Mixing
Given a field extension $Ksubset M$ with algebraic elements in x,y: Show that g(x,y)=0 for any g
Given a field-extension $Ksubset M$. Let $x,y in $ M be algebraic,
$ fin K[X,Y]$ and $z=f(x,y) in M$. Show that $z$ is algebraic.
So far I got this obvious step:
Since $x$ and $y$ are algebraic so we can find a $gin K[X,Y]$ with $g(x,y)=0$
I can't see a connection, can someone give me a hint?
I tried seeing it with a example, but i can't generalize it.
abstract-algebra extension-field
edited Nov 20 '18 at 19:23
dantopa
6,42932042
asked Nov 20 '18 at 19:17
dougle
155
add a comment |
Given a field-extension $Ksubset M$. Let $x,y in $ M be algebraic,
$ fin K[X,Y]$ and $z=f(x,y) in M$. Show that $z$ is algebraic.
So far I got this obvious step:
Since $x$ and $y$ are algebraic so we can find a $gin K[X,Y]$ with $g(x,y)=0$
I can't see a connection, can someone give me a hint?
I tried seeing it with a example, but i can't generalize it.
abstract-algebra extension-field
edited Nov 20 '18 at 19:23
dantopa
6,42932042
asked Nov 20 '18 at 19:17
dougle
155
add a comment |
Given a field-extension $Ksubset M$. Let $x,y in $ M be algebraic,
$ fin K[X,Y]$ and $z=f(x,y) in M$. Show that $z$ is algebraic.
So far I got this obvious step:
Since $x$ and $y$ are algebraic so we can find a $gin K[X,Y]$ with $g(x,y)=0$
I can't see a connection, can someone give me a hint?
I tried seeing it with a example, but i can't generalize it.
abstract-algebra extension-field
edited Nov 20 '18 at 19:23
dantopa
6,42932042
asked Nov 20 '18 at 19:17
dougle
155
Given a field-extension $Ksubset M$. Let $x,y in $ M be algebraic,
$ fin K[X,Y]$ and $z=f(x,y) in M$. Show that $z$ is algebraic.
So far I got this obvious step:
Since $x$ and $y$ are algebraic so we can find a $gin K[X,Y]$ with $g(x,y)=0$
I can't see a connection, can someone give me a hint?
I tried seeing it with a example, but i can't generalize it.
abstract-algebra extension-field
abstract-algebra extension-field
edited Nov 20 '18 at 19:23
dantopa
6,42932042
asked Nov 20 '18 at 19:17
dougle
155
edited Nov 20 '18 at 19:23
dantopa
6,42932042
asked Nov 20 '18 at 19:17
dougle
155
edited Nov 20 '18 at 19:23
dantopa
6,42932042
edited Nov 20 '18 at 19:23
dantopa
6,42932042
edited Nov 20 '18 at 19:23
dantopa
6,42932042
6,42932042
asked Nov 20 '18 at 19:17
dougle
155
asked Nov 20 '18 at 19:17
dougle
155
asked Nov 20 '18 at 19:17
dougle
155
155
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If $x, y in M$ are each algebraic over $K$, then of course
$K(x), K(y) subset M tag 1$
and
$[K(x):K] = n, [K(y):K] = m < infty; tag 2$
now any polynomial
$f(X, Y) in K[X, Y] tag 3$
may be written
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j, ; f_{ij} in K, forall i, j, tag 4$
which may be re-arranged as follows:
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j = sum_{j = 0}^{j = deg f} left ( sum_{i = 0}^{i = deg f - j} f_{ij} X^i right ) Y^j= sum_{j = 0}^{deg f} P_j(X)Y^j, tag 5$
where
$P_j(X) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} X^i in K[X]; tag 6$
in this way we express $f(X, Y)$ as a polynomial in $Y$ with coefficients in the ring $K[X]$; that is, as an element of $K[X][Y]$:
$f(X, Y) in K[X][Y]; tag 7$
in the light of these remarks, we see that
$z = f(x, y) = displaystyle sum_{j = 0}^{deg f} P_j(x)y^j in K(x)(y) = K(x, y), tag 8$
where
$P_j(x) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} x^i in K(x). tag 9$
By virtue of (8), we see that $z$ is algebraic over $K$ provided that
$[K(x)(y):K] = [K(x, y):K] < infty; tag{10}$
since
$[K(x):K] = n < infty, tag{11}$
we see that (10) will bind if
$[K(x)(y):K(x)] = [K(x, y):K(x)] < infty, tag{12}$
by virtue of
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K]; tag{13}$
now by (2), we know that $y$ satisfies some polynomial
$theta(X) in K[X]; ; theta(y) = 0, tag{14}$
with
$deg theta le m; tag{15}$
but since
$theta(X) in K[X] subset K(x)[X], tag{16}$
it follows that
$[K(x)(y):K(x)] le m; tag{17}$
therefore (13) yields
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K] le mn; tag{18}$
now (8) and (18) act together in collusion to prove that $z$ is algebraic over $K$.
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
add a comment |
Take a moment to think, what are you doing when you evaluate $f$ on $x$ and $y$? You're just adding and multiplying them together, right? So this just boils down to showing that the algebraic elements of a field extension are a subfield of the extension (in fact, you just need to show they are a subring for the purposes of this question).
Edit: My previous proof was really hasty and totally mistaken, but the fact that the algebraic elements of an extension are a subfield follows from basic facts about field extensions. Let me sketch an argument for you here.
Edit #2: I wasn't totally happy with the second attempt either. Here I give a more fleshed out proof.
Suppose $alpha$ and $beta$ are algebraic. Recall that $K(alpha) cong K[X]/langle min_alpha(x) rangle$, where $min_alpha(x)$ is the minimum polynomial of the element $alpha$ and $K(alpha)$ is the smallest subfield of $M$ that contains $alpha$. Now we have that $K(alpha)$ is a vector space over $K$ of dimension equal to the degree of $alpha$, and in particular it is spanned by the elements $1, alpha, alpha^2,ldots alpha^{n-1}$ where $n$ is the degree of $alpha$. What this means is that $alpha^{-1}$ is a polynomial expression in terms of $alpha$, and so if the algebraic numbers are closed under multplication and addition, they are closed under inversion (of nonzero numbers) as well. Since $alpha$ and $beta$ are algebraic, it follows that $[K(alpha,beta):K] = [K(alpha,beta):K(alpha)][K(alpha):K]$ is finite (do you see why each factor is finite, in particular the first one?), and so $K(alpha,beta)$ is algebraic. Therefore $alpha + beta$ and $alphabeta$ are algebraic over $K$, and we are done. For additional proofs of this fact you can look at this thread, from which I adapted the proof presented here.
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
1
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
2
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
1
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
1
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
1
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
|
show 6 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006762%2fgiven-a-field-extension-k-subset-m-with-algebraic-elements-in-x-y-show-that-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $x, y in M$ are each algebraic over $K$, then of course
$K(x), K(y) subset M tag 1$
and
$[K(x):K] = n, [K(y):K] = m < infty; tag 2$
now any polynomial
$f(X, Y) in K[X, Y] tag 3$
may be written
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j, ; f_{ij} in K, forall i, j, tag 4$
which may be re-arranged as follows:
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j = sum_{j = 0}^{j = deg f} left ( sum_{i = 0}^{i = deg f - j} f_{ij} X^i right ) Y^j= sum_{j = 0}^{deg f} P_j(X)Y^j, tag 5$
where
$P_j(X) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} X^i in K[X]; tag 6$
in this way we express $f(X, Y)$ as a polynomial in $Y$ with coefficients in the ring $K[X]$; that is, as an element of $K[X][Y]$:
$f(X, Y) in K[X][Y]; tag 7$
in the light of these remarks, we see that
$z = f(x, y) = displaystyle sum_{j = 0}^{deg f} P_j(x)y^j in K(x)(y) = K(x, y), tag 8$
where
$P_j(x) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} x^i in K(x). tag 9$
By virtue of (8), we see that $z$ is algebraic over $K$ provided that
$[K(x)(y):K] = [K(x, y):K] < infty; tag{10}$
since
$[K(x):K] = n < infty, tag{11}$
we see that (10) will bind if
$[K(x)(y):K(x)] = [K(x, y):K(x)] < infty, tag{12}$
by virtue of
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K]; tag{13}$
now by (2), we know that $y$ satisfies some polynomial
$theta(X) in K[X]; ; theta(y) = 0, tag{14}$
with
$deg theta le m; tag{15}$
but since
$theta(X) in K[X] subset K(x)[X], tag{16}$
it follows that
$[K(x)(y):K(x)] le m; tag{17}$
therefore (13) yields
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K] le mn; tag{18}$
now (8) and (18) act together in collusion to prove that $z$ is algebraic over $K$.
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
add a comment |
If $x, y in M$ are each algebraic over $K$, then of course
$K(x), K(y) subset M tag 1$
and
$[K(x):K] = n, [K(y):K] = m < infty; tag 2$
now any polynomial
$f(X, Y) in K[X, Y] tag 3$
may be written
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j, ; f_{ij} in K, forall i, j, tag 4$
which may be re-arranged as follows:
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j = sum_{j = 0}^{j = deg f} left ( sum_{i = 0}^{i = deg f - j} f_{ij} X^i right ) Y^j= sum_{j = 0}^{deg f} P_j(X)Y^j, tag 5$
where
$P_j(X) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} X^i in K[X]; tag 6$
in this way we express $f(X, Y)$ as a polynomial in $Y$ with coefficients in the ring $K[X]$; that is, as an element of $K[X][Y]$:
$f(X, Y) in K[X][Y]; tag 7$
in the light of these remarks, we see that
$z = f(x, y) = displaystyle sum_{j = 0}^{deg f} P_j(x)y^j in K(x)(y) = K(x, y), tag 8$
where
$P_j(x) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} x^i in K(x). tag 9$
By virtue of (8), we see that $z$ is algebraic over $K$ provided that
$[K(x)(y):K] = [K(x, y):K] < infty; tag{10}$
since
$[K(x):K] = n < infty, tag{11}$
we see that (10) will bind if
$[K(x)(y):K(x)] = [K(x, y):K(x)] < infty, tag{12}$
by virtue of
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K]; tag{13}$
now by (2), we know that $y$ satisfies some polynomial
$theta(X) in K[X]; ; theta(y) = 0, tag{14}$
with
$deg theta le m; tag{15}$
but since
$theta(X) in K[X] subset K(x)[X], tag{16}$
it follows that
$[K(x)(y):K(x)] le m; tag{17}$
therefore (13) yields
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K] le mn; tag{18}$
now (8) and (18) act together in collusion to prove that $z$ is algebraic over $K$.
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
add a comment |
If $x, y in M$ are each algebraic over $K$, then of course
$K(x), K(y) subset M tag 1$
and
$[K(x):K] = n, [K(y):K] = m < infty; tag 2$
now any polynomial
$f(X, Y) in K[X, Y] tag 3$
may be written
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j, ; f_{ij} in K, forall i, j, tag 4$
which may be re-arranged as follows:
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j = sum_{j = 0}^{j = deg f} left ( sum_{i = 0}^{i = deg f - j} f_{ij} X^i right ) Y^j= sum_{j = 0}^{deg f} P_j(X)Y^j, tag 5$
where
$P_j(X) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} X^i in K[X]; tag 6$
in this way we express $f(X, Y)$ as a polynomial in $Y$ with coefficients in the ring $K[X]$; that is, as an element of $K[X][Y]$:
$f(X, Y) in K[X][Y]; tag 7$
in the light of these remarks, we see that
$z = f(x, y) = displaystyle sum_{j = 0}^{deg f} P_j(x)y^j in K(x)(y) = K(x, y), tag 8$
where
$P_j(x) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} x^i in K(x). tag 9$
By virtue of (8), we see that $z$ is algebraic over $K$ provided that
$[K(x)(y):K] = [K(x, y):K] < infty; tag{10}$
since
$[K(x):K] = n < infty, tag{11}$
we see that (10) will bind if
$[K(x)(y):K(x)] = [K(x, y):K(x)] < infty, tag{12}$
by virtue of
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K]; tag{13}$
now by (2), we know that $y$ satisfies some polynomial
$theta(X) in K[X]; ; theta(y) = 0, tag{14}$
with
$deg theta le m; tag{15}$
but since
$theta(X) in K[X] subset K(x)[X], tag{16}$
it follows that
$[K(x)(y):K(x)] le m; tag{17}$
therefore (13) yields
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K] le mn; tag{18}$
now (8) and (18) act together in collusion to prove that $z$ is algebraic over $K$.
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
If $x, y in M$ are each algebraic over $K$, then of course
$K(x), K(y) subset M tag 1$
and
$[K(x):K] = n, [K(y):K] = m < infty; tag 2$
now any polynomial
$f(X, Y) in K[X, Y] tag 3$
may be written
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j, ; f_{ij} in K, forall i, j, tag 4$
which may be re-arranged as follows:
$f(X, Y) = displaystyle sum_{i, j = 0}^{i + j le deg f} f_{ij}X^i Y^j = sum_{j = 0}^{j = deg f} left ( sum_{i = 0}^{i = deg f - j} f_{ij} X^i right ) Y^j= sum_{j = 0}^{deg f} P_j(X)Y^j, tag 5$
where
$P_j(X) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} X^i in K[X]; tag 6$
in this way we express $f(X, Y)$ as a polynomial in $Y$ with coefficients in the ring $K[X]$; that is, as an element of $K[X][Y]$:
$f(X, Y) in K[X][Y]; tag 7$
in the light of these remarks, we see that
$z = f(x, y) = displaystyle sum_{j = 0}^{deg f} P_j(x)y^j in K(x)(y) = K(x, y), tag 8$
where
$P_j(x) = displaystyle sum_{i = 0}^{i = deg f - j} f_{ij} x^i in K(x). tag 9$
By virtue of (8), we see that $z$ is algebraic over $K$ provided that
$[K(x)(y):K] = [K(x, y):K] < infty; tag{10}$
since
$[K(x):K] = n < infty, tag{11}$
we see that (10) will bind if
$[K(x)(y):K(x)] = [K(x, y):K(x)] < infty, tag{12}$
by virtue of
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K]; tag{13}$
now by (2), we know that $y$ satisfies some polynomial
$theta(X) in K[X]; ; theta(y) = 0, tag{14}$
with
$deg theta le m; tag{15}$
but since
$theta(X) in K[X] subset K(x)[X], tag{16}$
it follows that
$[K(x)(y):K(x)] le m; tag{17}$
therefore (13) yields
$[K(x)(y):K] = [K(x)(y):K(x)][K(x):K] le mn; tag{18}$
now (8) and (18) act together in collusion to prove that $z$ is algebraic over $K$.
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
answered Nov 21 '18 at 8:19
Robert Lewis
43.8k22963
43.8k22963
add a comment |
add a comment |
Take a moment to think, what are you doing when you evaluate $f$ on $x$ and $y$? You're just adding and multiplying them together, right? So this just boils down to showing that the algebraic elements of a field extension are a subfield of the extension (in fact, you just need to show they are a subring for the purposes of this question).
Edit: My previous proof was really hasty and totally mistaken, but the fact that the algebraic elements of an extension are a subfield follows from basic facts about field extensions. Let me sketch an argument for you here.
Edit #2: I wasn't totally happy with the second attempt either. Here I give a more fleshed out proof.
Suppose $alpha$ and $beta$ are algebraic. Recall that $K(alpha) cong K[X]/langle min_alpha(x) rangle$, where $min_alpha(x)$ is the minimum polynomial of the element $alpha$ and $K(alpha)$ is the smallest subfield of $M$ that contains $alpha$. Now we have that $K(alpha)$ is a vector space over $K$ of dimension equal to the degree of $alpha$, and in particular it is spanned by the elements $1, alpha, alpha^2,ldots alpha^{n-1}$ where $n$ is the degree of $alpha$. What this means is that $alpha^{-1}$ is a polynomial expression in terms of $alpha$, and so if the algebraic numbers are closed under multplication and addition, they are closed under inversion (of nonzero numbers) as well. Since $alpha$ and $beta$ are algebraic, it follows that $[K(alpha,beta):K] = [K(alpha,beta):K(alpha)][K(alpha):K]$ is finite (do you see why each factor is finite, in particular the first one?), and so $K(alpha,beta)$ is algebraic. Therefore $alpha + beta$ and $alphabeta$ are algebraic over $K$, and we are done. For additional proofs of this fact you can look at this thread, from which I adapted the proof presented here.
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
1
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
2
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
1
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
1
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
1
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
|
show 6 more comments
Take a moment to think, what are you doing when you evaluate $f$ on $x$ and $y$? You're just adding and multiplying them together, right? So this just boils down to showing that the algebraic elements of a field extension are a subfield of the extension (in fact, you just need to show they are a subring for the purposes of this question).
Edit: My previous proof was really hasty and totally mistaken, but the fact that the algebraic elements of an extension are a subfield follows from basic facts about field extensions. Let me sketch an argument for you here.
Edit #2: I wasn't totally happy with the second attempt either. Here I give a more fleshed out proof.
Suppose $alpha$ and $beta$ are algebraic. Recall that $K(alpha) cong K[X]/langle min_alpha(x) rangle$, where $min_alpha(x)$ is the minimum polynomial of the element $alpha$ and $K(alpha)$ is the smallest subfield of $M$ that contains $alpha$. Now we have that $K(alpha)$ is a vector space over $K$ of dimension equal to the degree of $alpha$, and in particular it is spanned by the elements $1, alpha, alpha^2,ldots alpha^{n-1}$ where $n$ is the degree of $alpha$. What this means is that $alpha^{-1}$ is a polynomial expression in terms of $alpha$, and so if the algebraic numbers are closed under multplication and addition, they are closed under inversion (of nonzero numbers) as well. Since $alpha$ and $beta$ are algebraic, it follows that $[K(alpha,beta):K] = [K(alpha,beta):K(alpha)][K(alpha):K]$ is finite (do you see why each factor is finite, in particular the first one?), and so $K(alpha,beta)$ is algebraic. Therefore $alpha + beta$ and $alphabeta$ are algebraic over $K$, and we are done. For additional proofs of this fact you can look at this thread, from which I adapted the proof presented here.
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
1
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
2
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
1
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
1
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
1
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
|
show 6 more comments
Take a moment to think, what are you doing when you evaluate $f$ on $x$ and $y$? You're just adding and multiplying them together, right? So this just boils down to showing that the algebraic elements of a field extension are a subfield of the extension (in fact, you just need to show they are a subring for the purposes of this question).
Edit: My previous proof was really hasty and totally mistaken, but the fact that the algebraic elements of an extension are a subfield follows from basic facts about field extensions. Let me sketch an argument for you here.
Edit #2: I wasn't totally happy with the second attempt either. Here I give a more fleshed out proof.
Suppose $alpha$ and $beta$ are algebraic. Recall that $K(alpha) cong K[X]/langle min_alpha(x) rangle$, where $min_alpha(x)$ is the minimum polynomial of the element $alpha$ and $K(alpha)$ is the smallest subfield of $M$ that contains $alpha$. Now we have that $K(alpha)$ is a vector space over $K$ of dimension equal to the degree of $alpha$, and in particular it is spanned by the elements $1, alpha, alpha^2,ldots alpha^{n-1}$ where $n$ is the degree of $alpha$. What this means is that $alpha^{-1}$ is a polynomial expression in terms of $alpha$, and so if the algebraic numbers are closed under multplication and addition, they are closed under inversion (of nonzero numbers) as well. Since $alpha$ and $beta$ are algebraic, it follows that $[K(alpha,beta):K] = [K(alpha,beta):K(alpha)][K(alpha):K]$ is finite (do you see why each factor is finite, in particular the first one?), and so $K(alpha,beta)$ is algebraic. Therefore $alpha + beta$ and $alphabeta$ are algebraic over $K$, and we are done. For additional proofs of this fact you can look at this thread, from which I adapted the proof presented here.
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
Take a moment to think, what are you doing when you evaluate $f$ on $x$ and $y$? You're just adding and multiplying them together, right? So this just boils down to showing that the algebraic elements of a field extension are a subfield of the extension (in fact, you just need to show they are a subring for the purposes of this question).
Edit: My previous proof was really hasty and totally mistaken, but the fact that the algebraic elements of an extension are a subfield follows from basic facts about field extensions. Let me sketch an argument for you here.
Edit #2: I wasn't totally happy with the second attempt either. Here I give a more fleshed out proof.
Suppose $alpha$ and $beta$ are algebraic. Recall that $K(alpha) cong K[X]/langle min_alpha(x) rangle$, where $min_alpha(x)$ is the minimum polynomial of the element $alpha$ and $K(alpha)$ is the smallest subfield of $M$ that contains $alpha$. Now we have that $K(alpha)$ is a vector space over $K$ of dimension equal to the degree of $alpha$, and in particular it is spanned by the elements $1, alpha, alpha^2,ldots alpha^{n-1}$ where $n$ is the degree of $alpha$. What this means is that $alpha^{-1}$ is a polynomial expression in terms of $alpha$, and so if the algebraic numbers are closed under multplication and addition, they are closed under inversion (of nonzero numbers) as well. Since $alpha$ and $beta$ are algebraic, it follows that $[K(alpha,beta):K] = [K(alpha,beta):K(alpha)][K(alpha):K]$ is finite (do you see why each factor is finite, in particular the first one?), and so $K(alpha,beta)$ is algebraic. Therefore $alpha + beta$ and $alphabeta$ are algebraic over $K$, and we are done. For additional proofs of this fact you can look at this thread, from which I adapted the proof presented here.
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
edited Nov 21 '18 at 0:35
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
answered Nov 20 '18 at 19:37
Monstrous Moonshiner
2,25011337
2,25011337
1
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
2
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
1
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
1
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
1
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
|
show 6 more comments
1
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
2
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
1
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
1
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
1
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
1
1
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
that was pretty obvious, thank you very much! i already thought about what it means to evaluate f on x and y but i did not saw the connection to a subfield, mainly because the elements are in K but K is only a subfield so it is fine
– dougle
Nov 20 '18 at 19:46
2
2
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
@RobertLewis In fact you are right; I was way too hasty in writing out this proof. I'll update it accordingly.
– Monstrous Moonshiner
Nov 20 '18 at 20:59
1
1
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
Btw, if this answer was helpful, in addition to accepting it, might you also consider giving it an upvote? That's how we indicate that content is useful around here :)
– Monstrous Moonshiner
Nov 20 '18 at 21:11
1
1
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
@RobertLewis I can't tell whether or not you are being a little bit sarcastic... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:40
1
1
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
@RobertLewis Ah, I thought that the +1 had come from OP not you. For a while I had the impression that you might be teasing me for requesting upvotes... :)
– Monstrous Moonshiner
Nov 21 '18 at 4:45
|
show 6 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006762%2fgiven-a-field-extension-k-subset-m-with-algebraic-elements-in-x-y-show-that-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
