I am working through some problems in Field Theory on Hungerford's Algebra. However I have been struggling to understand the notation in some exercises:  1.14 In the field $k(x)$, let $u = x^3/(x+1)$. Show that $K(x)$ is a simple extension of the field $K(u)$. What is $[K(x):K(u)]$? 2.11 In the extension of $\mathbb{Q}$ by $\mathbb{Q}(x)$, the intermediate field $\mathbb{Q}(x^2)$ is closed but $\mathbb{Q}(x^3)$ is not.  I understand for a field $K$ what $K(x)$ means, this is the field of rational functions with indeterminate $x$. However, I do not understand what does $K(f(x))$ means. In class we worked with examples such as $K(u)$ where $u$ is a root of a polynomial in $K[x]$. In the linked questions the answers explain examples such as $\mathbb{Q}(\sqrt 2)$ which is the smallest field extension of $\mathbb{Q}$ that contains $\sqrt 2$. In general, for a field $K$ we have the extension $K(u)$ where $u\in F$ and $F$ is an extension of $K$. None of the linked answers indicates what it means to have $K(f)$ where $f\in K[x]$. To be more precise is $K(f)$ a quotient $k(x)/(f)$ or is it a shorthand notation for $K(u)$ where $f(u)=0$ or is it to replace the indeterminate $x$ for $f(x)$ in the rational function?
82449
I was wondering what the notation $R[a]$ really stands for, if $a\in K$, where $K$ is a ring  and $R$ is a subring of $K$.  In my book they define $\mathbb{Z}[\sqrt2]=\{a+b\sqrt2|a,b \in \mathbb{Z}\}$.  So, my guess is that $R[a] = \{P(a)\mid P \in R[X]\}$. Since for $\mathbb{Z}[\sqrt2]$ this is the case, or is this just a coincidence?
116701
From Berkeley Problems in Mathematics, Spring 1999, .  Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\ge 3$. Let $L$ be the splitting field of $f$, and let $\alpha\in L$ be a zero of $f$. Give that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^{4})=\mathbb{Q}(\alpha)$.   (BELOW IS NONSENSE, SHOULD BE IGNORED)  Following Gerry Myerson's advice, since $L$ is the splitting field of $f$, $L$ must be Galois over $\mathbb{Q}$. Thus the Galois group for $[L:\mathbb{Q}]=S_{n}$. Gerry now asserts that either $\mathbb{Q}(\alpha^{4})$ is of degree 2 over rationals or it must be $\mathbb{Q}$. But assuming this one can rule out the case $\mathbb{Q}$, since this would imply $\alpha$ is the root of a 4th or 2nd degree polynomial over $\mathbb{Q}$. But we know $n\ge 3$. So $\alpha$ must be the root of a 4th degree polynomial. We know $S_{n}$ has normal subgroups $A_{n}$ when $n\not=4$ and $V,A_{4}$ when $n=4$. I do not know how to proceed any further.   Now assuming $\mathbb{Q}(\alpha^{4})$ is a degree 2 abelian extension over $\mathbb{Q}$. We should have a chain of normal extensions $\mathbb{Q}(\alpha)\supset \mathbb{Q}(\alpha^{2})\supset \mathbb{Q}$. This would imply $S_{n}$ has a normal subgroup which has a normal subgroup. But we know $A_{n}$ for $n\ge 5$ are simple. So the only possibility is $\mathbb{Q}(\alpha)$ has a Galois group isomorphic to $V$. In this case $n=4$ as well. I can only proceed to here.
124213
Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?  ($\cong$ denotes an isomorphism that is the identity on restriction to $K$).  Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
117564
This is an exercise from Miles Reid, Undergraduate Algebraic Geometry. The proof has two parts, one of which I can do and one of which I can't. I could have some misunderstandings about notations here as well so anything people can point out will be helpful.  a) Let $k$ be an infinite field and $f\in k[X_1,...,X_n]$ be nonconstant. Prove  $$V(f)=\{P\in \mathbb{A}^n_k\mid f(P)=0\}\neq \mathbb{A}^n_k.$$  $A^n_k$ here is the point set $k^n$.   My proof (which follows from a hint of his) is write $f=\sum a_i(X_1,...,X_{n-1})X_n^i$. Then $V(f)$ consists of two kinds of points   $P=(p_1,...,p_{n-1},p_n=0)$ $P=(p_1,...,p_{n-1},p_n\neq 0 )$ with $\sum a_i(p_1,...,p_{n-1})p_{n}^i=0$   In the first case we have $V(f)\neq \mathbb{A}^n_k$. Assume we have $P$ like in the second case and now write $f=\sum b_i(X_1,...,\hat{X}_{n-1},X_n)X_{n-1}^i$. Here again, either $P=(p_1,...,p_{n-1}=0,p_n\neq 0)$ and $V(f)\neq \mathbb{A}^n_k$ or $p_{n-1}\neq 0$, and by induction we should get to $V(f)\neq \mathbb{A}^n_k$.  b) Now let $k$ be algebraically closed. Let $a_m(X_1,...,X_{n-1})X_n^m$ be the leading term of $f$. Show that when $a_m\neq 0$, there is a finite set of points of $V(f)$ corresponding to every value $(X_1,...,X_{n-1})$. Thus, $V(f)$ infinite for $n \geq 2$.  So I have kinda no idea what to do. A given value of $(X_1,...,X_{n-1})$ corresponds to points in $V(f)$ $P=(p_1,...,p_{n-1},p_n)$ where $f(P)=0$, but I don't know anything about the value of $p_n$; if it vanishes then clearly there is a correspondence $(X_1,...,X_n)\to (p_1,...,p_{n-1},0)$ but I guess I need to show there are a finite number of $p_n$ where this is true and I don't know how.
77911
I was asked to prove or disprove that if a field extension is not normal then it is separable. I can't see why this would be true so I want to disprove it with an example of a non-normal inseparable field extension. I am wondering if there is a simple example of such an extension, or even any examples at all. I know that any inseparable extension $L/K$ has to be such that $K$ is infinite and of positive characteristic, so a good place to start would be $K=\mathbb F_p(T)$, but I'm unsure how to prove that an extension of this is inseparable, and even unsure how to come up with a non-normal example, since my usual approach is to exploit that $K$ is a subfield of $\mathbb R$ (so e.g. $\mathbb Q(\sqrt[3]2)/\mathbb Q$ is not normal as $x^3-2$ has $\sqrt[3]2$ as a root but the other two roots are in $\mathbb C \setminus \mathbb R$ and $\mathbb Q(\sqrt[3]2)\subset \mathbb R$). However with $K$ as above, $K \nsubseteq \mathbb R. $
46247
I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet.  My question is as follows:  Suppose $p \in F[x]$ is the minimal polynomial of some matrix $A \in M_n(F)$ and $F \subseteq K$  I need to show that $p$ is also the minimal polynomial of $A$ over the field $K$. In other words: Show that the minimal polynomial remains the same when we move to a bigger field.  What I thought of:  Obviously $p$ is irreducible in $F$, suppose it is reducible in $K$, which means there are $q,t \in K, q,t \notin F$ such that $p=qt$, now we just need to show that $q(A) \neq 0$ and $t(A) \neq 0$...But why is that so?  Edit: Another way of presenting this question is "Show that if $p$ is the minimal polynomial of $A$ over the field $K$ while the entries of$~A$ are contained in a subfield $F \subseteq K$, then all the coefficients of $p$ are elements of $F$".
118804
I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
79826
My class notes has as theorem (without proof): "Let $K/F$ be finite field extension, with $K=F(\alpha_1,\ldots,\alpha_n)$ and $\alpha_k$ is separable for all $k$. Then $K/F$ is separable".  My question is: in either finite field, or field characteristic $0$, will the $\alpha_k$ always be separable? (by "$\alpha_k$ separable" I mean that its minimal polynomial is separable). Also, could somebody give me an example for which an $\alpha_k$ is not separable?
14228
How can I show given a field $K$ that $(K[x,y,z]/(x^2+y^3+z^7))_{(x,y,z)}$   is a UFD?   I found that statement in a Wikipedia page so i'm not 100% sure it's true, maybe it's true for some field only?
21801
Let $k$ be a finite field of order $q$ in characteristic $p$, let $n$ be a positive integer not divisible by $p$, and let $K$ be the splitting field of $X^n-1$ over $k$. Prove that $[K:k]$ equals the smallest positive integer $d$ such that $n\mid q^d-1$.  My approach: It's easy to show that if $k$ is a finite field of order $q$ with $\text{char} \ k=p$ then $k=\mathbb{F}_{q}$ with $q=p^m$ with $m\geq 1$. If $K$ is the splitting field of $X^n-1$ over $k$ then $K=\mathbb{F}_{q^r}$ for some suitable $r\geq 1$.  Let $\alpha \in K=\mathbb{F}_{q^r}$ be a root of $X^n-1$ then $\alpha^n-1=0$ and $\alpha\neq 0$.   Thus, $\alpha^n=1$ and $\alpha \in \mathbb{F}^{\times}_{q^r}$ where $\mathbb{F}^{\times}_{q^r}$ is the cyclic group of order $q^r-1$.  But I have some difficulties:  1) From $\alpha^n=1$ it follows that $o(\alpha)\mid n$. How to show that $o(\alpha)=n$? I was trying to show it using that $(n,p)=1$ but I failed.  Remark: Because if we can show that $o(\alpha)=n$ then the desired result follows immediately from cyclicity of multiplicative group of field $\mathbb{F}_{q^r}$.  I have spent one day in order to come up with this approach. So please do not duplicate this question and can anyone show how to answer my question, please.  Detailed explanation would be great!
112693
I have a question about calculating the degree of a finite field extension over $\mathbb{Q}$.  This is problem 18 in chapter 1 of Patrick Morandi's Field and Galois Theory. The problem asked to show that $$ \left[ \mathbb{Q}{ \left( \sqrt[4]{2},\sqrt{3} \right):\mathbb{Q} }\right] =8 $$ Here's my work : First, we have $$  \left[ \mathbb{Q} ( \sqrt[4]{2},\sqrt{3}):\mathbb{Q}\right] = \left[ \mathbb{Q}(\sqrt[4]{2},\sqrt{3}) : \mathbb{Q}({\sqrt[4]{2}}) \right] . \left[ \mathbb{Q}{(\sqrt[4]{2})}: \mathbb{Q} \right]$$ It's easy to show that $\left[ \mathbb{Q}{(\sqrt[4]{2})}: \mathbb{Q} \right] =4 $ because the minimal polynomial of $\sqrt[4]{2}$ over $\mathbb{Q}$ is $x^4 -2$. Next, I'm going to show that  $$ \left[ \mathbb{Q}(\sqrt[4]{2},\sqrt{3}) : \mathbb{Q}({\sqrt[4]{2}}) \right] =2$$ Consider the polynomial $f(x) = x^2 -3$. It's monic and $\sqrt{3}$ is a root of $f$, but I'm stuck in proving $\sqrt{3} \notin \mathbb{Q}(\sqrt[4]{2})$ (hence $f$ is irreducible over $\mathbb{Q}(\sqrt[4]{2}))$.  My question is: Is there any proof to show that $\sqrt{3} \notin \mathbb{Q}(\sqrt[4]{2})$ or another way to caculate the degree of this field extension?  Thank you for your help.
104050
My problem is "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is a simple extension"  I met a similar problem "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is a simple extension" and I showed that it is equal to Prove $\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}$. I let $u=\sqrt{2}+\sqrt{3}$ and deduce that $u-\sqrt{2}=\sqrt{3})$, then square it and prove $\sqrt{2} \in Q(\sqrt{2}+\sqrt{3})$. However, the same technique cannot be applied in the first problem since we have 3 square root.  I actually found a proof for the first problem, by letting $u=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and calculating $u,u^2,..., u^5$ and prove $\sqrt{2},\sqrt{3}\in Q(u)$. Though it's not so nice. Looking for a simpler answer
94168
Let $K$ be a field and $K(X)$ be the field of its rational functions.  Now let $\phi \in K(X)$ be a rational function such that $K(\phi) \neq K(X)$.  Now, since $\phi$ is transcendental over $K$, $K(\phi)$ is isomorphic to $K(X)$.  Is this a correct example of a field being isomorphic to its subfield?  Are there any other examples?
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.
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 $K[X]$.
111450
I want to show that $\mathbb{Q}(i,\sqrt[4]{1+i} )$ is normal and thus nonabelian extension of $\mathbb{Q}$, but got stuck. The main problem is that for the extension to be normal $\root4\of{1-i}$ should also be an element. I failed to show that. Thank you for your helps. ----------------------edit--------------- My question arose from , page 5, Remark 2.7. It is given as a counterexample of Kronecker and Weber theorem: When $K=\mathbb{Q}(i)$, $K(\root4\of{1-i})$ can never be any intermediate field of $K({ \zeta  }_{ n })/K$, where ${ \zeta  }_{ n }$ is a primitive $n$-th root of unity. I first expected that above statement could be proved by showing that $K(\root4\of{1-i})$ is not abelian over  $\mathbb{Q}$ since any cyclotomic extension of K is a cyclotomic extension of $\mathbb{Q}$ thus abelian. However from your advices, it seems that it's not even normal. But I can't still understand the method you showed me.. Is there any more easy explanation for this? When $K=\mathbb{Q}(i)$, $K(\root4\of{1-i})$ can never be any intermediate field of $K({ \zeta  }_{ n })/K$, where ${ \zeta  }_{ n }$ is a primitive $n$-th root of unity. I only know about elementary field and Galois theory.
81649
I found an argument online that the two fields are not isomorphic but I can't make sense of the argument. You find it .  These are the things I'm confused about when it comes to the argument.     It says that the field $\mathbb Q[x]/(x^2+1)$ contains a root of the polynomial $x^2+1$. I suppose this is the congruence class $[x]$? But this is not the same as the zero element of the field $\mathbb Q[x]/(x^2+1)$, right?   It says that every isomorfism takes $0$ to $0$. I interpret this as the zero-element in $\mathbb Q[x]/(x^2+1)$ (which I interpret to be $[0]$) is taken to the zero-element in $\mathbb Q[x]/(x^2+2)$ denoted $[0]_2$. But in the argument it says that this means that if $\mathbb Q[x]/(x^2+1)$ has a root of the polynomial, then also $\mathbb Q[x]/(x^2+2)$ must have a root of this same polynomial. Why is that?     Greatful for any help!