $(0)$.  I can't figure out where to start other that I need to prove some biconditional statement. Any help?
85158
Let F be a char 0 field, K be a normal extension of F and L be a normal extension of K. Can it be proved or disproved that L is normal extension of F ?
95086
Let $x$ be an element not algebraic over $F$, and $K \subset F(x)$ a subfield that strictly contains $F$. Why is $x$ algebraic over $K$?  Thanks a lot!
4951
It is a known fact that if $k$ is a field that is finitely generated as a ring, which is the same as having a surjective ring homomorphism $f:\mathbb{Z}[x_1,\dots,x_n]\to k$ for some $n\in \mathbb{N}$, then $k$ must be finite. Since finite generation as a ring implies finite generation over the prime field ($\mathbb Q$ or $\mathbb F_p$), by Noether normalization it follows that $k$ must be a finite extension of its prime field. In positive characteristic this finishes the job and in zero characteristic, should lead quickly to a contradiction, though I don't see immediately how. Is there an elementary/slick proof of this fact?
93537
Let $L/K$ be a field extension. Let be $\alpha, \beta \in \mathbb{C}$, such that $[\mathbb{Q}(\alpha):\mathbb{Q}] = p$, and $[\mathbb{Q}(\beta):\mathbb{Q}] = q$, for some prime numbers $p$ and $q$. Assume $p \neq q$. Prove that:  $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\mathbb{Q}(\alpha,\beta):Q] = [\mathbb{Q}(\alpha):\mathbb{Q}] \cdot [\mathbb{Q}(\beta):\mathbb{Q}]$  I have no idea how to proceed on this, besides proving that the left hand side is $pq$.
27600
Suppose that $E$ is an extension of $F$ of prime degree. Show that $~~\forall~ a \in E : ~ F(a)=F$ or $F(a)=E$   Attempt: Suppose that $E$ is an extension of a field $F$ of prime degree, $p$. Therefore $p = [E :F] = [E : F(a)][F(a) : F]$. Since $p$ is a prime number, we see that either $[E : F(a)] = 1$ or $[F(a) : F] = 1$.   Now, $[E : F(a)] = 1 \implies $ there is only one element $x \in E$ which forms a basis and every element in $E$ is generated by $x$ i.e. $E = \{x~c~|~ c \in F(a)\}$  $[F(a) : F] = 1 \implies $ there is only one element $y \in F(a)$ which forms a basis and every element in $F(a)$ is generated by $y$ i.e. $F(a) = \{y~d~|~ d \in F\}$  Have I inferred it correctly? How do I move ahead?  Thank you for your help.
35458
Let $A$ be a ring with $0 \neq 1 $, which has $2^n-1$ invertible elements and less non-invertible elements. Prove that $A$ is a field.
14742
If $R$ is a number ring with number field $K$ and $f$ is a monic polynomial over $R$, then I want to show that any root of $f$ is an algebraic integer.