Undergrad Algebraically Closed Fields

[Math Amateur]

Dummit and Foote in their book Abstract Algebra give the following definition of an algebraically closed field ... ...

From the remarks following the definition it appears that the definition only applies to $K[x]$ ...

Does it also apply to $K[x_1, x_2], K[x_1, x_2, x_3], \ ... \ ... \ , K[x_1, x_2, \ ... \ ... \ , x_n] , \ ... \ ... \ ...$ ?

That is ... when we say K is an algebraically closed field does it imply that every polynomial in $K[x_1, x_2, \ ... \ ... \ , x_n]$ has a root in $K$ ... ... ?

... ... or maybe it is better if I say ... how does the definition of algebraically closed generalise to $K[x_1, x_2, \ ... \ ... \ , x_n]$ ... ... ?

Hope someone can clarify this issue ... ...

Peter

 

[mathwonk]

take a polynomial in x and y and set x equal to anything. then what do you have?

 

[andrewkirk]

It seems to me that extending the notion to multivariate polynomials makes no difference to the field.

If every univariate polynomial over $K$ has a root, so must any multivariate polynomial $P(x_1,...,x_m)$, since the univariate polynomial in $x_1$ that is obtained by setting all other $x_j$ to 0 will have a root $a$, whence $(a,0,...,0)$ will be a root of $P$.

Conversely, if every multivariate polynomial has a root then univariate polynomials must all have roots, since they are just the case where $m=1$.

Edit: Oh, ah, er Jinx! I suppose.

 

[Math Amateur]

take a polynomial in x and y and set x equal to anything. then what do you have?

Thanks for the help ... I see that if you put x equal to something, then you have a polynomial in y, which, since K is algebraically closed has a root in K ...

So we have a point $(x, y) \in K^2$ for which the polynomial is zero ...

Is that what you are getting at? ...

Peter

 

[Math Amateur]

It seems to me that extending the notion to multivariate polynomials makes no difference to the field.

If every univariate polynomial over $K$ has a root, so must any multivariate polynomial $P(x_1,...,x_m)$, since the univariate polynomial in $x_1$ that is obtained by setting all other $x_j$ to 0 will have a root $a$, whence $(a,0,...,0)$ will be a root of $P$.

Conversely, if every multivariate polynomial has a root then univariate polynomials must all have roots, since they are just the case where $m=1$.

Edit: Oh, ah, er Jinx! I suppose.

Thanks for the help, Andrew ... that appears to clarify my problem/issue ...!

Peter

 

[fresh_42]

From the remarks following the definition it appears that the definition only applies to $K[x]$ ...

Algebraic closure is a property of the field $K$. It means that all polynomial roots, which are per definition zeros, i.e. "numbers", are already in the field. E.g. $\mathbb{C}$ in which the additional element $i$ provides all possible polynomial zeros over the real numbers. It is not a property of the polynomial ring, so the question about multivariate polynomial rings doesn't make sense.

Does it also apply to $K[x_1, x_2], K[x_1, x_2, x_3], \ ... \ ... \ , K[x_1, x_2, \ ... \ ... \ , x_n] , \ ... \ ... \ ...$ ?

No. $K[x_1, x_2] = K[x_1][x_2]=(K[X_1])[X_2]$. But $(K[X_1])$ is a ring and thus fails the condition of the definition.
One would have to consider its quotient field $K(X_1)$ instead, which isn't an algebraic extension over $K$ and thus again fails the definition. So to make sense at all, one has to consider $K(X_1)$ as the "starting field". But this reduces the entire question to a situation $K' \, [X_2]$ as described anyway.
So the only question which might be left is, whether the quotient field of a polynomial ring over an algebraic closed field is algebraic closed again, which is a different issue. However, $p(y) = y^2 -x \in \mathbb{C}(x)[y]$ has a root $y=\sqrt{x}$ which is not an element of $\mathbb{C}(x)$. Hence $\mathbb{C}(x)$ is not algebraic closed although $\mathbb{C}$ is.

That is ... when we say K is an algebraically closed field does it imply that every polynomial in $K[x_1, x_2, \ ... \ ... \ , x_n]$ has a root in $K$ ... ... ?

No. $X_1 - X_2 = 0$ as polynomial in $X_2$ over $\mathbb{C}[X_1]$ has no root in $\mathbb{C}$.
Only if you consider all variables as being replaced by elements of $K$ then you have a chance to end up in $K$.
But this situation is not distinguishable from $K[a_1,...,a_{n-1}][X_n] = K[X_n]$, which is exactly the situation in the definition.

Edit : The reason for it doesn't make sense to speak of an algebraic closure of rings $R$ in general - such as polynomial rings $R = K[X_1]$ - is: Even the division, i.e. the zeros of $a \cdot X_1 - 1 = 0$ for an element $a \in R - \{0\}$ isn't a priori guaranteed. Therefore one would have to assume the existence of and pass to a quotient field of $R$ and ending up with a field again.

 

[micromass]

Edit : The reason for it doesn't make sense to speak of an algebraic closure of rings $R$ in general

Correct, but: https://en.wikipedia.org/wiki/Integral_element

 

[Math Amateur]

Algebraic closure is a property of the field $K$. It means that all polynomial roots, which are per definition zeros, i.e. "numbers", are already in the field. E.g. $\mathbb{C}$ in which the additional element $i$ provides all possible polynomial zeros over the real numbers. It is not a property of the polynomial ring, so the question about multivariate polynomial rings doesn't make sense.

No. $K[x_1, x_2] = K[x_1][x_2]=(K[X_1])[X_2]$. But $(K[X_1])$ is a ring and thus fails the condition of the definition.
One would have to consider its quotient field $K(X_1)$ instead, which isn't an algebraic extension over $K$ and thus again fails the definition. So to make sense at all, one has to consider $K(X_1)$ as the "starting field". But this reduces the entire question to a situation $K' \, [X_2]$ as described anyway.
So the only question which might be left is, whether the quotient field of a polynomial ring over an algebraic closed field is algebraic closed again, which is a different issue. However, $p(y) = y^2 -x \in \mathbb{C}(x)[y]$ has a root $y=\sqrt{x}$ which is not an element of $\mathbb{C}(x)$. Hence $\mathbb{C}(x)$ is not algebraic closed although $\mathbb{C}$ is.

No. $X_1 - X_2 = 0$ as polynomial in $X_2$ over $\mathbb{C}[X_1]$ has no root in $\mathbb{C}$.
Only if you consider all variables as being replaced by elements of $K$ then you have a chance to end up in $K$.
But this situation is not distinguishable from $K[a_1,...,a_{n-1}][X_n] = K[X_n]$, which is exactly the situation in the definition.

Edit : The reason for it doesn't make sense to speak of an algebraic closure of rings $R$ in general - such as polynomial rings $R = K[X_1]$ - is: Even the division, i.e. the zeros of $a \cdot X_1 - 1 = 0$ for an element $a \in R - \{0\}$ isn't a priori guaranteed. Therefore one would have to assume the existence of and pass to a quotient field of $R$ and ending up with a field again.

Thanks for the extensive help fresh_42 ... really appreciate your assistance in clarifying this matter ...

Still reflecting on what you have written ... ...

Thanks again,

Peter

 

[Math Amateur]

Correct, but: https://en.wikipedia.org/wiki/Integral_element

Thanks for the extra input micromass ... appreciate your help ..

Peter